www. elsolucionario.blogspot .com libros … the period t on the moon would be 2(2.45 ) or 4.90 s. 2

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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 Mathematics

Signed Numbers

2-1. +7 2-8. -17 2-15. +2 2-22. +12

2-2. +4 2-9. +6 2-16. -2 2-23. +8

2-3. +2 2-10. -32 2-17. -4 2-24. -4

2-4. -2 2-11. -36 2-18. -3 2-25. 0

2-5. -10 2-12. +24 2-19. +2 2-26. +220

2-6. -33 2-13. -48 2-20. -4 2-27. +32

2-7. -5 2-14. +144 2-21. -3 2-28. -32

2-29. (a) -60C; (b) -170C; (c) 360C

2-30. 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 = +7

2-33. 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 3

22 3

212

( ) ;

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 22 2 1

34 4

3( )( )( ) ( ); 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 4 4 4 22 3

8; ( ) ; ( )( )

;

2-45. 3 2 3 2 23

2 33 2

1ax abc

cx b x bc

x

; ; ( )( )

;

2-46. 4 2 16 4 2 16 4 162

4 2 2 16 32

32acb

xb

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

2-47. 5m 16 = 3m 4 2-48. 3p = 7p - 16

5m 3m = -4 + 16 3p 7p = -16

2m = 12; m = 6 -4p = - 16; p = +4

2-49. 4m = 2(m 4) 2-50. 3(m 6) = 6

4m = 2m 8 3m 18 = 6

2m = -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. 12

1 16

62

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 T PVnR

;

2-59. F ma a Fm

; 2-60. s = vt + d; d = s vt

2-61. F mvR

FR mv R mvF

2

22

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

22


3

2-63. 22

202

202

as v v av v

sff

; 2-64. C QV

V QC

2 2

2 2;

2-65. 1 1 11 2

2 1 1 2R R RR R R R R R ; 2-66. mv Ft mv

Ft ;

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

1 2

1 2

t mvF

2-67. mv mv Ft mv Ft mv2 1 2 1 ; 2-68.PVT

PVT

PV T PV T1 11

2 2

21 1 2 2 2 1 ;

v Ft mvm2

1 T PV T

PV22 2 1

1 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 Radicals

2-71. 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. 3

2-87. x3 2-88. a2b3 2-89. 2 x 102 2-90. 2 x 10-9

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

Scientific Notation

2-93. 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. 140,000

2-105. 0.0367 2-106. 0.400 2-107. 0.006

2-108. 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

Graphs

2-133. 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.

Geometry

2-137. 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 Trigonometry

2-141. 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 lb

2-175. = 1800 - 1200 = 600; y = 300 sin 600 = 260 m; x = 300 cos 600 = 150 m, left


6

Challenge Problems

2-176. 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 = -140C

2-180. 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 = -1

xb + 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 1

2 274

74

( ) ( )( )( ) ( )

;

2-184. c b a b c a b2 2 2 2 2 2 250 20 53 9 ; . b = 53.9

2-185. F Gm mR

F 1 226 67 5 00( . ; . 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 Problems

2-189. 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

T Lg

L gT 2 44

2 22

2

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 of1

1 66 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 L0

P = 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 Vectors

Unit Conversions

3-1. 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 this

distance 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 area

in square meters.

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

A = 0.0371 m2

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

USCS units? .

3 3

3 3 32.54 cm 1 m(5 in.) 125 in. 0.00205 m1 in. 100 cm

V

V = 0.00205 m3

33 31 ft(125 in. ) 0.0723 ft

12 in.V

V = 0.0723 ft3


9

3-6. The speed limit on an interstate highway is posted at 75 mi/h. (a) What is this speed in

kilometers per hour? (b) In feet per second?

(a) mi 1.609 km75h 1 mi

121 km/h (b) mi 1 h 5280 ft75h 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 84

mm. Express these measurements in cubic inches and inches. Ans. 97.6 in.3, 3.31 in.

(a) 3

3 1 in.1600 cm2.54 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 a

tank 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 3

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

= 8890 kg/m3

Addition of Vectors by Graphical Methods

3-11. 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 then

turns and moves a distance of 66 m at an angle of 2700. What is the displacement from

the starting position?

