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DaladierTypewritten textLIBROS UNIVERISTARIOS Y SOLUCIONARIOS DE MUCHOS DE ESTOS LIBROS

LOS SOLUCIONARIOS CONTIENEN TODOS LOS EJERCICIOS DEL LIBRORESUELTOS Y EXPLICADOSDE FORMA CLARA

VISITANOS PARADESARGALOS GRATIS.

Chapter 2 Technical Mathematics Physics, 6th Edition

1

Chapter 2. Technical MathematicsSigned Numbers2-1. +7 2-8. -17 2-15. +2 2-22. +122-2. +4 2-9. +6 2-16. -2 2-23. +82-3. +2 2-10. -32 2-17. -4 2-24. -42-4. -2 2-11. -36 2-18. -3 2-25. 02-5. -10 2-12. +24 2-19. +2 2-26. +2202-6. -33 2-13. -48 2-20. -4 2-27. +322-7. -5 2-14. +144 2-21. -3 2-28. -322-29. (a) -60C; (b) -170C; (c) 360C2-30. L = 2 mm[(-300C) (-50C)] = 2 mm(-25) = -50 mm; Decrease in length.

Algebra Review

2-31. x = (2) + (-3) + (-2) = -3; x = -3 2-32. x = (2) (-3) (-2) = +7; x = +72-33. x = (-3) + (-2) - (+2) = -7; x = -7 2-34. x = -3[(2) (-2)] = -3(2 + 2) = -12; x = -12

2-35. x b ca x 3 22

3 22

12

( ) ; 2-36. x x

2 32

2 32

12

( ) ;

2-37. x = (-3)2 (-2)2 = 9 4 = 5; x = 5 2-38. x bac x

( )( )( )

32 2

34

34;

2-39. x x 2

3 2 2 213 4

43( )( ) ( ) ( ); 2-40. x = (2)

2 + (-3)2 + (-2)2; x = 17

2-41. x a b c 2 2 2 17 2-42. x = (2)(-3)[(-2) (+2)]2; x = -6(-4)2 = -96

2-43. Solve for x: 2ax b = c; 2ax = b + c; x b ca x 2

3 22 2

54( ) ;


2

2-44. ax bx c a b x c x ca b x

4 44 4 2

2 3 8; ( ) ;( )( ) ;

2-45. 3 2 3 2 232 33 2 1ax

abc cx b x

bc x

; ;( )( ) ;

2-46. 4 2 16 4 2 16 4 1624 2 2 16 3

2 32acb

xb ac x b x

ac b x ; ; ( )( ) ( ) ;

2-47. 5m 16 = 3m 4 2-48. 3p = 7p - 165m 3m = -4 + 16 3p 7p = -162m = 12; m = 6 -4p = - 16; p = +4

2-49. 4m = 2(m 4) 2-50. 3(m 6) = 64m = 2m 8 3m 18 = 62m = -8; m = -4 3m = 24; m = +8

2-51. x x3 4 3 12 36 ( )( ) ; 2-52.p p3

26

13 1 ;

2-53. 96 48 9648 2x x ; 2-54. 14 = 2(b 7); 14 = 2b 14; b = 14

2-55. R2 = (4)2 + (3)2 = 16 + 9 2-56. 121 1

662

6 66 p

p pp

p;

R R2 25 5 3p = 6 + p; p = 3

2-57. V = IR; R VI 2-58. PV nRT TPVnR ;

2-59. F ma a Fm ; 2-60. s = vt + d; d = s vt

2-61. F mvR FR mv RmvF

22

2; ; 2-62. s = at2; 2s = at2; a st

22


3

2-63. 2 22

02

202

as v v a v vsff ; 2-64. C QV V

QC

2 2

2 2;

2-65. 1 1 11 2

2 1 1 2R R R R R R R R R ; 2-66. mv FtmvF t ;

( ) ;R R R R R R R RR R1 2 1 21 2

1 2 t

mvF

2-67. mv mv Ft mv Ft mv2 1 2 1 ; 2-68. PVTPVT PV T PV T

1 11

2 22

1 1 2 2 2 1 ;

v Ft mvm21 T PV TPV2

2 2 11 1

2-69. v = vo + at; v v0 = at 2-70. c2 = a2 + b2 ; b2 = c2 - a2

a v vt 0 b c a 2 2

Exponents and Radicals2-71. 212 2-72. 3523 2-73. x10 2-74. x5

2-75. 1/a 2-76. a/b2 2-77. 1/22 2-78. a2/b2

2-79. 2x5 2-80. 1/a2b2 2-81. m6 2-82. c4/n6

2-83. 64 x 106 2-84. (1/36) x 104 2-85. 4 2-86. 32-87. x3 2-88. a2b3 2-89. 2 x 102 2-90. 2 x 10-9

2-91. 2a2 2-92. x + 2

Scientific Notation2-93. 4.00 x 104 2-94. 6.70 x 101 2-95. 4.80 x 102

2-96. 4.97 x 105 2-97. 2.10 x 10-3 2-98. 7.89 x 10-1

2-99. 8.70 x 10-2 100. 9.67 x 10-4 2-101. 4,000,000


4

2-102. 4670 2-103. 37.0 2-104. 140,0002-105. 0.0367 2-106. 0.400 2-107. 0.0062-108. 0.0000417 2-109. 8.00 x 106 2-110. 7.40 x 104

2-111. 8.00 x 102 2-112. 1.80 x 10-8 2-113. 2.68 x 109

2-114. 7.40 x 10-3 2-115. 1.60x 10-5 2-116. 2.70 x 1019

2-117. 1.80 x 10-3 2-118. 2.40 x 101 2-119. 2.00 x 106

2-120. 2.00 x 10-3 2-121. 2.00 x 10-9 2-122. 5.71 x 10-1

2-123. 2.30 x 105 2-124. 6.40x 102 2-125. 2.40 x 103

2-126. 5.60 x 10-5 2-127. 6.90 x 10-2 2-128. 3.30 x 10-3

2-129. 6.00 x 10-4 2-130. 6.40 x 106 2-131. 8.00x 106

2-132. -4.00 x 10-2

Graphs2-133. Graph of speed vs. time: When t = 4.5 s, v = 144 ft/s; When v = 100 m/s, t = 3.1 s.2-134. Graph of advance of screw vs. turns: When screw advances 2.75 in., N = 88 turns.2-135. Graph of wavelength vs. frequency: 350 kHz 857 m; 800 kHz 375 m.2-136. Electric Power vs. Electric Current: 3.20 A 10.4 W; 8.0 A 64.8 W.

Geometry2-137. 900. 1800, 2700, and 450 2-138.

2-139a. A = 170, B = 350, C = 380 2-139b. A = 500 Rule 2; B = 400 Rule 2.

2-140a. A = 500 Rule 3; B = 1300 2-140b. B = 700, C = 420 Rule 2

A

C

B

D


5

Right Triangle Trigonometry2-141. 0.921 2-147. 19.3 2-153. 684 2-159. 54.20 2-165. 36.90

2-142. 0.669 2-148. 143 2-154. 346 2-160. 6.730 2-166. 76.00

2-143. 1.66 2-149. 267 2-155. 803 2-161. 50.20 2-167. 31.20

2-144. 0.559 2-150. 32.4 2-156. 266 2-162. 27.10

2-145. 0.875 2-151. 235 2-157. 2191 2-163. 76.80

2-146. 0.268 2-152. 2425 2-158. 1620 2-164. 6.370

Solve triangles for unknown sides and angles (Exercises 168 175):

2-168. tan = 18/35, = 35.80 ; R 18 252 2 R = 30.8 ft

2-169. tan = 600/400, = 56.30 ; R 40 802 2 R = 721 m.2-170. y = 650 sin 210 = 233 m; x = 650 cos 210 = 607 m.2-171. sin = 200/500, = 23.60; 500 2002 2 2 x , x = 458 km.2-172. sin = 210/400, = 31.70; 500 2002 2 2 m , m = 340 m.2-173. x = 260 cos 510 = 164 in.; y = 260 sin 510 = 202 in.

2-174. tan = 40/80, = 26.60; R 40 802 2 R = 89.4 lb2-175. = 1800 - 1200 = 600; y = 300 sin 600 = 260 m; x = 300 cos 600 = 150 m, left


6

Challenge Problems2-176. 30.21 0.59 in. = 29.62 in.2-178. T = Tf T0 = -150C (290C); T = -44 C0.2-179. Tf T0 = -340C; Tf - 200C = -340C; Tf = -140C2-180. Six pieces @ 4.75 in. = 6(4.75 in.) = 28.5 in.; Five cuts @ 1/16 = 5/16 = 0.3125 in.

Original length = 28.5 in. + 0.3125 in. = 28.8 in.

2-181. V = r2h; Solve for h: h Vr 2

2-182.2 2;mv mvF RR F

2-183. Solve for x and evaluate: a = 2, b = -2, c = 3, and d = -1xb + cd = a(x + 2) xb + cd = ax + 2a xb ax = 2a cd (b - a)x = 2a cd

x a cdb a

2 ; x a cdb a x

2 2 2 3 12 2

74

74

( ) ( )( )( ) ( ) ;

2-184. c b a b c a b2 2 2 2 2 2 250 20 539 ; . b = 53.9

2-185. F GmmR F 1 22

6 67 500( . ; . x 10 )(4 x 10 )(3 x 10 )(4 x 10 ) x 10-11 -8 -7

-2 2-22

2-186. L = L0 + L0(t t0); L = 21.41 cm + (2 x 10-3/C0)( 21.41 cm )(1000C - 200C);L = 24.84 cm.

2-187. Construct graph of y = 2x and verify that x = 3.5 when y = 7 (from the graph).2-188. (a) A + 600 = 900; A = 300. A + C = 900; C = 600. B = 600 by rule 2.

(b) D + 300 = 900; D = 600. A = 600 (alt. int. angles); B = 300; C = 1200.


7

Critical Thinking Problems2-189. A = (-8) (-4) = -4; B = (-6) + (14) = 8; C = A B = (-4) (8) = -12; C = - 12 cm.

B A = (8) (-4) = +12. There is a difference of 24 cm between B A and A B.

2-190. T Lg TLg L

gT 2 4 42 2

22

Let L = 4Lo; Since 4 2 , the period will be doubled when the length is quadrupled.

Let gm = ge /6, Then, T would be changed by a factor of 11 6 6 2 45/ .

Thus, the period T on the moon would be 2(2.45) or 4.90 s.2-191. (a) Area = LW = (3.45 x 10-4 m)(9.77 x 10-5 m); Area = 3.37 x 10-8 m2.

Perimeter (P) = 2L + 2W = 2(L + W); P = 2(3.45 x 10-4 + 9.77 x 10-5) = 8.85 x 10-4 m.(b) L = L0/2 and W = 2W0: A = (L0/2)(2W0) = L0W0; No change in area.

P P0 = [2(L0/2) + 2(2W0)] - [2L0 + 2W0] = 2W0 L0P = 2(9.77 x 10-5) 3.45 x 10-4 P = -1.50 x 10-4 m.The area doesnt change, but the perimeter decreases by 0.150 mm.

2-192. Graph shows when T = 420 K, P = 560 lb/in.2; when T = 600 K, P = 798 lb/in.2

2-193. Graph shows when V = 26 V, I = 377 mA; when V = 48 V, I = 696 mA.

Chapter 3 Technical Measurement and Vectors Physics, 6th Edition

8

Chapter 3. Technical Measurement and VectorsUnit Conversions3-1. A soccer field is 100 m long and 60 m across. What are the length and width of the field in

feet?100 cm 1 in. 1 ft(100 m) 328 ft1 m 2.54 cm 12 in.

L = 328 ft

100 cm 1 in. 1 ft(60 m) 197 ft1 m 2.54 cm 12 in. W = 197 ft

3-2. A wrench has a handle 8 in. long. What is the length of the handle in centimeters?2.54 cm(8 in.) 20.3 cm1 in.

L = 20.3 cm

3-3. A 19-in. computer monitor has a viewable area that measures 18 in. diagonally. Express thisdistance in meters. .

2.54 cm 1 m(18 in.) 0.457 m1 in. 100 cm L = 0.457 m

3-4. The length of a notebook is 234.5 mm and the width is 158.4 mm. Express the surface areain square meters.