Choose a scale, e.g., 1 cm = 10 m

Draw each vector to scale as shown.

Measure R = 7.62 cm or R = 76.2 m

Measure 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 net

displacement from the starting point? .

Choose a scale, 1 cm = 100 m

Draw 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 force

using 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

R200 N

500 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 S

of 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 between

the two ropes. The tension in rope A is 80 lb and the tension in rope B is 120 lb. Use the

parallelogram 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 N

of 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 Vectors

3-19. 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

C

B

A

B

A

R

36.90

40 lb

FR

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 the

horizontal 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. What

is 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 the

resultant 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/h

in still water. In the river, at maximum throttle, the boat heads due west. What is the

resultant speed and direction of the boat?

R (50) ( ) .

;

2 220 53 9 km / h;

tan = 2050

= 21.8 S of W0

* 3-24. A rope, making an angle of 300 with the horizontal, drags a crate along the floor. What

must be the tension in the rope, if a horizontal force of 40 lb is required to drag the crate?

Fx = F cos 300; FFx

cos3040

0

lbcos30

;0 F = 46.2 N

540 N

400

(a)

F

R

50 km/h

20 km/hR

300

Fx

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; FFy

sin sin;

3440

340 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 are

the 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 Addition

3-27. 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 N

Ay = 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 lb

Ay = 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, 116.60

F80 N340

B

400 N200 N

300

R

CB

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 N

Rx = 120 N - 100 N; Rx = -220 N

Ay = 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 parameters

are 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 N

Rx = 120 N + 100 N; Rx = -20 N

Ay = 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

83 40 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 lb

Rx = 0 37.6 lb 25.0 lb; Rx = -62.6 lb

Ay = +60 lb; By = 40 sin 200 = 13.7 lb; Cy = 50 sin 60 = -43.3 lb

Ry = 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

600

B

A

A

600

B

600200

A = 60 lb

B = 40 lb

C = 50 lb


15

*3-32. Determine the resultant of the following forces by the component method of vector

addition: A = (200 N, 300); B = (300 N, 3300; and C = (400 N, 2500).

Ax = 200 cos 300 = 173 N; Bx = 300 cos 300 = 260 N

Cx = -400 cos 700 = -137 N; Rx = Fx = 296 N

Ay = 200 sin 300 = 100 N; By = 300 sin 300 = -150 N

Cy = -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 of

these three forces.

Ax = 420 cos 600 = +210 N; Cx = -500 cos 400 = -383 N

Bx = 0; Rx = 210 N + 0 383 N; Rx = -173 N

Ay = 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, 78.30 N of W

Challenge Problems

3-34. 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 N

Bx = 90 cos 400 = 68.9 N; By = 90 sin 400 = 57.9

Cx = -70 cos 300 = -60.6 N; Cy = 70 sin 300 = 25.0 N

700 300

300

CB

A

400 600

C

BA500 N

420 N

150 N

370

90 N

300

400

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 horizontal

is needed to produce an effective horizontal force of 200 N?

P cos 400 = 200 N; P = 261 N

3-36. 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, = 62.60 S of W

(b) R ( ) ( )15 182 2 ; tan ; 1815

R = 23.4 N, 50 .20 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 angle

of 100 with the horizontal. The tension in each rope segment is 200 N. If the resultant

force 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)

R

B

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. 3-30

Rx = 420 N 200 cos 700 410 cos 530 = 105 lb

Ry = 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 N

Ry = 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 vertically

downward, what are the components of the weight down the plane and perpendicular to

the 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 m

Bx = 125 cos 2100 = -108 m; By = 125 sin 2100 = -62.5 m

Cx = 175 cos 3400 = 164.4 m; Cy = 175 sin 3400 = -59.9 m

Rx = 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 30 ; R = 180 m, = 22.30

300 200600

A

B C

55030045

0

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 Questions

3-44. 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 does

not matter, i.e., A + B + C = C + B + A. Is this also true for subtracting vectors? Show

graphically 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 is

the maximum resultant force? What is the minimum resultant force? Can the resultant

force be zero?

Maximum resultant force occurs when A and B are in same direction.

A + B = 30 N + 90 N = 120 N

Minimum resultant force occurs when A and B are in opposite directions.

B - A = 90 N 30 N = 60 N

No combination gives R = 0.

*3-46. Consider two forces A = 40 N and B = 80 N. What must be the angle between these two

forces 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 of

cosines 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 is

zero: A = 120 N, 1100 and B = 60 N, 2000?

Components of A: Ax = 120 Cos 1100 = -40.0 N; Ay = 120 Sin 1100 = 113 N

Components of B: Bx = 60 Cos 2000 = -56.4 N; By = 60 Sin 2000 = -20.5 N

Rx = 0; Rx = Ax + Bx + Fx = 0; Rx = -40.0 N 56.4 N + Fx = 0; Or Fx = +97.4 N

Ry = 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

43 30 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 in

still air. If the wind has a speed of 40 km/h and blows in a direction of 300 S of W, what

direction 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/h

600 sin - 20 = 0; 600 sin = 20

sin ; . 20600

1910 N 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/h

R = -599.7 km/h 34.6 km/h; R = -634 km/h. Thus, the speed of the plane relative to the

ground 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 lb

200 E

F

*3-49. What are the magnitude F and direction of the force needed to pull the car of Fig. 3-32

directly east with a resultant force of 400 lb?

Rx = 400 lb and Ry = 0; Rx = Ax + Fx = 400 lb

200 Cos 200 + Fx = 400 lb

Fx = 400 lb 200 Cos 200 = 212 lb

Ry = 0 = Ay + Fy; Ay = -200 sin 200 = -68.4 lb

Fy = -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

BA

300600

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 Diagrams

4-1. 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 the

reference 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 tension

in 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 are

then connected with a cord that passes over the pulley. What is the tension in the

supporting 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 N

Bx 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 maximum

weight W that can be supported?

Fy = 0; By W = 0; W = B sin 400; B = 200 N

W = (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 in

Fig. 18b? What is the tension in rope B?

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

A = (600 N) cos 600 = 300 N

Fy = 0; B Wy = 0; B = Wy = W sin 600

B = (600 N) sin 600 = 520 N

A = 300 N; B = 520 N

Wy

Wx

80 N80 N

40 N

ByBx

B400

A

W

Bx

B400

A

W


23

W

B = 800 NA

600

300

W

F N

*4-8. If the rope B in Fig. 18a will break if its tension exceeds 400 N, what is the maximum

weight W? Fy = By - W = 0; By = W

B sin 400 = 400 N ; B = 622 N Fx = 0

Bx 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 of

only 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 friction

force 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 the

line causing it to sag vertically a distance of 50 cm. If the tension in each line segment is

2000 N, what is the weight of the sign? (h = 0.50 m)

tan = (0.5/5) or = 5.710 ; 2(2000 N) sin = W

W = 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 to

poles. 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

B400

A

W

By


24

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

6010 N

2 N

sin .; T = 601 N

*4-13. The ends of three 8-ft studs are nailed together forming a tripod with an apex that is 6ft

above the ground. What is the compression in each of these studs if a 100-lb weight is hung

from 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 33 3 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 an

angle 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 N

Ty = T sin 600; So T sin 600 = 10 N, and T = 11.5 N.

Friction

4-15. 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 the

coefficients of static and kinetic friction.

s k 40 10 N600 N

0.0667 N600 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 needed

to 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 start

a 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 across

the deck at constant speed. What is the coefficient of kinetic friction?

k FN

; k 60 lb

150 lb 0.400 k = 0.400

4-19. The dockworker in Problem 4-16 finds that a smaller crate of similar material can be

dragged at constant speed with a horizontal force of only 40 lb. What is the weight of this

crate?

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 friction

force 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 FN

32 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 for

an 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 friction

between 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 FN

44 1. 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 weight

acting 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

Fk

N

280

W

W

F

NP

430


27

*4-28. If the block in Fig. 4-21 is released, it will overcome static friction and slide rapidly down

the plane. What push P directed up the incline will retard the downward motion until the

block 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 N

The 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 be

rolled up the plane at constant speed. Ignore friction.

Fx = 0; P - Wx = 0; P = (90 N) sin 370

P = 54.2 N

W

F

NP

430

B

400 N

A600 By

A

W

200 N

B 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 N

Fy = 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

6 840

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 the

maximum weight that can be supported at the midpoint?