1 m 1 mArea = (234.5 mm)(158.4 mm) 0.0371000 mm 1000 mm A = 0.0371 m

2

3-5. A cube has 5 in. on a side. What is the volume of the cube in SI units and in fundamentalUSCS units? .

3 33 3 32.54 cm 1 m(5 in.) 125 in. 0.00205 m1 in. 100 cmV V = 0.00205 m33

3 31 ft(125 in. ) 0.0723 ft12 in.V V = 0.0723 ft

3


9

3-6. The speed limit on an interstate highway is posted at 75 mi/h. (a) What is this speed inkilometers per hour? (b) In feet per second?

(a) mi 1.609 km75 h 1 mi 121 km/h (b)

mi 1 h 5280 ft75 h 3600 s 1 mi = 110 ft/s

3-7. A Nissan engine has a piston displacement (volume) of 1600 cm3 and a bore diameter of 84mm. Express these measurements in cubic inches and inches. Ans. 97.6 in.3, 3.31 in.

(a) 33 1 in.1600 cm 2.54 cm = 97.6 in.3 (b) 1 in.84 mm = 25.4 mm = 3.31 in.3-8. An electrician must install an underground cable from the highway to a home located 1.20

mi into the woods. How many feet of cable will be needed?5280 ft1.2 mi 6340 ft1 mi

L = 6340 ft

3-9. One U.S. gallon is a volume equivalent to 231 in.3. How many gallons are needed to fill atank that is 18 in. long, 16 in. wide, and 12 in. high? Ans. 15.0 gal.V = (18 in.)(16 in.)(12 in.) = 3456 in.3

33

1 gal(3456 in. ) 15.0 gal231 in.V V = 15.0 gal

3-10. The density of brass is 8.89 g/cm3. What is the density in kg/m3?3

3 3g 1 kg 100 cm kg8.89 8890cm 1000 g 1 m m

= 8890 kg/m3

Addition of Vectors by Graphical Methods3-11. A woman walks 4 km east and then 8 km north. (a) Use the polygon method to find her

resultant displacement. (b) Verify the result by the parallelogram method.Let 1 cm = 1 km; Then: R = 8.94 km, = 63.40 4 km

8 kmR


10

3-12. A land-rover, on the surface of Mars, moves a distance of 38 m at an angle of 1800. It thenturns and moves a distance of 66 m at an angle of 2700. What is the displacement fromthe starting position?Choose a scale, e.g., 1 cm = 10 mDraw each vector to scale as shown.Measure R = 7.62 cm or R = 76.2 mMeasure angle = 60.10 S of W = 1800 + 60.10 = 240.10 R = 76.2 m, 240.10

3-13. A surveyor starts at the southeast corner of a lot and charts the following displacements:A = 600 m, N; B = 400 m, W; C = 200 m, S; and D = 100 m, E. What is the netdisplacement from the starting point? .Choose a scale, 1 cm = 100 mDraw each vector tail to tip until all are drawn.Construct resultant from origin to finish.R = 500 m, = 53.10 N of E or = 126.90.

3-14. A downward force of 200 N acts simultaneously with a 500-N force directed to the left.Use the polygon method to find the resultant force.Chose scale, measure: R = 539 N, = 21.80 S. of E.

3-15. The following three forces act simultaneously on the same object. A = 300 N, 300 N of E;B = 600 N, 2700; and C = 100 N due east. Find the resultant forceusing the polygon method. Choose a scale, draw and measure:

R = 576 N, = 51.40 S of E

67 m

38 m, 1800

B

D R

CA

R

200 N500 N

BC

R

A


11

3-16. A boat travels west a distance of 200 m, then north for 400 m, and finally 100 m at 300 Sof E. What is the net displacement? (Set 1 cm = 100 N)Draw and measure: R = 368 N, = 108.00

3-17. Two ropes A and B are attached to a mooring hook so that an angle o 60 exists betweenthe two ropes. The tension in rope A is 80 lb and the tension in rope B is 120 lb. Use theparallelogram method to find the resultant force on the hook.Draw and measure: R = 174 lb

3-18. Two forces A and B act on the same object producing a resultant force of 50 lb at 36.90 Nof W. The force A = 40 lb due west. Find the magnitude and direction of force B?Draw R = 50 lb, 36.90 N of W first, then draw 40 lb, W.

F = 30 lb, 900

Trigonometry and Vectors3-19. Find the x and y-components of: (a) a displacement of 200 km, at 340. (b) a velocity of 40

km/h, at 1200; and (c) A force of 50 N at 330o.(a) Dx = 200 cos 340 = 166 km

Dy= 200 sin 340 = 112 km(b) vx = -40 cos 600 = -20.0 km/h

vy = 40 sin 600 = +34.6 km/h(c) Fx = 50 cos 300 = 43.3 N; Fy = - 50 sin 300 = -25.0 N

R

CB

A

B

A

R

36.90

40 lbFR

300

340 600

(a) (b) (c)


12

3-20. A sled is pulled with a force of 540 N at an angle of 400 with the horizontal. What are thehorizontal and vertical components of this force?Fx = 540 cos 400 = 414 N Fy= 540 sin 400 = 347 N

3-21. The hammer in Fig. 3-26 applies a force of 260 N at an angle of 150 with the vertical. Whatis the upward component of the force on the nail?F = 260 lb, = 750; Fxy = 260 sin Fy = 251 N.

3-22. A jogger runs 2.0 mi west and then 6.0 mi north. Find the magnitude and direction of theresultant displacement.

R ( ) ( )2 62 2 6.32 mi tan = 62 ; = 71.60 N of W

* 3-23. A river flows south with a velocity of 20 km/h. A boat has a maximum speed of 50 km/hin still water. In the river, at maximum throttle, the boat heads due west. What is theresultant speed and direction of the boat?

R (50) ( ) .;

2 220 539 km / h;tan = 2050 = 21.8 S of W

0

* 3-24. A rope, making an angle of 300 with the horizontal, drags a crate along the floor. Whatmust be the tension in the rope, if a horizontal force of 40 lb is required to drag the crate?

Fx = F cos 300; F Fx cos3040

0 lb

cos30 ;0 F = 46.2 N

540 N400

(a)

F

R

50 km/h20 km/hR

300Fx

F

R = 53.9 km/h, 21.80 Sof E


13

* 3-25. An vertical lift of 80 N is needed to lift a window. A long pole is used to lift the window.What force must be exerted along the pole if it makes an angle of 340 with the wall?

Fy = F sin 300; F Fy sin sin ;3440340 lb

0 F = 96.5 N

* 3-26. The resultant of two forces A and B is 400 N at 2100. If force A is 200 N at 2700, what arethe magnitude and direction of force B? ( = 210 - 1800 = 300)B = -400 N cos 300 = -346 N: B = 346 N, 1800

The Component Method of Vector Addition3-27. Find the resultant of the following perpendicular forces: (a) 400 N, 00; (b) 820 N, 2700;

and (b) 500 N, 900. Draw each vector, then find R:Ax= +400 N; Bx = 0; Cx= 0: Rx = +400 NAy = 0; By = -820 N; Cy = +500 N; Ry= 0 820 N + 500 N = -320 N

2 2400 320R ; 320tan ;400 R = 512 N, 38.70 S of E

3-28. Four ropes, all at right angles to each other, pull on a ring. The forces are A = 40 lb, E;B = 80 lb, N; C = 70 lb, W; and D = 20 lb, S. Find the resultant force on the ring.Ax= +40 lb; Bx = 0; Cx= -70 lb Dx = 0:

Rx = +40 lb 70 lb = -30 lbAy = 0; By = +80 lb; Cy = 0; Dy = -20 lb ;

Ry= 0 + 80 lb - 20 lb = +60 lb

2 230 60R ; tan ; 6030 R = 67.1 N, 116.6

0

F80 N 340

B

400 N200 N300

RC B

A

A = 40 lb, E


14

*3-29. Two forces act on the car in Fig. 3-27. Force A is 120 N, west and force B is 200 N at 600

N of W. What are the magnitude and direction of the resultant force on the car?Ax= -120; Bx = - (200 N) cos 600 = -100 NRx = 120 N - 100 N; Rx = -220 NAy= 0, By = 200 sin 600 = +173 N; Ry= 0 + 173 N = 173 N;

Thus, Rx = -100 N, Ry = +173 N and 2 2(220) (173) 280 NR

Resultant direction: 0173tan ; 38.2 N of W210 ; R = 280 N, 141.80

*3-30. Suppose the direction of force B in Problem 3-29 is reversed (+1800) and other parametersare unchanged. What is the new resultant? (This result is the vector difference A B).The vector B will now be 600 S of E instead of N of E.Ax= -120 N; Bx = +(200 N) cos 600 = +100 NRx = 120 N + 100 N; Rx = -20 NAy= 0, By = -200 sin 600 = -173 N; Ry= 0 - 173 N = -173 N;

Thus, Rx = -20 N, Ry = -173 N and 2 2( 20) (173) 174 NR

Resultant direction: tan ; . 17320 834

0 S of W ; R = 174 N, 253.40

*3-31. Determine the resultant force on the bolt in Fig. 3-28. ( Ax = 0 )Bx = -40 cos 200 = -37.6 lb; Bx = -50 cos 600 = -25.0 lbRx = 0 37.6 lb 25.0 lb; Rx = -62.6 lbAy = +60 lb; By = 40 sin 200 = 13.7 lb; Cy = 50 sin 60 = -43.3 lbRy = 60 lb 13.7 lb 43.3 lb; Ry = +30.4 lb

2 2( 62.6) (30.4)R R = 69.6 lb 30.4tan ;62.6 = 25.90 N of W

600B

A

A600B

600200

A = 60 lbB = 40 lb

C = 50 lb


15

*3-32. Determine the resultant of the following forces by the component method of vectoraddition: A = (200 N, 300); B = (300 N, 3300; and C = (400 N, 2500).Ax = 200 cos 300 = 173 N; Bx = 300 cos 300 = 260 NCx = -400 cos 700 = -137 N; Rx = Fx = 296 NAy = 200 sin 300 = 100 N; By = 300 sin 300 = -150 NCy = -400 sin 700 = -376 N; Ry = Fy = -430 N

2 2(296) ( 430)R ; 426tan ;296 R = 519 N, 55.20 S of E

*3-33. Three boats exert forces on a mooring hook as shown in Fig. 3-29. Find the resultant ofthese three forces.Ax = 420 cos 600 = +210 N; Cx = -500 cos 400 = -383 NBx = 0; Rx = 210 N + 0 383 N; Rx = -173 NAy = 420 sin 600 = 364 N; By = 150;Cy = 500 sin 400 = 321 N Ry = Fy = 835 N;

R ( ) (835)173 2 2 ; tan ; 835173 R = 853 N, 78.3

0 N of W

Challenge Problems3-34. Find the horizontal and vertical components of the following vectors: A = (400 N, 370);

B = 90 m, 3200); and C = (70 km/h, 1500).Ax = 400 cos 370 = 319 N; Ay = 400 sin 370 = 241 NBx = 90 cos 400 = 68.9 N; By = 90 sin 400 = 57.9Cx = -70 cos 300 = -60.6 N; Cy = 70 sin 300 = 25.0 N

700 300300

C B

A

400 600C

B A500 N420 N

150 N

370

90 N

300400

C

B

A70 N 400 N


16

3-35. A cable is attached to the end of a beam. What pull at an angle of 400 with the horizontalis needed to produce an effective horizontal force of 200 N?

P cos 400 = 200 N; P = 261 N3-36. A fishing dock runs north and south. What must be the speed of a boat heading at an

angle of 400 E of N if its velocity component along the dock is to be 30 km/h?v cos 400 = 30 km/h; v = 39.2 km/h

3-37. Find the resultant R = A + B for the following pairs of forces: A = (520 N, south); B =269 N, west; (b) A = 18 m/s, north; B = 15 m/s, west.

(a) R ( ) (520)269 5852 2 N

tan ; 520295 R = 585 N, = 62.6

0 S of W

(b) R ( ) ( )15 182 2 ; tan ; 1815 R = 23.4 N, 50 .2

0 N of W

*3-38. Determine the vector difference (A B) for the pairs of forces in Problem 3-37.