2Tsin 6.840 = 250 N; 2(1200 N) sin 6.840 = W W = 289 N

Wy

WyWx 300

W = 160 N

B

A500

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 lb

Fy = 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 A

Fy = A sin 450 B sin 300 420 N = 0; 0.707 A 0.5 B = 420 N

Substituting B = 0.816A: 0.707 A (0.5)(0.816 A) = 420 N

Solving for A, we obtain: A = 1406 N; and B = 0.816A = 0.816(1406) or B = 1148 N

Thus 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 under

tension or compression. ( Note: A = 900 - 300 = 600 )

Fx = Ax Bx = 0; A cos 600 = B cos 450; A = 1.414 B

Fy = B sin 450 + A sin 600 46 lb = 0; 0.707 B + 0.866 A = 46 lb

Substituting A = 1.414B: 0.707 B + (0.866)(1.414 B) = 46 lb

Solving for B: B = 23.8 lb; and A = 1.414B = 01.414 (23.8 lb) or A = 33.7 lb

A = 33.7 lb, tension; B = 23.8 lb, compression

W = 26 lb

A

B370

B

W 68 lb

A 480 By

420 N

A

B300

450

W

46 lb

AB

600450

W


30


4-41. 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 ends

of the poles? What is the direction of the forces exerted BY the poles at that point? Draw

the appropriate free-body diagram. Imagine that the poles are bolted together at their

upper 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.


Fy = A sin 600 B sin 300 500 N = 0; 0.866 A 0.5 B = 500 N

Substituting B = 0.577 A: 0.866 A (0.5)( 0.577 A) = 500 N

Solving for A, we obtain: A = 866 N; and B = 0.577 A = 0.577(866) or B = 500 N

Thus the forces are : A = 866 N; B = 500 N

Can 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 force W 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

Br

300

600

Forces BY Bolt at Ends (Reaction Forces):

The force Wr 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. If

s = 0.25, find the horizontal force required to start motion parallel to the floor? What if

you want to start its motion upward or downward? Find the vertical forces required to just

start 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 N

For vertical motion, P 2 N Fk = 0

P = 2 N + 3 N; P = 5.00 N

For 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 lawn

mower at constant speed. The handle of the mower makes an angle of 400 with the

ground. What push along the handle will move the mower at constant speed? Is the

normal 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 200 552

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 N

F

N

2 N

FF P

P

Fk

N

400

W

12 NF

2 N

P

N


32

W = 70N

NF

400

*4-45. Suppose the lawn mower of Problem 4-40 is to be moved backward. What pull along the

handle 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. When

the 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 lb

T = 58.5 lb

*4-47. Suppose a force of 900 N is required to remove the move the truck in Fig. 4-28. What

force 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 the

maximum 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

N

400

W

F

h T T


33

*4-49. Determine the compression in the center strut B and the tension in the rope A for the

situation described by Fig. 4-29. Distinguish clearly the difference between the

compression force in the strut and the force indicated on your free-body diagram.


Fy = B sin 500 A sin 200 500 N = 0; 0.766 B 0.342 A = 500 N

Substituting B = 1.46 A: 0.766 (1.46 A) (0.342 A) = 500 N

Solving for A, we obtain: A = 644 N; and B = 1.46 A = 1.46 (644) or B = 940 N

Thus 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 the

plane? 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 is

actually helping to prevent slipping. You might think at first that the push P (to stop

downward slipping) would then need to be GREATER than before, due to the lesser

friction 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. You

should verify this value by reworking the problem.

WA

B

200500

x

W600

600F

N

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 = 0

A = 476 N; Fx = A cos 300 C cos 600 B = 0; 476 cos 300 275 cos 600 B = 0

B = 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 pull

along a pole that makes a 300 angle with the ground (k = 0.4). Now find the force

required if you push along the pole at the same angle. What is the major factor that

changes 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 C

600 600 B

C 275 N

600300

300

P

Fk

N

300

W

PFk

N

W


35

**4-53. Two weights are hung over two frictionless pulleys as shown in Fig. 4-31. What weight

W 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 lb

Fy = B sin 200 W = 0; W = B sin 200 = (63.9 lb) sin 200; W = 21.8 lb

F

40 lbN

W450 300

300 lb

W

200F

BA A

Chapter 5 Torque and Rotational Equilibrium Physics, 6th Edition

36

Chapter 5. Torque and Rotational Equilibrium

Unit Conversions

5-1. 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 ft

5-2. Find the moment arm about axis B in Fig. 11a. (See figure above.)

rB = (3 ft) sin 250 rB = 1.27 ft

5-3. 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

Torque

5-5. 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 the

resultant 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 ft

2 ft rB

BA

250

F

rA

250

2 m

5 m

rBrA

600 B

300

A

F


37

5-7. A leather belt is wrapped around a pulley 20 cm in diameter. A force of 60 N is applied to

the belt. What is the torque at the center of the shaft?

r = D = 10 cm; = (60 N)(0.10 m) = +6.00 N m

5-8. 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. If

the entire weight acts on each downward moving pedal, what is the maximum torque?

= (250 N)(0.40 m) = 260 N m

5-10. 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 at

the edge of each pulley, what are the input and output torques?

Input torque = (50 N)(0.10 m) = 5 N m

Output torque = (50 N)(0.20 m) = 10 N m

Resultant Torque

5-11. 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

A200 N

r

60 cm

30 N

2 m

15 N

20 NA

4 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 torque

about point B equal to zero when the force F = 80 N?

= P (2 m) (80 N)(5 m) (sin 300) = 0

2 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 axis

as 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 new

weight 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 m

5-17. Find the resultant torque about point C in Fig. 5-15.

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

30 N

2 m

15 N

20 N

A 4 m 3 m

2 m

5 m

rB

B

300

P

F = 80 N

C

B

A

80 N

40020 cm

60 cm

r400

160 N

C

80 N

40020 cm

60 cmr

160 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

Equilibrium

5-19. 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 the

system? (The 60-N weight is 20 cm from the axis)

= 0; (60 N)(20 cm) (40 N)x = 0

40 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 60

cm marks, respectively. The meterstick is balanced by a single support at its midpoint. At

what 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 cm

The 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 a

50-N weight is attached. What downward force at the must be exerted at the left end to

produce equilibrium?

F (6 m) (50 N)(2 m) = 0

6 F = 100 N m or F = 16.7 N

Fx

FyB

80 N

40020 cm

60 cm

160 N

20 cm x

40 N60 N

10 cm30 cm

10 N 20 N

x

5 N30 N

50 N

F6 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 a

point 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 N

Fy = 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 provided

the system is in equilibrium.

= (80 N)(1.20 m) F (0.90 m) = 0; F = 107 N

Fy = F A 80 N = 0; A = 107 N 80 N = 26.7 N

F = 107 N, A = 26.7 N

5-24. 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 = 0

F1 = 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 the

supports 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 = 0

A = 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 is

required, what force must be applied along the belt?

R = (16 in.) = 8 in. R = (8/12 ft) = 0.667 ft

F (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 cmAxis

20 lbF2

5 ft

Axis

1 ft

90 lb

F14 ft

B3 m

Axis

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 the

forces 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 = 0

A = 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 painter

is 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 = 0

A = 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. A

cable is attached at a point 4.5 m away from the wall, and a 1200-N weight is attached to

the 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 the

boom? 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 N

Fy = V + T sin 530 400 N 1200 N = 0; V = 1600 N (2340 N) sin 530 = -269 N

Thus, 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

Axis

4500 N

2 m

1600 N

A8 m

B4 ft

Axis

180 lb

1 ft

40 lb

A5 ft

1.5 mH

Ty

Ty B1.5 m

Axis

1200 N400 N

V3 m


42

Center of Gravity

5-31. 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 single

upward 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 far

from 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 m

from 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 N

Fx (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 and

weight.