(a) R ( ) (520)269 5852 2 N

tan ; 520295 R = 585 N, 62.60 S of E

(b) R ( ) ( )15 182 2 ; tan ; 1815 R = 23.4 N, 50.20 N of E

*3-39. A traffic light is attached to the midpoint of a rope so that each segment makes an angleof 100 with the horizontal. The tension in each rope segment is 200 N. If the resultantforce at the midpoint is zero, what must be the weight of the traffic light?

Rx = Fx = 0; T sin 100 + T sin 100 W = 0;2(200) sin 100 = W: W = 69.5 N

P400

(a)

(b)

RB

A

15 m/s

18 m/s

R

BA

269 N

520 N

(a)

(b)RA

-B =15 m/s

18 m/s

RA

-B = 269 N

520 N

Change B into the vector B, then ADD:

W

TT


17

*3-40. Determine the resultant of the forces shown in Fig. 3-30Rx= 420 N 200 cos 700 410 cos 530 = 105 lbRy = 0 + 200 sin 700 410 sin 700 = -139.5 lb

R R Rx y 2 2 R = 175 lb; = 306.90

*3-41. Determine the resultant of the forces shown in Fig. 3-31.Rx= 200 cos 300 300 cos 450 155 cos 550 = 128 NRy = 0 + 200 sin 700 410 sin 700 = -185 N;

R R Rx y 2 2 R = 225 N; = 124.60

*3-42. A 200-N block rests on a 300 inclined plane. If the weight of the block acts verticallydownward, what are the components of the weight down the plane and perpendicular tothe plane? Choose x-axis along plane and y-axis perpendicular.Wx= 200 sin 300; Wx = 173 N, down the plane.Wy= 200 sin 600; Wx = 100 N, normal to the plane.

*3-43. Find the resultant of the following three displacements: A = 220 m, 600; B = 125 m, 2100;and C = 175 m, 3400.Ax = 220 cos 600 = 110 m; Ay = 220 sin 600 = 190.5 mBx = 125 cos 2100 = -108 m; By = 125 sin 2100 = -62.5 mCx = 175 cos 3400 = 164.4 m; Cy = 175 sin 3400 = -59.9 mRx = 110 m 108 m + 164.4 m; Ry = 190.5 m 62.5 m 59.9 m ;

Rx= 166.4 m; Ry = 68.1 m R ( . ) ( . )166 4 681 1802 2 m

tan .. ; . 681166 4 22 3

0 ; R = 180 m, = 22.30

300 200600

A

B C

550300450

C = 155 N

B = 300 N A = 200 N

530700

C = 410 lb

B = 200 lb

A = 420 lb

300

W300


18

C = 60 N B = 40 N

A = 80 N

Critical Thinking Questions3-44. Consider three vectors: A = 100 m, 00; B = 400 m, 2700; and C = 200 m, 300. Choose an

appropriate scale and show graphically that the order in which these vectors is added doesnot matter, i.e., A + B + C = C + B + A. Is this also true for subtracting vectors? Showgraphically how A C differs from C A.

3-45. Two forces A = 30 N and B = 90 N can act on an object in any direction desired. What isthe maximum resultant force? What is the minimum resultant force? Can the resultantforce be zero?Maximum resultant force occurs when A and B are in same direction.A + B = 30 N + 90 N = 120 NMinimum resultant force occurs when A and B are in opposite directions.B - A = 90 N 30 N = 60 NNo combination gives R = 0.

*3-46. Consider two forces A = 40 N and B = 80 N. What must be the angle between these twoforces in order to produce a resultant force of 60 N?Since R = C = 60 N is smaller than 80 N, the angle between A and B must be > 900. Applying the law ofcosines to the triangle, we can find and then .

A + B + C C + B + A A C C - A


19

C2 = A2 + B2 2AB Cos ; (60)2 = (80)2 + (40)2 2(80)(40) Cos ; = 46.60.The angle between the direction of A and the direction of B is = 1800 - ; = 133.40.

*3-47. What third force F must be added to the following two forces so that the resultant force iszero: A = 120 N, 1100 and B = 60 N, 2000?Components of A: Ax = 120 Cos 1100 = -40.0 N; Ay= 120 Sin 1100 = 113 NComponents of B: Bx = 60 Cos 2000 = -56.4 N; By= 60 Sin 2000 = -20.5 NRx = 0; Rx= Ax + Bx + Fx = 0; Rx = -40.0 N 56.4 N + Fx = 0; Or Fx = +97.4 NRy = 0; Ry= Ay + By + Fy = 0; Ry = 113 N 20.5 N + Fy = 0; Or Fy = -92.2 N

F ( . ) ( . )97 4 92 2 1312 2 N tan .. ; . 92 297 4 433

0 And = 3600 43.40

Thus, the force F has a magnitude and direction of: F = 134 N, = 316.60

*3-48. An airplane needs a resultant heading of due west. The speed of the plane is 600 km/h instill air. If the wind has a speed of 40 km/h and blows in a direction of 300 S of W, whatdirection should the aircraft be pointed and what will be its speed relative to the ground?From the diagram, Rx= R, Ry= 0, So that Ay + By= 0.Ay = 600 sin y = -40 sin 300 = -20 km/h600 sin - 20 = 0; 600 sin = 20

sin ; . 20600 1910N of W (direction aircraft should be pointed)

Noting that R = Rx and that Ax+Bx = Rx, we need only find sum of x-components.Ax = -600 cos 599.7 km/hx = 40 Cos 300 = -34.6 km/hR = -599.7 km/h 34.6 km/h; R = -634 km/h. Thus, the speed of the plane relative to theground is 634 km/h, 00; and the plane must be pointed in a direction of 1.910 N of W.

R

B = 40 km/h

A = 600 km/h


20

A = 200 lb200

EF

*3-49. What are the magnitude F and direction of the force needed to pull the car of Fig. 3-32directly east with a resultant force of 400 lb?Rx= 400 lb and Ry = 0; Rx = Ax + Fx = 400 lb200 Cos 200 + Fx = 400 lbFx = 400 lb 200 Cos 200 = 212 lbRy = 0 = Ay + Fy; Ay = -200 sin 200 = -68.4 lbFy = -Ay = +68.4 lb; So, Fx = 212 lb and Fy = +68.4 lb

F ( ) ( . ) ;212 68 42 2 ; tan = 68.4212 R = 223 lb, 17.90 N of E

Chapter 4. Translational Equilibrium and Friction Physics, 6th Ed.

21

BA

W

B400

Bx

By

W

B A300600

W

B A

600

W B

A

300600

Chapter 4. Translational Equilibrium and Friction.Note: For all of the problems at the end of this chapter, the rigid booms or struts are consideredto be of negligible weight. All forces are considered to be concurrent forces.

Free-body Diagrams4-1. Draw a free-body diagram for the arrangements shown in Fig. 3-18. Isolate a point where

the important forces are acting, and represent each force as a vector. Determine thereference angle and label components.

(a) Free-body Diagram (b) Free-body with rotation of axes to simplify work.4-2. Study each force acting at the end of the light strut in Fig. 3-19. Draw the appropriate free-

body diagram.There is no particular advantage to rotating axes.Components should also be labeled on diagram.

Solution of Equilibrium Problems:4-3. Three identical bricks are strung together with cords and hung from a scale that reads a total

of 24 N. What is the tension in the cord that supports the lowest brick? What is the tensionin the cord between the middle brick and the top brick?Each brick must weight 8 N. The lowest cord supports only one brick,whereas the middle cord supports two bricks. Ans. 8 N, 16 N.


22

W

B A

600

4-4. A single chain supports a pulley whose weight is 40 N. Two identical 80-N weights arethen connected with a cord that passes over the pulley. What is the tension in thesupporting chain? What is the tension in each cord?Each cord supports 80 N, but chain supports everything.T = 2(80 N) + 40 N = 200 N. T = 200 N

*4-5. If the weight of the block in Fig. 4-18a is 80 N, what are the tensions in ropes A and B?By - W = 0; B sin 400 80 N = 0; B = 124.4 NBx A = 0; B cos 400 = A; A = (124.4 N) cos 400

A = 95.3 N; B = 124 N.

*4-6. If rope B in Fig. 4-18a will break for tensions greater than 200 lb, what is the maximumweight W that can be supported?Fy = 0; By W = 0; W = B sin 400; B = 200 NW = (200 N) sin 400; W = 129 lb

*4-7. If W = 600 N in Fig. 18b, what is the force exerted by the rope on the end of the boom A inFig. 18b? What is the tension in rope B?

Fx = 0; A Wx = 0; A = Wx = W cos 600

A = (600 N) cos 600 = 300 NFy = 0; B Wy = 0; B = Wy = W sin 600

B = (600 N) sin 600 = 520 NA = 300 N; B = 520 N

Wy

Wx

80 N80 N

40 N

ByBx

B400A

W

Bx

B400A

W


23

W

B = 800 N A

600

300W

F N

*4-8. If the rope B in Fig. 18a will break if its tension exceeds 400 N, what is the maximumweight W? Fy = By - W = 0; By = W

B sin 400 = 400 N ; B = 622 N Fx = 0Bx A = 0; B cos 400 = A; A = (622 N) cos 400 A = 477 N.

*4-9. What is the maximum weight W for Fig. 18b if the rope can sustain a maximum tension ofonly 800 N? (Set B = 800 N).Draw diagram, then rotate x-y axes as shown to right.Fy = 0; 800 N W Sin 600 = 0; W = 924 N.The compression in the boom is A = 924 Cos 600 A = 462 N.

*4-10. A 70-N block rests on a 300 inclined plane. Determine the normal force and find the frictionforce that keeps the block from sliding. (Rotate axes as shown.)

Fx =N Wx = 0; N = Wx = (70 N) cos 300; N = 60.6 N

Fx = F Wy = 0; F = Wy = (70 N) sin 300; F = 35.0 N

*4-11. A wire is stretched between two poles 10 m apart. A sign is attached to the midpoint of theline causing it to sag vertically a distance of 50 cm. If the tension in each line segment is2000 N, what is the weight of the sign? (h = 0.50 m)tan = (0.5/5) or = 5.710 ; 2(2000 N) sin = WW = 4000 sin 5.71; W = 398 N.

*4-12. An 80-N traffic light is supported at the midpoint of a 30-m length of cable between topoles. Find the tension in each cable segment if the cable sags a vertical distance of 1 m.h = 1 m; Tan = (1/15); = 3.810

T sin + T sin = 80 N; 2T sin = 80 N

15 m

5 m

W = ?

h 2000 N 2000 N

5 m

15 m

W = 80 N

h T T

Bx

B400A

WBy


24

Solution to 4-12 (Cont.): T 80381 6010 N

2 Nsin . ; T = 601 N

*4-13. The ends of three 8-ft studs are nailed together forming a tripod with an apex that is 6ftabove the ground. What is the compression in each of these studs if a 100-lb weight is hungfrom the apex?Three upward components Fy hold up the 100 lb weight:3 Fy = 100 lb; Fy = 33.3 lb sin = (6/8); = 48.90

F sin 48.90 = 33.3 lb; F 333 44 4. . lbsin 48.9 lb0 F = 44.4 lb, compression

*4-14. A 20-N picture is hung from a nail as in Fig. 4-20, so that the supporting cords make anangle of 600. What is the tension of each cord segment?According to Newtons third law, the force of frame on nail (20 N)is the same as the force of the nail on the rope (20 N , up).

Fy = 0; 20 N = Ty + Ty; 2Ty = 20 N; Ty= 10 NTy = T sin 600; So T sin 600 = 10 N, and T = 11.5 N.

Friction4-15. A horizontal force of 40 N will just start an empty 600-N sled moving across packed snow.

After motion is begun, only 10 N is needed to keep motion at constant speed. Find thecoefficients of static and kinetic friction.

s k 40 10 N600 N 0.0667 N

600 N 0.0167 s = 0.0667; k = 0.016

4-16. Suppose 200-N of supplies are added the sled in Problem 4-13. What new force is neededto drag the sled at constant speed?N= 200 N + 600 N = 800 N; Fk = kN = (0.0167)(800 N); Fk = 13.3 N

F Fy

h

600 600

T T

20 N


25

4-17. Assume surfaces where s = 0.7 and k = 0.4. What horizontal force is needed to just starta 50-N block moving along a wooden floor. What force will move it at constant speed?Fs = sN = (0.7)(50 N) = 35 N ; Fk = sN = (0.4)(50 N) = 20 N

4-18. A dockworker finds that a horizontal force of 60 lb is needed to drag a 150-lb crate acrossthe deck at constant speed. What is the coefficient of kinetic friction?

k FN ; k 60 lb150 lb 0.400 k = 0.400

4-19. The dockworker in Problem 4-16 finds that a smaller crate of similar material can bedragged at constant speed with a horizontal force of only 40 lb. What is the weight of thiscrate?