Fy = F 2 lb 12 lb = 0; F = 14 lb

Fx (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

x

3 m3 m

F

12 N40 N200 mm

x

10 N5 N 8 N2 N

2 m

2 m 2 m

2 m

2 mx F

F

16 in. 16 in.

x

2 lb12 lb


43

Challenge Problems

5-35. 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 the

bar 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 N

5-37. 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 lb

Fy = A + B 100 lb 200 lb 500 lb = 0

A = 800 lb B = 800 lb 425 lb; A = 375 lb

The 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 a

9000-N weight is placed on the right end, what force must be exerted at the left end to

balance 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

400500

F

200 N

60 cm

40 cmr 400

40080 N

200 N60 cm40 cm

r 40080 N

Axis

100 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


*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 center

of the board does NOT contribute to torque about

the 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 then

calculate the weight W from equilibrium methods.0.5 m

F 4 N W

ba

b

a70 N

50 N

B

5 cm

16 cm

500

550

r

r

70 N

50 N

B5 cm

A

16 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) = 0

F1 = 213 lb Now, Fy = 0 gives: 213 lb + R 300 lb 50 lb = 0; R = 138 lb = R

Next 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 lb

Fy = 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 in

Fig. 5-24?. (b) What would be the tension in the rope if W = 400 N? Neglect the weight

of 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 = 0

T = 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 components

of the force exerted by the floor hinge on the base of the boom?

Fx = H 1169 N = 0; or H = 1169 N

Fy = 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 its

length is unknown. (Select axis at wall, L cancels.)

TL N L Lsin ( ) sin sin75 3002

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 by

the 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 N

Fy = 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 weight

rests 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 = 0

But F = W: 1.36 W W x = 0

x = 1.36 m from front axle

V

H

100 NAxis

300

4 m

2 m1169 N

400 N

300

600

300

450

T = 360 N

T

H

546 N

L

r

750

300 N

450

300

V

0.4W

xF

0.6W3.4 m

Axis

Chapter 6 Uniform Acceleration Physics, 6th Edition

47

Chapter 6. Uniform Acceleration

Problems:

Speed and Velocity

6-1. 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 s

t

t = 2.99 h

6-2. Sound travels at an average speed of 340 m/s. Lightning from a distant thundercloud is

seen almost immediately. If the sound of thunder reaches the ear 3 s later, how far away is

the storm?

t st

20 m340 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 before

returning to the earth five seconds after it was launched. What was the average velocity for

the trip?

v st

40 80 m + 40 m5 s

m5 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 the

magnitude of the average velocity?

Average speed: v st

400 m30 s

v = 13.3 m/s

Average velocity: v Dt

m30 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 she

walks 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 h

s1 = v1t1 = (6 km/h)(0.0667 h) = 0.400 km

s1 = v2t2 = (4 km/h)(0.167 h) = 0.667 km

v s st t1 2

1 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 ..

0 667 0 40 667

km) + (0.400 km) km km

2 2 D = 0.778 km, 31.00

v 0 778 3 33. . 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 5

m below its starting point. The entire trip took only 2 s. What was the average speed

and 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 m2 s

v = 2.5 m/s, down plane.

D

s2 s1

6 km/h,4 min

4 km/h, 10 min

D

s1

s2CB

A E


49

Uniform Acceleration

6-10. 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 sa = 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?

av v

tf o

40 m / s - 00.5 s

a = 80.0 m/s2

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

final speed?

vo = 50 km/h = 13.9 m/s; vf = vo + at

vf = (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 acceleration

and stopping time?

vo = 60 mi/h = 88.0 ft/s 2as = vf2 vo2

2 2 20 (88.0 ft/s)2 2(180 ft)

f ov vas

a = 21.5 ft/s2

0

0

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

f

f

v v xx t tv v

t = 4.09 s


50

6-14. An arresting device on a carrier deck stops an airplane in 1.5 s. The average acceleration

was 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/s

s = 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 the

acceleration 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 2

fv 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 the

barrel? (s = 28 in. = 2.33 ft)

2as = vo2 - vf2 ; av v

sf

2 02

2(2700 ft / s) 0

2(2.33 ft)

2

; a = 1.56 x 106 m/s2

sv v

t sv v

f

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 inclined

plane. Two seconds later it is still moving up the plane, but with a velocity of only 4

m/s. What is the acceleration?

vf = vo + at; av v

tf0

4 m / s - (16 m / s)2 s

; a = -6.00 m/s2

6-18. For Problem 6-17, what is the maximum displacement from the bottom and what is the

velocity 4 s after leaving the bottom? (Maximum displacement occurs when vf = 0)

2as = vo2 - vf2; sv v

af2 0

2

20 (16 m / s)