Fk = sN = (0.4)W = 40 lb; W = (40 lb/0.4) = 100 lb; W = 100 lb.4-20. A steel block weighing 240 N rests on level steel beam. What horizontal force will move

the block at constant speed if the coefficient of kinetic friction is 0.12?Fk = sN = (0.12)(240 N) ; Fk = 28.8 N.

4-21. A 60-N toolbox is dragged horizontally at constant speed by a rope making an angle of 350

with the floor. The tension in the rope is 40 N. Determine the magnitude of the frictionforce and the normal force.Fx = T cos 350 Fk = 0; Fk = (40 N) cos 350 = 32.8 N

Fy =N + Ty W = 0; N = W Ty = 60 N T sin 350

N = 60 N (40 N) sin 350; N = 37.1 N Fk = 32.8 N

4-22. What is the coefficient of kinetic friction for the example in Problem 4-19?

k FN32 8. ; N37.1 N k = 0.884

FN T350

W


26

4-23. The coefficient of static friction for wood on wood is 0.7. What is the maximum angle foran inclined wooden plane if a wooden block is to remain at rest on the plane?

Maximum angle occurs when tan = s; s = tan = 0.7; = 35.00

4-24. A roof is sloped at an angle of 400. What is the maximum coefficient of static frictionbetween the sole of the shoe and the roof to prevent slipping?

Tan = k; k = Tan 400 =0.839; k = 0.839*4-25. A 200 N sled is pushed along a horizontal surface at constant speed with a 50-N force that

makes an angle of 280 below the horizontal. What is the coefficient of kinetic friction?Fx = T cos 280 Fk = 0; Fk = (50 N) cos 280 = 44.1 N

Fy =N - Ty W = 0; N = W + Ty = 200 N + T sin 280

N = 200 N + (50 N) sin 350; N = 223 N

k FN441. N223 N k = 0.198

*4-26. What is the normal force on the block in Fig. 4-21? What is the component of the weightacting down the plane?

Fy =N - W cos 430 = 0; N = (60N) cod 430 = 43.9 N

Wx = (60 N) sin 350; Wx = 40.9 N*4-27. What push P directed up the plane will cause the block in Fig. 4-21 to move up the plane

with constant speed? [From Problem 4-23:N = 43.9 N and Wx = 40.9 N]

Fk = kN = (0.3)(43.9 N); Fk = 13.2 N down plane.

Fx = P - Fk Wx = 0; P = Fk + Wx; P = 13.2 N + 40.9 N; P = 54.1 N

P

FkN

280W

WF

N P

430


27

*4-28. If the block in Fig. 4-21 is released, it will overcome static friction and slide rapidly downthe plane. What push P directed up the incline will retard the downward motion until theblock moves at constant speed? (Note that F is up the plane now.)

Magnitudes of F , Wx, andN are same as Prob. 4-25.

Fx = P +Fk Wx = 0; P = Wx - Fk; P = 40.9 N - 13.2 N

P = 27.7 N directed UP the inclined plane

Challlenge Problems*4-29. Determine the tension in rope A and the compression B in the strut for Fig. 4-22.

Fy = 0; By 400 N = 0; B 400 462 N Nsin600

Fx = 0; Bx A = 0; A = B cos 600

A = (462 N) cos 600; A = 231 N and B = 462 N*4-30. If the breaking strength of cable A in Fig. 4-23 is 200 N, what is the maximum weight that

can be supported by this apparatus?Fy = 0; Ay W = 0; W = (200 N) sin 400 = 129 NThe maximum weight that can be supported is 129 N.

*4-31. What is the minimum push P parallel to a 370 inclined plane if a 90-N wagon is to berolled up the plane at constant speed. Ignore friction.Fx = 0; P - Wx = 0; P = (90 N) sin 370

P = 54.2 N

WF

N P

430

B

400 N

A 600 By

A

W

200 NB 400 Ay

N P

370

W = 90 N


28

340 N

A B

W

4-32. A horizontal force of only 8 lb moves a cake of ice slides with constant speed across a floor(k = 0.1). What is the weight of the ice?

Fk = kN = (0.3) W; Fk = 8 lb; (0.1)W = 8 lb; W = 80 lb.

*4-33. Find the tension in ropes A and B for the arrangement shown in Fig. 4-24a.Fx = B Wx = 0; B = Wx = (340 N) cos 300; B = 294 NFy = A Wx = 0; A = Wy = (340 N) sin 300; A = 170 N

A = 170 N; B = 294 N

*4-34. Find the tension in ropes A and B in Fig. 4-24b.Fy = By 160 N = 0; By = 160 N ; B sin 500 = 294 N

B 160500 N

sin ; B = 209 N

Fx = A Bx = 0; A = Bx = (209 N) cos 500; A = 134 N*4-35. A cable is stretched horizontally across the top of two vertical poles 20 m apart. A 250-N

sign suspended from the midpoint causes the rope to sag a vertical distance of 1.2 m.What is the tension in each cable segment?.

h = 1.2 m; tan . ; . 1210 6840

2Tsin 6.840 = 250 N; T = 1050 N

*4-36. Assume the cable in Problem 4-31 has a breaking strength of 1200 N. What is themaximum weight that can be supported at the midpoint?2Tsin 6.840 = 250 N; 2(1200 N) sin 6.840 = W W = 289 N

WyWyWx 30

0

W = 160 N

BA500

W = 250 N

h T T

10 m 10 m


29

*4-37. Find the tension in the cable and the compression in the light boom for Fig. 4-25a.Fy = Ay 26 lb = 0; Ay = 26 lb ; A sin 370 = 26 lb

A 26 lbsin ;370 A = 43.2 lb

Fx = B Ax = 0; B = Ax = (43.2 lb) cos 370; B = 34.5 lb*4-38. Find the tension in the cable and the compression in the light boom for Fig. 4-25b.

First recognize that = 900 - 420 = 480, Then W = 68 lbFy = By 68 lb = 0; By = 68 lb ; B sin 480 = 68 lb

B 68 lbsin ;480 A = 915 lb

Fx = Bx A = 0; A = Bx = (91.5 lb) cos 480; B = 61.2 lb*4-39. Determine the tension in the ropes A and B for Fig. 4-26a.

Fx = Bx Ax = 0; B cos 300 = A cos 450; B = 0.816 AFy = A sin 450 B sin 300 420 N = 0; 0.707 A 0.5 B = 420 NSubstituting B = 0.816A: 0.707 A (0.5)(0.816 A) = 420 NSolving for A, we obtain: A = 1406 N; and B = 0.816A = 0.816(1406) or B = 1148 NThus the tensions are : A = 1410 N; B = 1150 N

*4-40. Find the forces in the light boards of Fig. 4-26b and state whether the boards are undertension or compression. ( Note: A = 900 - 300 = 600 )Fx = Ax Bx = 0; A cos 600 = B cos 450; A = 1.414 BFy = B sin 450 + A sin 600 46 lb = 0; 0.707 B + 0.866 A = 46 lbSubstituting A = 1.414B: 0.707 B + (0.866)(1.414 B) = 46 lbSolving for B: B = 23.8 lb; and A = 1.414B = 01.414 (23.8 lb) or A = 33.7 lbA = 33.7 lb, tension; B = 23.8 lb, compression

W = 26 lb

AB370

B

W 68 lb

A 480 By

420 N

A

B300

450

W

46 lb

AB

600450

W


30

Critical Thinking Questions4-41. Study the structure drawn in Fig. 4-27 and analyze the forces acting at the point where the

rope is attached to the light poles. What is the direction of the forces acting ON the endsof the poles? What is the direction of the forces exerted BY the poles at that point? Drawthe appropriate free-body diagram. Imagine that the poles are bolted together at theirupper ends, then visualize the forces ON that bolt and BY that bolt.

*4-42. Determine the forces acting ON the ends of the poles in Fig 3-27 if W = 500 N.Fx = Bx Ax = 0; B cos 300 = A cos 600; B = 0.577 AFy = A sin 600 B sin 300 500 N = 0; 0.866 A 0.5 B = 500 NSubstituting B = 0.577 A: 0.866 A (0.5)( 0.577 A) = 500 NSolving for A, we obtain: A = 866 N; and B = 0.577 A = 0.577(866) or B = 500 NThus the forces are : A = 866 N; B = 500 NCan you explain why B = W? Would this be true for any weight W?Try another value, for example W = 800 N and solve again for B.

W

A

B300

600

Forces ON Bolt at Ends (Action Forces):The forceW is exerted ON the bolt BY the weight.The force B is exerted ON bolt BY right pole. Theforce A is exerted ON bolt BY the middle pole. Tounderstand these directions, imagine that the polessnap, then what would be the resulting motion.

Wr Ar

Br300

600

Forces BY Bolt at Ends (Reaction Forces):The forceWr is exerted BY the bolt ON the weight.The force Br is exerted ON bolt BY right pole. Theforce Ar is exerted BY bolt ON the middle pole. Donot confuse action forces with the reaction forces.

W

A

B300

600


31

*4-43. A 2-N eraser is pressed against a vertical chalkboard with a horizontal push of 12 N. Ifs = 0.25, find the horizontal force required to start motion parallel to the floor? What ifyou want to start its motion upward or downward? Find the vertical forces required to juststart motion up the board and then down the board? Ans. 3.00 N, up = 5 N, down = 1 N.For horizontal motion, P = Fs = sN

P = 0.25 (12 N); P = 3.00 NFor vertical motion, P 2 N Fk = 0

P = 2 N + 3 N; P = 5.00 NFor down motion: P + 2 N Fs = 0; P = - 2 N + 3 N; P = 1.00 N

*4-44. It is determined experimentally that a 20-lb horizontal force will move a 60-lb lawnmower at constant speed. The handle of the mower makes an angle of 400 with theground. What push along the handle will move the mower at constant speed? Is thenormal force equal to the weight of the mower? What is the normal force?

k 20 0 333 lb60 lb . Fy = N Py - W= 0; W = 60 lb

N = P sin 400 + 60 lb; Fk = kN = 0.333N

Fy = Px - Fk = 0; P cos 400 0.333N = 0

P cos 400 0.333 (P sin 400 + 60 lb) = 0; 0.766 P = 0.214 P + 20 lb;

0.552 P = 20 lb; P 200552 36 2 lb lb. . ; P = 36.2 lb

The normal force is: N = (36.2 lb) sin 400 + 60 lb N = 83.3 lb

12 NP

2 NF

N

2 N

FF P

P

FkN

400W

12 NF

2 NP

N


32

W = 70N

N F

400

*4-45. Suppose the lawn mower of Problem 4-40 is to be moved backward. What pull along thehandle is required to move with constant speed? What is the normal force in this case?Discuss the differences between this example and the one in the previous problem.

k 20 0 333 lb60 lb . Fy = N + Py - W= 0; W = 60 lb

N = 60 lb - P sin 400; Fk = kN = 0.333N

Fy = Px - Fk = 0; P cos 400 0.333N = 0

P cos 400 0.333 (60 lb - P sin 400) = 0; 0.766 P - 20 lb + 0.214 P = 0;

0.980 P = 20 lb; P 200 980 20 4 lb lb. . ; P = 20.4 lb

The normal force is: N = 60 lb (20.4 lb) sin 400 N = 46.9 lb

*4-46. A truck is removed from the mud by attaching a line between the truck and the tree. Whenthe angles are as shown in Fig. 4-28, a force of 40 lb is exerted at the midpoint of the line.What force is exerted on the truck? = 200

T sin 200 + T sin 200 = 40 lb 2 T sin 200 = 40 lbT = 58.5 lb

*4-47. Suppose a force of 900 N is required to remove the move the truck in Fig. 4-28. Whatforce is required at the midpoint of the line for the angles shown?.

2 T sin 200 = F; 2(900 N) sin 200 = F; F = 616 N*4-48. A 70-N block of steel is at rest on a 400 incline. What is the static

friction force directed up the plane? Is this necessarily themaximum force of static friction? What is the normal force?F = (70 N) sin 400 = 45.0 N N = (70 N) cos 400 = 53.6 N

PFk

N400

W

F

h T T


33

*4-49. Determine the compression in the center strut B and the tension in the rope A for thesituation described by Fig. 4-29. Distinguish clearly the difference between thecompression force in the strut and the force indicated on your free-body diagram.Fx = Bx Ax = 0; B cos 500 = A cos 200; B = 1.46 AFy = B sin 500 A sin 200 500 N = 0; 0.766 B 0.342 A = 500 NSubstituting B = 1.46 A: 0.766 (1.46 A) (0.342 A) = 500 NSolving for A, we obtain: A = 644 N; and B = 1.46 A = 1.46 (644) or B = 940 NThus the tensions are : A = 644 N; B = 940 N

*4-50. What horizontal push P is required to just prevent a 200 N block from slipping down a 600

inclined plane where s = 0.4? Why does it take a lesser force if P acts parallel to theplane? Is the friction force greater, less, or the same for these two cases?(a) Fy =N Wy Py = 0; Wy = (200 N) cos 600 = 100 N

Py = P sin 600 = 0.866 P; N = 100 N + 0.866 P

F = N = 0.4(100 N + 0.866 P); F = 40 N + 0.346 P

Fx = Px Wx + F = 0; P cos 600 - (200 N) sin 600 + (40 N + 0.346 P) = 0

0.5 P 173.2 N + 40 N + 0.346 P = 0 Solving for P gives: P = 157 N(b) If P were parallel to the plane, the normal force would be LESS, and therefore the

friction force would be reduced. Since the friction force is directed UP the plane, it isactually helping to prevent slipping. You might think at first that the push P (to stopdownward slipping) would then need to be GREATER than before, due to the lesserfriction force. However, only half of the push is effective when exerted horizontally.If the force P were directed up the incline, a force of only 133 N is required. Youshould verify this value by reworking the problem.

WA

B

200500

x

W600

600FN

P600


34

*4-51. Find the tension in each cord of Fig. 4-30 if the suspended weight is 476 N.Consider the knot at the bottom first since more information is given at that point.Cy + Cy = 476 N; 2C sin 600 = 476 N

C 476 275 N2sin60 N0

Fy = A sin 300 - (275 N) sin 600 = 0A = 476 N; Fx = A cos 300 C cos 600 B = 0; 476 cos 300 275 cos 600 B = 0B = 412 N 137 N = 275 N; Thus: A = 476 N, B = 275 N, C = 275 N

*4-52. Find the force required to pull a 40-N sled horizontally at constant speed by exerting a pullalong a pole that makes a 300 angle with the ground (k = 0.4). Now find the forcerequired if you push along the pole at the same angle. What is the major factor thatchanges in these cases?(a) Fy = N + Py - W= 0; W = 40 N

N = 40 N - P sin 300; Fk = kN

Fx = P cos 300 - kN = 0; P cos 400- 0.4(40 N - P sin 300) =0;

0.866 P 16 N + 0.200 P = 0; P = 15.0 N(b) Fy = N - Py - W= 0; N = 40 N + P sin 300; Fk = kN

Fx = P cos 300 - kN = 0; P cos 400- 0.4(40 N + P sin 300) =0;

0.866 P 16 N - 0.200 P = 0; P = 24.0 N Normal force is greater!

476 N

AC C600 600 B

C 275 N600

300

300

P

FkN

300W

PFk

N

W


35

**4-53. Two weights are hung over two frictionless pulleys as shown in Fig. 4-31. What weightW will cause the 300-lb block to just start moving to the right? Assume s = 0.3. Note:The pulleys merely change the direction of the applied forces.Fy =N + (40 lb) sin 450 + W sin 300 300 lb = 0

N = 300 lb 28.3 lb 0.5 W; F = sN

Fx = W cos 300 - sN (40 lb) cos 450 = 0

0.866 W 0.3(272 lb 0.5 W) 28.3 lb = 0; W = 108 lb**4-54. Find the maximum weight than can be hung at point O in Fig. 4-32 without upsetting the

equilibrium. Assume that s = 0.3 between the block and table.We first find F max for the block

F = sN = 0.3 (200 lb) = 60 lb

Now set A = F = 60 lb and solve for W:

Fx = B cos 200 A = 0; B cos 200 = 60 lb; B = 63.9 lbFy = B sin 200 W = 0; W = B sin 200 = (63.9 lb) sin 200; W = 21.8 lb

F

40 lb NW450 300

300 lb

W

200FB

A A

Chapter 5 Torque and Rotational Equilibrium Physics, 6th Edition

36

Chapter 5. Torque and Rotational EquilibriumUnit Conversions5-1. Draw and label the moment arm of the force F about an axis at point A in Fig. 5-11a. What

is the magnitude of the moment arm?Moment arms are drawn perpendicular to action line:

rA = (2 ft) sin 250 rA = 0.845 ft5-2. Find the moment arm about axis B in Fig. 11a. (See figure above.)

rB = (3 ft) sin 250 rB = 1.27 ft5-3. Determine the moment arm if the axis of rotation is at point A in Fig. 5-11b. What is the

magnitude of the moment arm?rB = (2 m) sin 600 rB = 1.73 m

5-4. Find the moment arm about axis B in Fig. 5-11b.rB = (5 m) sin 300 rB = 2.50 m

Torque5-5. If the force F in Fig. 5-11a is equal to 80 lb, what is the resultant torque about axis A

neglecting the weight of the rod. What is the resultant torque about axis B?Counterclockwise torques are positive, so that A is - and B is +.(a) A = (80 lb)(0.845 ft) = -67.6 lb ft (b) B = (80 lb)(1.27 ft) = +101 lb ft

5-6. The force F in Fig. 5-11b is 400 N and the angle iron is of negligible weight. What is theresultant torque about axis A and about axis B?

Counterclockwise torques are positive, so that A is + and B is -.(a) A = (400 N)(1.732 m) = +693 N m; (b) B = (400 N)(2.50 m) = -1000 N m

3 ft2 ft rB

BA

250

F

rA250

2 m

5 m

rBrA

600 B300

A

F


37

5-7. A leather belt is wrapped around a pulley 20 cm in diameter. A force of 60 N is applied tothe belt. What is the torque at the center of the shaft?

r = D = 10 cm; = (60 N)(0.10 m) = +6.00 N m5-8. The light rod in Fig. 5-12 is 60 cm long and pivoted about point A. Find the magnitude and

sign of the torque due to the 200 N force if is (a) 900, (b) 600, (c) 300, and (d) 00. = (200 N) (0.60 m) sin for all angles:(a) = 120 N m (b) = 104 N m(b) = 60 N m (d) = 0

5-9. A person who weighs 650 N rides a bicycle. The pedals move in a circle of radius 40 cm. Ifthe entire weight acts on each downward moving pedal, what is the maximum torque?

= (250 N)(0.40 m) = 260 N m5-10. A single belt is wrapped around two pulleys. The drive pulley has a diameter of 10 cm,

and the output pulley has a diameter of 20 cm. If the top belt tension is essentially 50 N atthe edge of each pulley, what are the input and output torques?

Input torque = (50 N)(0.10 m) = 5 N mOutput torque = (50 N)(0.20 m) = 10 N m

Resultant Torque5-11. What is the resultant torque about point A in Fig. 5-13. Neglect weight of bar.

= +(30 N)(6 m) - (15 N)(2 m) - (20 N)(3 m) = 90.0 N m, Counterclockwise.

F

A 200 N

r 60 cm

30 N

2 m15 N

20 NA4 m 3 m


38

5-12. Find the resultant torque in Fig. 5-13, if the axis is moved to the left end of the bar. = +(30 N)(0) + (15 N)(4 m) - (20 N)(9 m) = -120 N m, counterclockwise.

5-13. What horizontal force must be exerted at point A in Fig 5-11b to make the resultant torqueabout point B equal to zero when the force F = 80 N?

= P (2 m) (80 N)(5 m) (sin 300) = 02 P = 200 N; P = 100 N

5-14. Two wheels of diameters 60 cm and 20 cm are fastened together and turn on the same axisas in Fig. 5-14. What is the resultant torque about a central axis for the shown weights?

r1 = (60 cm) = 0.30 m ; r2 = (30 cm) = 0.15 m = (200 N)(0.30 m) (150 N)(0.15 m) = 37.5 N m; = 37.5 N m, ccw

5-15. Suppose you remove the 150-N weight from the small wheel in Fig. 5-14. What newweight can you hang to produce zero resultant torque? = (200 N)(0.30 m) W (0.15 m) = 0; W = 400 N

5-16. Determine the resultant torque about the corner A for Fig. 5-15. = +(160 N)(0.60 m) sin 400 - (80 N)(0.20 m) = 61.7 N m 16.0 N m = 45.7 N m

R = 45.7 N m5-17. Find the resultant torque about point C in Fig. 5-15.

= - (80 N)(0.20 m) = -16 N m

30 N

2 m15 N

20 N

A 4 m 3 m

2 m

5 m

rB

B300

P

F = 80 N

C

B

A

80 N

40020 cm

60 cm

r400

160 N

C

80 N400

20 cm60 cm

r160 N


39

*5-18. Find the resultant torque about axis B in Fig. 5-15.Fx = 160 cos 400; Fy = 160 sin 400

= (123 N)(0.2 m) + (103 N)(0.6 m) = 37.2 N m

Equilibrium5-19. A uniform meter stick is balanced at its midpoint with a single support. A 60-N weight is

suspended at the 30 cm mark. At what point must a 40-N weight be hung to balance thesystem? (The 60-N weight is 20 cm from the axis)

= 0; (60 N)(20 cm) (40 N)x = 040 x = 1200 N cm or x = 30 cm: The weight must be hung at the 80-cm mark.

5-20. Weights of 10 N, 20 N, and 30 N are placed on a meterstick at the 20 cm, 40 cm, and 60cm marks, respectively. The meterstick is balanced by a single support at its midpoint. Atwhat point may a 5-N weight be attached to produce equilibrium.

= (10 N)(30 cm) + (20 N)(10 cm) (30 N)(10 cm) (5 N) x = 0

5 x = (300 + 200 300) or x = 40 cmThe 5-N weight must be placed at the 90-cm mark

5-21. An 8-m board of negligible weight is supported at a point 2 m from the right end where a50-N weight is attached. What downward force at the must be exerted at the left end toproduce equilibrium?

F (6 m) (50 N)(2 m) = 06 F = 100 N m or F = 16.7 N

Fx

FyB

80 N400

20 cm60 cm

160 N

20 cm x

40 N60 N

10 cm30 cm

10 N 20 N

x

5 N30 N

50 N

F 6 m 2 m


40

5-22. A 4-m pole is supported at each end by hunters carrying an 800-N deer which is hung at apoint 1.5 m from the left end. What are the upward forces required by each hunter? = A (0) (800 N)(1.5 m) + B (4.0 m) = 0

4B = 1200 N or B = 300 NFy = A + B 800 lb = 0; A = 500 N

5-23. Assume that the bar in Fig. 5-16 is of negligible weight. Find the forces F and A providedthe system is in equilibrium. = (80 N)(1.20 m) F (0.90 m) = 0; F = 107 NFy = F A 80 N = 0; A = 107 N 80 N = 26.7 N

F = 107 N, A = 26.7 N5-24. For equilibrium, what are the forces F1 and F2 in Fig. 5-17. (Neglect weight of bar.) = (90 lb)(5 ft) F2 (4 ft) (20 lb)(5 ft) = 0;

F2 = 87.5 lb Fy = F1 F2 20 lb 90 lb = 0F1 = F2 +110 lb = 87.5 lb + 110 lb, F1 = 198 lb

5-25. Consider the light bar supported as shown in Fig. 5-18. What are the forces exerted by thesupports A and B?

= B (11 m) (60 N)(3 m) (40 N)( 9 m) = 0;B = 49.1 N Fy = A + B 40 N 60 N = 0A= 100 N B = 100 N 49.1 N; B = 50.9 N

5-26. A V-belt is wrapped around a pulley 16 in. in diameter. If a resultant torque of 4 lb ft isrequired, what force must be applied along the belt?R = (16 in.) = 8 in. R = (8/12 ft) = 0.667 ftF (0.667 ft) = 4 lb ft; F = 6.00 lb

F

800 N

BA2.5 m1.5 m

Axis

80 N

F

A

90 cm30 cm Axis

20 lbF2

5 ft

Axis

1 ft

90 lb

F14 ft

B3 mAxis

40 N

2 m

60 N

A6 m


41

5-27. A bridge whose total weight is 4500 N is 20 m long and supported at each end. Find theforces exerted at each end when a 1600-N tractor is located 8 m from the left end.