2(-6 m / s )

2

2 ; 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 average

acceleration is required and what is the stopping time? ( vo = 80 km/h = 22.2 m/s)

2as = vo2 - vf2; av v

sf2 0

2

20 (22.2 m / s)

2(40 m)

2

; a = -6.17 m/s2

sv v

t sv v

f

f

0

022 2 40; t = m)

22.2 m / s + 0( ; t = 3.60 m/s

Gravity and Free-Falling Bodies

6-20. 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 m

vf = 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 of

release? What is its velocity at that time?

s = vot + at2; (-18 m) = (0)(t) + (-9.8 m/s2)t2 ; t = 1.92 s

vf = 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 time

to 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 after

falling 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. What

was 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 / s

2

2 ; s = 30.6 m

6-25. An arrow is shot vertically upward with an initial velocity of 80 ft/s. What is its

maximum height? (At maximum height, vf = 0; a = g = -32 ft/s2)

2as = vo2 - vf2; sv v

af

202

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

2

2 ); 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/s

s = vot + at2 = (80 ft/s)(6 s) + (-32 ft/s2)(6 s)2 ; s = -96 ft

vf = 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 minimum

initial velocity was required?

2as = vo2 - vf2 ; v v asf02 2 16 (0) 2(-9.8 m / s m)2 2 )( ; vo = 17.7 m/s

Horizontal Projection

6-28. 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 the

box travel before striking the ground 340 m below?

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

g

2 29 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 is

20 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 m

6-31. 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 ; vxtx0

50 5

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. Find

the horizontal and vertical displacements after 3 s.

x = vox t = (400 m/s)(3 s) ; x = 1200 m

y = 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. Find

the 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 Trajectories

6-34. 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 m

y = 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 the

horizontal and vertical components of its velocity after 3 s?


vx = 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

tvg

toy

30 309 8

1530sin

.; .

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 m

The 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 the

horizontal. What are the horizontal and vertical components of is displacement two

seconds 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/s

x = 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 a

green located 10 m higher than the tee, what was the time of flight, and what was the

horizontal distance to the tee?


y = 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 s

The 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 s

Horizontal 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 the

maximum height attained.


ymax occurs when vy = 0, or when: vy = voy + gt = 0 and t = - voy/g

tvg

toy

18 55

9 8189

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 above

the ground. What was the time of flight and how far did it travel horizontally.


y = 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 s

The 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 s

Horizontal distance to tee: x = voxt = (29.7 m/s)(3.29 s); x = 97.7 m

Challenge Problems

6-42. 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 in

this time?

av v

tf0

(140 m / s) - (60 m / s)8 s

; a = 10 m/s2

sv v

tf

FHG IKJ0 2140 8 m / s + 60 m / s

2 sbg; t = 800 s

6-43. A railroad car starts from rest and coasts freely down an incline. With an average

acceleration 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/s

s = 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 located

12 m down range is dropped from rest. When will they collide and how far are they

located below the release point?

A: vox = 20 m/s, voy = 0; B: vox = voy = 0

Ball 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 s

y = 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 the

acceleration of the car and what was the stopping distance?

av v

tf0

0 - 30 m / s10 s

; a = -3.00 m/s2

sv v

tf

FHG IKJ0 230 10 m / s + 0

2 sb g; s = 150 m

6-46. 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

y

BA

12 m


58

6-47. A stone is thrown vertically downward from the top of a bridge. Four seconds later it

strikes the water below. If the final velocity was 60 m/s. What was the initial velocity of

the 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/s

s = 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 its

position and velocity after (a) 1 s; (b) 3 s; and (c) 6 s

Apply 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, the

package 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 the

components of its final velocity?

vo = 500 mi/h = 733 ft/s; vx = vox = 733 ft/s; voy = 0; t = 4 s

x = 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. What

must be the magnitude and direction of the initial velocity if a ball is to strike the green

at this location after a time of 4 s?

x = voxt; 240 ft = vox (4 s); vox = 60 m/s

s = 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 / s

60 ft / sv = 100 ft/s, = 53.10


6-52. 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:

D

www. elsolucionario.blogspot .com libros … the period t on the moon would be 2(2.45 ) or 4.90 s. 2

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