= B (20 m) (1600 N)(8 m) (4500 N)( 10 m) = 0;B = 2890 N Fy = A + B 1600 N 4500 N = 0A= 6100 N B = 6100 N 2890 N; B = 3210 N

5-28. A 10-ft platform weighing 40 lb is supported at each end by stepladders. A 180-lb painteris located 4 ft from the right end. Find the forces exerted by the supports.

= B(10 ft) (40 lb)(5 ft) (180 lb)( 6 ft) = 0;B = 128 lb Fy = A + B 40 lb 180 lb = 0A= 220 lb B = 220 lb 128 lb; A = 92.0 lb

*5-29. A horizontal, 6-m boom weighing 400 N is hinged at the wall as shown in Fig. 5-19. Acable is attached at a point 4.5 m away from the wall, and a 1200-N weight is attached tothe right end. What is the tension in the cable?

= 900 370 = 530; Ty= T sin 530

= (T sin 530)(4.5 m) (400 N)(3 m) (1200 N)(6 m) = 0;3.59 T = 1200 N + 7200 N; T = 2340 N

*5-30. What are the horizontal and vertical components of the force exerted by the wall on theboom? What is the magnitude and direction of this force?Fx = H Tx = 0; H T cos 530 = 0; H = (2340 N) cos 530; H = 1408 NFy = V + T sin 530 400 N 1200 N = 0; V = 1600 N (2340 N) sin 530 = -269 NThus, the components are: H = 1408 N and V = -269 N. The resultant of these is:

R H V 2 2 1434 N; tan = -2691408 = 10.8 S of E0 R = 1434 N, 349.20

B10 m

Axis4500 N

2 m

1600 N

A8 m

B4 ft

Axis180 lb

1 ft

40 lb

A5 ft

1.5 mHTy

Ty B1.5 m

Axis1200 N400 N

V3 m


42

Center of Gravity5-31. A uniform 6-m bar has a length of 6 m and weighs 30 N. A 50-N weight is hung from the

left end and a 20-N force is hung at the right end. How far from the left end will a singleupward force produce equilibrium?Fy = F 50 N 30 N 20 N = 0; F = 100 N = F x (30 N)(3 m) (20 N)(6 m) = 0(100 N) x = 210 N m; x = 2.10 m

5-32. A 40-N sphere and a 12-N sphere are connected by a light rod 200 mm in length. How farfrom the middle of the 40-N sphere is the center of gravity?Fy = F 40 N 12 N = 0; F = 52 N = F x (40 N)(0) (12 N)(0.20 m) = 0(52 N) x = 2.40 N m; x = 0.0462 m or x = 46.2 mm

5-33. Weights of 2, 5, 8, and 10 N are hung from a 10-m light rod at distances of 2, 4, 6, and 8 mfrom the left end. How far from the left in is the center of gravity?Fy = F 10 N 8 N 5 N 2 N = 0; F = 25 NFx (2 N)(2 m) (5 N)(4 m) (8 N)(6 m) (10 N)(8 m) = 0(25 N) x = 152 N m; x = 6.08 m

5-34. Compute the center of gravity of sledgehammer if the metal head weighs 12 lb and the 32-in. supporting handle weighs 2 lb. Assume that the handle is of uniform construction andweight.Fy = F 2 lb 12 lb = 0; F = 14 lbFx (12 lb)(0) (2 lb)(16 in.) = 0; Fx = 32 lb in.(14 lb) x = 32 lb in.; x = 2.29 in. from head.

Axis

F

20 N30 N50 N

x3 m3 m

F

12 N40 N200 mmx

10 N5 N 8 N2 N

2 m2 m 2 m

2 m2 m

x F

F16 in. 16 in.x

2 lb12 lb


43

Challenge Problems5-35. What is the resultant torque about the hinge in Fig. 4-20? Neglect weight of the curved bar.

= (80 N)(0.6 m) (200 N)(0.4 m) sin 400

= 48.0 N m 51.4 N m; = 3.42 N m

5-36. What horizontal force applied to the left end of thebar in Fig. 4-20 will produce rotational equilibrium?

From Prob. 5-33: = - 3.42 N m.Thus, if = 0, then torque of +3.42 N m must be added.

F (0.6 m) cos 400 = +3.45 N m; F = 7.45 N5-37. Weights of 100, 200, and 500 lb are placed on a light board resting on two supports as

shown in Fig. 4-21. What are the forces exerted by the supports? = (100 lb)(4 ft) + B(16 ft)

(200 lb)(6 ft) (500 lb)(12 ft) = 0; B = 425 lbFy = A + B 100 lb 200 lb 500 lb = 0A = 800 lb B = 800 lb 425 lb; A = 375 lbThe forces exerted by the supports are : A = 375 N and B = 425 N

5-38. An 8-m steel metal beam weighs 2400 N and is supported 3 m from the right end. If a9000-N weight is placed on the right end, what force must be exerted at the left end tobalance the system?

= A (5 m) + (2400 N)(1 m) (9000 N)( 3 m) = 0;A = 4920 N Fy = A + B 2400 N 9000 N = 0

B= 11,400 N A = 11,400 N 4920 N; A = 6480 N

400500F

200 N60 cm

40 cmr 400

40080 N

200 N60 cm 40 cm

r 40080 N

Axis100 lb 200 lb 500 lb

A B6 ft6 ft 4 ft4 ft

A 9000 N

F4 m 3 m1 m

2400 N


44

*5-39. Find the resultant torque about point A in Fig. 5-22. = (70 N)(0.05 m) sin 500 (50 N)(0.16 m) sin 550

= 2.68 N m 6.55 N m = 3.87 N m = 3.87 N m

*5-40. Find the resultant torque about point B in Fig. 5-22. = (70 N)(0) (50 N)(a + b) ; First find a and b.a = (0.05 m) cos 500 = 0.0231 m; b = (0.16 m) sin 550 = 0.131 m = (50 N)(0.0231 m + 0.131 m) = 8.16 N m

= 8.16 N m

Critical Thinking Questions*5-41. A 30-lb box and a 50-lb box are on opposite ends of a 16-ft board supported only at its

midpoint. How far from the left end should a 40-lb box be placed to produce equilibrium?Would the result be different if the board weighed 90 lb? Why, or why not?

= (30 lb)(8 ft) + (40 lb)(x) (50 lb)(8 ft) = 0;x= 4.00 ft Note that the weight acting at the centerof the board does NOT contribute to torque aboutthe center, and therefore, the balance point is not affected, regardless of the weight.

5-42. On a lab bench you have a small rock, a 4-N meterstick and a single knife-edge support.Explain how you can use these three items to find the weight of the small rock.Measure distances a and b; determine F and thencalculate the weight W from equilibrium methods.

0.5 mF 4 N W

ba

b

a70 N

50 N

B5 cm

16 cm

500

550

r

r

70 N

50 N

B5 cm

A16 cm

500

550

x

F

W40 lb

8 ft8 ft

50 lb30 lb


45

*5-43. Find the forces F1, F2, and F3 such that the system drawn in Fig. 5-23 is in equilibrium.Note action-reaction forces R and R.First, lets work with top board: (about R) = 0; Force R is upward.R = (300 lb)(6 ft) (50 lb)(2 ft) F1(8 ft) = 0F1 = 213 lb Now, Fy = 0 gives: 213 lb + R 300 lb 50 lb = 0; R = 138 lb = RNext we sum torques about F2 with R = 138 lb is directed in a downward direction:F = (138 lb)(3 ft) + F3(7 ft) (200 lb)(5 ft) = 0; From which: F3 = 83.9 lbFy = 0 = F2 + 83.9 lb 138 lb 200 lb; F2 = 254 lb

The three unknown forces are: F1 = 213 lb, F2 = 254 lb, F3 = 83.9 lb

*5-44. (a) What weight W will produce a tension of 400 N in the rope attached to the boom inFig. 5-24?. (b) What would be the tension in the rope if W = 400 N? Neglect the weightof the boom in each case.(a) m) sin 300) W (6 m) cos 300 = 0

W = 154 N(b) = T(4 m) sin 300 (400 N)(6 m) cos 300 = 0

T = 600 N*5-45. Suppose the boom in Fig. 5-24 has a weight of 100 N and the suspended weight W is

equal to 400 N. What is the tension in the cord? m) sin 300) (400 N)(6 m) cos 300

(100 N)(3 m) cos 300 = 0T = 1169 N

50 lb2 ft

5 ft

2 ft6 ft

3 ft 2 ft300 lb

200 lb

F3F2

F1

R

R

Axis

300

4 m

2 m400 N

W300

100 NAxis

300

4 m

2 mT

W300


46

*5-46. For the conditions set in Problem 5-5, what are the horizontal and vertical componentsof the force exerted by the floor hinge on the base of the boom?Fx = H 1169 N = 0; or H = 1169 NFy = V 100 N 400 N = 0; or V = 500 N

H = 1169 N and V = 500 N

**5-47. What is the tension in the cable for Fig. 5-25. The weight of the boom is 300 N but itslength is unknown. (Select axis at wall, L cancels.)

TL N L Lsin ( ) sin sin75 300 2 30 546 30 00 0 0

T sin 750 = 75.0 N + 273 N; T = 360 N

**5-48. What are the magnitude and direction of the force exerted bythe wall on the boom in Fig. 5-25? Again assume that the weight of the board is 300 N.Refer to the figure and data given in Problem 5-7 and recall that T = 360 N.Fx = H - (360 N) cos 450 = 0; H = 255 NFy = V + (360 N) sin 450 300 N 546 N = 0; V = 591 N

H = 255 N and V = 591 N

*5-49. An car has a distance of 3.4 m between front and rear axles. If 60 percent of the weightrests on the front wheels, how far is the center of gravity located from the front axle?

= 0.6W(0) + 0.4W(3.4 m) F x = 0But F = W: 1.36 W W x = 0

x = 1.36 m from front axle

V

H100 NAxis

300

4 m

2 m1169 N400 N

300

600300

450

T = 360 N

T

H546 N

Lr

750

300 N

450

300V

0.4W

xF

0.6W3.4 m

Axis

Chapter 6 Uniform Acceleration Physics, 6th Edition

47

Chapter 6. Uniform AccelerationProblems:Speed and Velocity6-1. A car travels a distance of 86 km at an average speed of 8 m/s. How many hours were

required for the trip?

s vt 86,000 m 1 h10,750 s8 m/s 3600 st t = 2.99 h

6-2. Sound travels at an average speed of 340 m/s. Lightning from a distant thundercloud isseen almost immediately. If the sound of thunder reaches the ear 3 s later, how far away isthe storm?

t st 20 m

340 m / s 0.0588 s t = 58.8 ms

6-3. A small rocket leaves its pad and travels a distance of 40 m vertically upward beforereturning to the earth five seconds after it was launched. What was the average velocity forthe trip?

v st 40 80 m + 40 m

5 s m

5 s v = 16.0 m/s

6-4. A car travels along a U-shaped curve for a distance of 400 m in 30 s. Its final location,however is only 40 m from the starting position. What is the average speed and what is themagnitude of the average velocity?

Average speed: v st 400 m30 s v = 13.3 m/s

Average velocity: v Dt m

30 s40 v = 1.33 m/s, E

s = 400 m

D = 40 m


48

6-5. A woman walks for 4 min directly north with a average velocity of 6 km/h; then shewalks eastward at 4 km/h for 10 min. What is her average speed for the trip?t1 = 4 min = 0.0667 h; t2 = 10 min = 0.167 hs1 = v1t1 = (6 km/h)(0.0667 h) = 0.400 kms1 = v2t2 = (4 km/h)(0.167 h) = 0.667 km

v s st t

1 21 2

0.4 km + 0.667 km0.0667 h + 0.167 h v = 4.57 km/h

6-6. What is the average velocity for the entire trip described in Problem 6-5?

D ( . ; tan ..0 6670 4

0 667 km) + (0.400 km) km km

2 2 D = 0.778 km, 31.00

v 0 778 333. . km0.0667 h + 0.167 h km / h v = 3.33 km/h, 31.00

6-7. A car travels at an average speed of 60 mi/h for 3 h and 20 min. What was the distance?

t = 3 h + 0.333 h = 3.33 h; s = vt = (60 mi/h)(3.33 h); s = 200 mi

6.8 How long will it take to travel 400 km if the average speed is 90 km/h?

t st 400 km90 km / h t = 4.44 h

*6-9. A marble rolls up an inclined ramp a distance of 5 m, then stops and returns to a point 5m below its starting point. The entire trip took only 2 s. What was the average speedand what was the average velocity? (s1 = 5 m, s2 = -10 m)

speed = 5 m + 10 m2 s v = 7.50 m/s

velocity = Dt 5 m - 10 m

2 s v = 2.5 m/s, down plane.

D

s2 s1

6 km/h,4 min

4 km/h, 10 min

D

s1s2

CB

A E


49

Uniform Acceleration6-10. The tip of a robot arm is moving to the right at 8 m/s. Four seconds later, it is moving to

the left at 2 m/s. What is the change in velocity and what is the acceleration.v = vf - vo = (2 m/s) (8 m/s) v = 10 m/s

a vt 10 m / s

4 s a = 2.50 m/s2

6-11. An arrow accelerates from zero to 40 m/s in the 0.5 s it is in contact with the bow string.What is the average acceleration?

a v vtf o 40 m / s - 00.5 s a = 80.0 m/s

2

6-12. A car traveling initially at 50 km/h accelerates at a rate of 4 m/s2 for 3 s. What is thefinal speed?

vo = 50 km/h = 13.9 m/s; vf = vo + atvf = (13.9 m/s) + (4 m/s2)(3 s) = 25.9 m/s; vf = 25.9 m/s

6-13. A truck traveling at 60 mi/h brakes to a stop in 180 ft. What was the average accelerationand stopping time?

vo = 60 mi/h = 88.0 ft/s 2as = vf2 vo22 2 20 (88.0 ft/s)2 2(180 ft)

f ov va s a = 21.5 ft/s2

0

0

2 2(180 ft);2 88.0 ft/s + 0f

f

v v xx t t v v

t = 4.09 s


50

6-14. An arresting device on a carrier deck stops an airplane in 1.5 s. The average accelerationwas 49 m/s2. What was the stopping distance? What was the initial speed?

vf = vo + at; 0 = vo + ( 49 m/s2)(1.5 s); vo = 73.5 m/ss = vf t - at2 ; s = (0)(1.5 s) (-49 m/s2)(1.5 s)2; s = 55.1 m

6-15. In a braking test, a car traveling at 60 km/h is stopped in a time of 3 s. What was theacceleration and stopping distance? ( vo = 60 km/h = 16.7 m/s)

vf = vo + at; (0) = (16.7 m/s) + a (3 s); a = 5.56 m/s2

0 16.6 m/s + 0 3 s2 2fv vs t ; s = 25.0 m

6-16. A bullet leaves a 28-in. rifle barrel at 2700 ft/s. What was its acceleration and time in thebarrel? (s = 28 in. = 2.33 ft)

2as = vo2 - vf2 ; a v vsf 2

02

2(2700 ft / s) 0

2(2.33 ft)2

; a = 1.56 x 106 m/s2

s v v t sv vf

f

0

022 2 2 33; t = ft)0 + 2700 ft / s

( . ; t = 1.73 ms

6-17. The ball in Fig. 6-13 is given an initial velocity of 16 m/s at the bottom of an inclinedplane. Two seconds later it is still moving up the plane, but with a velocity of only 4m/s. What is the acceleration?

vf = vo + at; a v vtf 0 4 m / s - (16 m / s)2 s ; a = -6.00 m/s

2

6-18. For Problem 6-17, what is the maximum displacement from the bottom and what is thevelocity 4 s after leaving the bottom? (Maximum displacement occurs when vf = 0)

2as = vo2 - vf2; s v vaf 2

02

20 (16 m / s)2(-6 m / s )

22 ; s = +21.3 m

vf = vo + at = 16 m/s = (-6 m/s2)(4 s); vf = - 8.00 m/s, down plane


51

6-19. A monorail train traveling at 80 km/h must be stopped in a distance of 40 m. What averageacceleration is required and what is the stopping time? ( vo = 80 km/h = 22.2 m/s)

2as = vo2 - vf2; a v vsf 2

02

20 (22.2 m / s)

2(40 m)2; a = -6.17 m/s2

s v v t sv vf

f

0

022 2 40; t = m)22.2 m / s + 0

( ; t = 3.60 m/s

Gravity and Free-Falling Bodies6-20. A ball is dropped from rest and falls for 5 s. What are its position and velocity?

s = vot + at2; s = (0)(5 s) + (-9.8 m/s2)(5 s)2 ; s = -122.5 mvf = vo + at = 0 + (-9.8 m/s2)(5 s); v = -49.0 m/s

6-21. A rock is dropped from rest. When will its displacement be 18 m below the point ofrelease? What is its velocity at that time?

s = vot + at2; (-18 m) = (0)(t) + (-9.8 m/s2)t2 ; t = 1.92 svf = vo + at = 0 + (-9.8 m/s2)(1.92 s); vf = -18.8 m/s

6-22. A woman drops a weight from the top of a bridge while a friend below measures the timeto strike the water below. What is the height of the bridge if the time is 3 s?

s = vot + at2 = (0) + (-9.8 m/s2)(3 s)2; s = -44.1 m

6-23. A brick is given an initial downward velocity of 6 m/s. What is its final velocity afterfalling a distance of 40 m?

2as = vo2 - vf2 ; v v asf 02 2 40(-6 m / s) 2(-9.8 m / s m)2 2 )( ;

v = 28.6 m/s; Since velocity is downward, v = - 28.6 m/s


52

6-24. A projectile is thrown vertically upward and returns to its starting position in 5 s. Whatwas its initial velocity and how high did it rise?

s = vot + at2; 0 = vo(5 s) + (-9.8 m/s2)(5 s)2 ; vo = 24.5 m/s

It rises until vf = 0; 2as = vo2 - vf2 ; s 0 ( )24.5 m / s)

2(-9.8 m / s2

2 ; s = 30.6 m

6-25. An arrow is shot vertically upward with an initial velocity of 80 ft/s. What is itsmaximum height? (At maximum height, vf = 0; a = g = -32 ft/s2)

2as = vo2 - vf2; s v vaf 2

02

20 - (80 ft / s)2(-32 ft / s

22 ) ; s = 100 ft

6-26. In Problem 6-25, what are the position and velocity of the arrow after 2 s and after 6 s?s = vot + at2 = (80 ft/s)(2 s) + (-32 ft/s2)(2 s)2 ; s = 96 ft

vf = vo + at = (80 ft/s) + (-32 ft/s2)(2 s); vf = 16 ft/ss = vot + at2 = (80 ft/s)(6 s) + (-32 ft/s2)(6 s)2 ; s = -96 ftvf = vo + at = (80 ft/s) + (-32 ft/s2)(6 s); vf = -112 ft/s

6-27. A hammer is thrown vertically upward to the top of a roof 16 m high. What minimuminitial velocity was required?

2as = vo2 - vf2 ; v v asf0 2 2 16 (0) 2(-9.8 m / s m)2 2 )( ; vo = 17.7 m/s

Horizontal Projection6-28. A baseball leaves a bat with a horizontal velocity of 20 m/s. In a time of 0.25 s, how far

will it have traveled horizontally and how far has it fallen vertically?x = vox t = (20 m/s)(2.5 s) ; x = 50.0 m

y = voy + gt2 = (0)(2.5 s) + (-9.8 m/s2)(0.25 s)2 y = -0.306 m


53

0

6-29. An airplane traveling at 70 m/s drops a box of supplies. What horizontal distance will thebox travel before striking the ground 340 m below?

First we find the time to fall: y = voy t + gt2 t yg

2 2

9 8(.340 m) m / s2

t = 8.33 s ; x = vox t = (70 m/s)(8.33 s) ; x = 583 m

6-30. At a lumber mill, logs are discharged horizontally at 15 m/s from a greased chute that is20 m above a mill pond. How far do the logs travel horizontally?

y = gt2; t yg

2 2

9 8(.20 m) m / s2 ; t = 2.02 s

x = vox t = (15 m/s)(8.33 s) ; x = 30.3 m6-31. A steel ball rolls off the edge of a table top 4 ft above the floor. If it strikes the floor 5 ft

from the base of the table, what was its initial horizontal speed?

First find time to drop 4 ft: t yg

2 2

32( 4 ft) ft / s2 ; t = 0.500 s

x = vox t ; v xtx0505 ft s. ; vox = 10.0 ft/s

6-32. A bullet leaves the barrel of a weapon with an initial horizontal velocity of 400 m/s. Findthe horizontal and vertical displacements after 3 s.

x = vox t = (400 m/s)(3 s) ; x = 1200 my = voy + gt2 = (0)(3 s) + (-9.8 m/s2)(3 s)2 y = -44.1 m

6-33. A projectile has an initial horizontal velocity of 40 m/s at the edge of a roof top. Findthe horizontal and vertical components of its velocity after 3 s.

vx = vox = 40 m/s vy = voy t + gt = 0 + (-9.8 m/s2)(3s); vy = -29.4 m/s


54

The More General Problem of Trajectories6-34. A stone is given an initial velocity of 20 m/s at an angle of 580. What are its horizontal

and vertical displacements after 3 s?vox = (20 m/s) cos 580 = 10.6 m/s; voy = (20 m/s) sin 580 = 17.0 m/s

x = voxt = (10.6 m/s)(3 s); x = 31.8 my = voyt + gt2 = (17.0 m/s)(3 s) +(-9.8 m/s2)(3 s)2; y = 6.78 m

6-35. A baseball leaves the bat with a velocity of 30 m/s at an angle of 300. What are thehorizontal and vertical components of its velocity after 3 s?

vox = (30 m/s) cos 300 = 26.0 m/s; voy = (30 m/s) sin 300 = 15.0 m/svx = vox = 26.0 m/s ; vx = 26.0 m/s

vy = voy + gt = (15 m/s) + (-9.8 m/s2)(3 s) ; vy = -14.4 m/s

6-36. For the baseball in Problem 6-33, what is the maximum height and what is the range?ymax occurs when vy = 0, or when: vy = voy + gt = 0 and t = - voy/g

t vg toy

30 309 8 153

0sin. ; . m / s s ; Now we find ymax using this time.

ymax = voyt + gt2 = (15 m/s)(1.53 s) + (-9.8 m/s2)(1.53 s)2; ymax = 11.5 mThe range will be reached when the time is t = 2(1.53 s) or t = 3.06 s, thus

R = voxt= (30 m/s) cos 300 (3.06 s); R = 79.5 m

6-37. An arrow leaves the bow with an initial velocity of 120 ft/s at an angle of 370 with thehorizontal. What are the horizontal and vertical components of is displacement twoseconds later?

vox = (120 ft/s) cos 370 = 104 ft/s; voy = (120 ft/s) sin 300 = 60.0 ft/s


55

6-37. (Cont.) The components of the initial velocity are: vox = 104 ft/s; voy = 60.0 ft/sx = voxt = (104 ft/s)(2 s); x = 208 ft

y = voyt + gt2 = (60.0 m/s)(2 s) +(-32 ft/s2)(2 s)2; y = 56.0 ft

*6-38. In Problem 6-37, what are the magnitude and direction of arrows velocity after 2 s?vx = vox = 104 ft/s ; vx = 104 ft/s

vy = voy + gt = (60 m/s) + (-32 ft/s2)(2 s) ; vy = -4.00 ft/s

*6-39. A golf ball in Fig. 6-14 leaves the tee with a velocity of 40 m/s at 650. If it lands on agreen located 10 m higher than the tee, what was the time of flight, and what was thehorizontal distance to the tee?

vox = (40 m/s) cos 650 = 16.9 m/s; voy = (40 m/s) sin 650 = 36.25 m/sy = voyt + gt2: 10 ft = (36.25 m/s) t + (-9.8 m/s2)t2

Solving quadratic (4.9t2 36.25t + 10 = 0) yields: t1 = 0.287 s and t2 = 7.11 sThe first time is for y = +10 m on the way up, the second is y = +10 m on the way down.

Thus, the time from tee to green was: t = 7.11 sHorizontal distance to tee: x = voxt = (16.9 m/s)(7.11 s); x = 120 m

*6-40. A projectile leaves the ground with a velocity of 35 m/s at an angle of 320. What is themaximum height attained.

vox = (35 m/s) cos 320 = 29.7 m/s; voy = (35 m/s) sin 320 = 18.55 m/symax occurs when vy = 0, or when: vy = voy + gt = 0 and t = - voy/g

t vg toy

18559 8 189

0.. ; . m / s s2 ; Now we find ymax using this time.

ymax = voyt + gt2 = (18.55 m/s)(1.89 s) + (-9.8 m/s2)(1.89 s)2; ymax = 17.5 m


56

*6-41. The projectile in Problem 6-40 rises and falls, striking a billboard at a point 8 m abovethe ground. What was the time of flight and how far did it travel horizontally.

vox = (35 m/s) cos 320 = 29.7 m/s; voy = (35 m/s) sin 320 = 18.55 m/sy = voyt + gt2: 8 m = (18.55 m/s) t + (-9.8 m/s2)t2

Solving quadratic (4.9t2 18.55t + 8 = 0) yields: t1 = 0.497 s and t2 = 3.36 sThe first time is for y = +8 m on the way up, the second is y = +8 m on the way down.

Thus, the time from tee to green was: t = 3.29 sHorizontal distance to tee: x = voxt = (29.7 m/s)(3.29 s); x = 97.7 m

Challenge Problems6-42. A rocket travels in space at 60 m/s before it is given a sudden acceleration. Its velocity

increases to 140 m/s in 8 s, what was its average acceleration and how far did it travel inthis time?

a v vtf 0 (140 m / s) - (60 m / s)8 s ; a = 10 m/s

2

s v v tf FHG IKJ0 2 140 8 m / s+ 60 m / s2 sbg; t = 800 s

6-43. A railroad car starts from rest and coasts freely down an incline. With an averageacceleration of 4 ft/s2, what will be the velocity after 5 s? What distance does it travel?

vf = vo + at = 0 + (4 ft/s2)(5 s); vf = 20 ft/ss = vot + at2 = 0 + (4 ft/s2)(5 s)2; s = 50 ft


57

*6-44. An object is projected horizontally at 20 m/s. At the same time, another object located12 m down range is dropped from rest. When will they collide and how far are theylocated below the release point?A: vox = 20 m/s, voy = 0; B: vox = voy= 0Ball B will have fallen the distance y at the same time t as ball A. Thus,

x = voxt and (20 m/s)t = 12 m; t = 0.600 sy = at2 = (-9.8 m/s2)(0.6 s)2 ; y = -1.76 m

6-45. A truck moving at an initial velocity of 30 m/s is brought to a stop in 10 s. What was theacceleration of the car and what was the stopping distance?

a v vtf 0 0 - 30 m / s10 s ; a = -3.00 m/s

2

s v v tf FHG IKJ0 2 30 10 m / s + 02 sb g; s = 150 m6-46. A ball is thrown vertically upward with an initial velocity of 23 m/s. What are its

position and velocity after 2s, after 4 s, and after 8 s?Apply s = vot + at2 and vf = vo + at for time of 2, 4, and 8 s:(a) s = (23 m/s)(2 s) + (-9.8 m/s2)(2 s)2 ; s = 26.4 m

vf = (23 m/s) + (-9.8 m/s2)(2 s) ; vf = 3.40 m/s(b) s = (23 m/s)(4 s) + (-9.8 m/s2)(4 s)2 ; s = 13.6 m

vf = (23 m/s) + (-9.8 m/s2)(4 s) ; vf = -16.2 m/s(c) s = (23 m/s)(8 s) + (-9.8 m/s2)(8 s)2 ; s = -130 m

vf = (23 m/s) + (-9.8 m/s2)(8 s) ; vf = -55.4 m/s

yBA

12 m


58

6-47. A stone is thrown vertically downward from the top of a bridge. Four seconds later itstrikes the water below. If the final velocity was 60 m/s. What was the initial velocity ofthe stone and how high was the bridge?

vf = vo + at; v0 = vf at = (-60 m/s) - (-9.8 m/s)(4 s); vo = -20.8 m/ss = vot + at2 = (-20.8 m/s)(4 s) + (-9.8 m/s)(4 s)2; s = 162 m

6-48. A ball is thrown vertically upward with an initial velocity of 80 ft/s. What are itsposition and velocity after (a) 1 s; (b) 3 s; and (c) 6 sApply s = vot + at2 and vf = vo + at for time of 2, 4, and 8 s:(a) s = (80 ft/s)(1 s) + (-32 ft/s2)(1 s)2 ; s = 64.0 ft

vf = (80 ft/s) + (-32 ft/s2)(2 s) ; vf = 16.0 ft/s(b) s = (80 ft/s)(3 s) + (-32 ft/s2)(3 s)2 ; s = 96.0 ft

vf = (80 ft/s) + (-32 ft/s2)(3 s) ; vf = -16.0 ft/s(c) s = (80 ft/s)(6 s) + (-32 ft/s2)(6 s)2 ; s = 64.0 ft

vf = (80 ft/s) + (-32 ft/s2)(6 s) ; vf = -96.0 ft/s

6-49. An aircraft flying horizontally at 500 mi/h releases a package. Four seconds later, thepackage strikes the ground below. What was the altitude of the plane?

y = gt2 = (-32 ft/s2)(4 s)2; y = -256 ft

*6-50. In Problem 6-49, what was the horizontal range of the package and what are thecomponents of its final velocity?

vo = 500 mi/h = 733 ft/s; vx = vox = 733 ft/s; voy = 0; t = 4 sx = vxt = (733 ft/s)(4 s); x = 2930 ft

vy= voy + at = 0 + (-32 ft/s)(4 s); vy = -128 ft/s; vx= 733 m/s


59

*6-51. A putting green is located 240 ft horizontally and 64 ft vertically from the tee. Whatmust be the magnitude and direction of the initial velocity if a ball is to strike the greenat this location after a time of 4 s?

x = voxt; 240 ft = vox (4 s); vox = 60 m/ss = vot + at2; 64 ft = voy(4 s) + (-32 ft/s2)(4 s)2; voy = 80 ft/s

v v vx y 2 2 60( (80 ft / s) ft / s)2 2 ; tan 80 ft / s60 ft / s v = 100 ft/s, = 53.10

Critical Thinking Questions6-52. A long strip of pavement is marked off in 100-m intervals. Students use stopwatches to

record the times a car passes each mark. The following data is listed:

Distance, m 0 10 m 20 m 30 m 40 m 50 mTime, s 0 2.1 s 4.3 s 6.4 s 8.4 s 10.5 s

Plot a graph with distance along the y-axis and time along the x-axis. What is thesignificance of the slope of this curve? What is the average speed of the car? At whatinstant in time is the distance equal to 34 m? What is the acceleration of the car?Data taken directly from the graph (not drawn): Ans. Slope is v, 4.76 m/s, 7.14 s, 0.

6-53. An astronaut tests gravity on the moon by dropping a tool from a height of 5 m. Thefollowing data are recorded electronically.

Height, m 5.00 m 4.00 m 3.00 m 2.00 m 1.00 m 0 mTime, s 0 1.11 s 1.56 s 1.92 s 2.21 s 2.47 s


60

6-53. (Cont.) Plot the graph of this data. Is it a straight line? What is the average speed for theentire fall? What is the acceleration? How would you compare this with gravity on earth?Data taken directly from the graph (not drawn): Ans. Slope is v, 4.76 m/s, 7.14 s, 0.

*6-54. A car is traveling initially North at 20 m/s. After traveling a distance of 6 m, the carpasses point A where it's velocity is still northward but is reduced to 5 m/s. (a) Whatare the magnitude and direction of the acceleration of the car? (b) What time wasrequired? (c) If the acceleration is held constant, what will be the velocity of the carwhen it returns to point A?(a) vo = 20 m/s, vf = 5 m/s, x = 6 m

2as = vo2 - vf2;2 2 2 20 (5 m/s) (20 m/s)2 2(6 m)

fv va s ; a = -31.2 m/s2

(b) 00

2 2(6 m);2 20 m/s + 5 m/sf

f

v v ss t t v v ; t = 0.480 s

(c) Starts at A with vo = + 5 m/s then returns to A with zero net displacement (s = 0):

2as = vo2 - vf2; 0 = (5 m/s)2 vf2; v f (5 m / s) m / s2 5 ; vf = - 5 m/s

*6-55. A ball moving up an incline is initially located 6 m from the bottom of an incline and hasa velocity of 4 m/s. Five seconds later, it is located 3 m from the bottom. Assumingconstant acceleration, what was the average velocity? What is the meaning of a negativeaverage velocity? What is the average acceleration and final velocity?vo = + 4 m/s; s = -3 m; t = 5 s Find vavg

s = vavg t; v 3 m5 s ; vavg = -0.600 m/s

Negative average velocity means that the velocity was down the plane most of the time.

x = 6 mx = 0

A v = 5 m/sv = 20 m/s

4 m/s6 m3 m

s = 0


61

*6-55. (Cont.) s = vot + at2; -3 m = (4 m/s)(5 s) + a (5 s)2; a = -1.84 m/s2

vf = vo + at = 4 m/s + (-1.84 m/s2)(5 s); vf = -5.20 m/s

*6-56. The acceleration due to gravity on an distant planet is determined to be one-fourth itsvalue on the earth. Does this mean that a ball dropped from a height of 4 m above thisplanet will strike the ground in one-fourth the time? What are the times required on theplanet and on earth?

The distance as a function of time is given by: s = at2 so thatone-fourth the acceleration should result in twice the drop time.

t sge e 2 2(4 m)9.8 m / s2 te = 0.904 s t

sgp p

2 2(4 m)2.45 m / s2 tp = 1.81 s

*6-57. Consider the two balls A and B shown in Fig. 6-15. Ball A has a constant acceleration of4 m/s2 directed to the right, and ball B has a constant acceleration of 2 m/s2 directed tothe left. Ball A is initially traveling to the left at 2 m/s, while ball B is traveling to the leftinitially at 5 m/s. Find the time t at which the balls collide. Also, assuming x = 0 at theinitial position of ball A, what is their common displacement when they collide?Equations of displacement for A and B:s = so + vot + at2 (watch signs)

For A: sA = 0 + (-2 m/s)t + (+4 m/s2) t2

For B: sB = 18 m + (-5 m/s)t + (-2 m/s2) t2; Next simplify and set sA = sB- 2t + 2t2 = 18 5t - t2 3t2 + 3t 18 = 0 t1 = - 3 s, t2 = +2 sAccept t = +3 s as meaningful answer, then substitute to find either sA or sB:

sA = -2(2 s) + 2(2 s)2; x = + 4 m

v = - 5 m/s2v = - 2 m/s

+

aa = +4 m/s2

x = 18 mx = 0

A Bab = -2 m/s2


62

*6-58. Initially, a truck with a velocity of 40 ft/s is located distance of 500 ft to the right of acar. If the car begins at rest and accelerates at 10 ft/s2, when will it overtake the truck?How far is the point from the initial position of the car?Equations of displacement for car and truck:

s = so + vot + at2 (watch signs)For car: sC = 0 + (+10 ft/s2) t2 ; Truck: sT = 500 ft + (40 ft/s)t + 0;Set sC = sT 5t2 = 500 + 40t or t2 8t 100 = 0; t1= -6.77 s; t2 = +14.8 sSolve for either distance: sC = (10 ft/s2)(14.8 s)2; s = 1092 ft

*6-59. A ball is dropped from rest at the top of a 100-m tall building. At the same instant asecond ball is thrown upward from the base of the building with an initial velocity of 50m/s. When will the two balls collide and at what distance above the street?For A: sA = 100 m + v0At + gt2 = 100 m + 0 + (-9.8 m/s2) t2

For B: sB = 0 + (50 m/s)t + (-9.8 m/s2) t2 Set sA = sB100 4.9 t2 = 50 t 4.9 t2; 50 t = 100; t = 2.00 sSolve for s: sA = 100 m (4.9 m/s2)(2 s)2; s = 80.4 m

*6-60. A balloonist rising vertically with a velocity of 4 m/s releases a sandbag at the instantwhen the balloon is 16 m above the ground. Compute the position and velocity of

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