ENGR0135 - Statics and Mechanics of Materials 1 (2161)Homework #1

Solution Set

1. The masses of the two bodies are

m1 =W1

g=

500 lb

32.2 ft/s2= 15.5280

lb · s2ft

= 15.5280 slug

m2 =W2

g=

1500 lb

32.2 ft/s2= 46.5839

lb · s2ft

= 46.5839 slug

Then, the gravitational force of attraction between the two bodies is

F = Gm1m2

r2

=

(3.439× 10−8

ft3

slug · s2)

(15.5280 slug)(46.5839 slug)

(5 ft)2

= 9.9505× 10−7slug · ft

s2

Thus,

F = 9.95× 10−7 lb

2. The weight in SI units is

W = (150 lb)(4.448 N/lb) = 667.2 N

Then, the mass in SI units is

m =W

g=

(667.2 N)

(9.81 m/s2)= 68.0122 kg

Thus,m = 68.0 kg

3. Substituting the following dimensions

Re = 1 , ρ =M

L3, v =

L

T, L = L

into the equations for the Reynolds number gives

1 =(M/L3)(L/T )(L)

µ=

(L/T )(L)

ν

Then, solving for the dimensions of µ and ν gives

µ =M

LT, ν =

L2

T

4. (a) The rounded-off number is 0.00165 and the percentage difference is

%D =0.00165− 0.00164893

0.00164893(100) = 0.0649%

(b) The rounded-off number is 6.00 and the percentage difference is

%D =6.00− 5.99511

5.99511(100) = 0.0816%

(c) The rounded-off number is 2, 370, 000 and the percentage difference is

%D =2, 370, 000− 2, 367, 001

2, 367, 001(100) = 0.127%

5. To find the resultant R of the three forces, first find the resultant R12 of two of theforces and then R is the resultant of R12 and the third force. Let R12 be the resultantof the 10 kN and 5 kN forces:

x

y

10 kN

5 kNα

R12R2

12 = 102 + 52 =⇒ R12 = 11.1803 kN

sinα

10=

sin 90

R12=⇒ α = 63.4354◦

Then, R is the resultant of R12 and the 7 kN force:

x

y

α R1230◦7 kN

Rβ = 180− 30− α = 86.5646◦

R2 = R212 + 72 − 2R12(7) cosβ

}=⇒ R = 12.8304 kN

β

β

γ sin γ

7=

sinβ

R=⇒ γ = 32.9973◦

Thus, the magnitude of the resultant R and the angle its line of action forms with thex-axis are

x

yR

θx

R = 12.83 kN

θx = γ + (90− α) = 59.6◦

6. Using the law of sines and the law of cosines, in conjunction with a sketch of the forcetriangle:

35◦

35◦

45◦

45◦

αα

F1

F2

Rα = 180− 35− 45 = 100◦

F1

sin 35◦=

F2

sin 45◦=

400 lb

sin 100◦

F1 = 233.0 lb , F2 = 287.2 lb

7. (a) The x- and y-components of the force are

Fx = 950 cos 30◦ = 822.7 N

Fy = −950 sin 30◦ = −475.0 N

(b) The x′- and y′-components of the force are

Fx′ = 950 cos(45◦ − 30◦) = 917.6 N

Fy′ = 950 sin(45◦ − 30◦) = 245.9 N

8. (a) The vector from D to B is

rDB = (xB − xD)i + (yB − yD)j + (zB − zD)k

= (5− 0)i + (−5− 0)j + (0− 8)k

= 5i− 5j− 8k (m)

The distance from D to B is

rDB =√

(5)2 + (−5)2 + (−8)2 = 10.6771 m

The unit vector in the direction of TB, i.e., in the direction from D to B, is

eDB =rDB

rDB

= 0.4683i− 0.4683j− 0.7493k

Thus, TB is expressed in Cartesian vector form as

TB = 2230 eDB = 1044i− 1044j− 1671k (N)

(b) The vector from D to A is

rDA = (xA − xD)i + (yA − yD)j + (zA − zD)k

= (10− 0)i + (5− 0)j + (0− 8)k

= 10i + 5j− 8k (m)

The distance from D to A is

rDA =√

(10)2 + (5)2 + (−8)2 = 13.7477 m

The unit vector in the direction from D to A, is

eDA =rDA

rDA

= 0.7274i + 0.3637j− 0.5819k

The scalar product of TB and eDA is

TB · eDA = (1044)(0.7274) + (−1044)(0.3637) + (−1671)(−0.5819)

= 1352.06 N

Thus, the magnitude of the rectangular component of the force TB along DA is

TB · eDA = 1352 N

(c) Since TB · eDA = |TB||eDA| cosα, |TB| = 2230 N, and |eDA| = 1, it follows that

α = cos−1(

1352.06

2230

)= 52.7◦

Alternatively, the angle could be determined from eDB · eDA = cosα.

9. The x- and y-components of the resultant R are

Rx = 10− 7 sin 30◦ = 6.5 kN , Ry = 5 + 7 cos 30◦ = 11.0622 kN

Thus, the magnitude of the resultant force is

R =√R2

x +R2y = 12.83 kN

and the angle between its line of action and the x-axis is

θx = cos−1(Rx

R

)= 59.6◦

10. The x-component of the resultant is the sum of the x-components of the three forces:

Rx = 300 cos 40◦ cos 60◦ + 450 cos 35◦ cos 45◦ − 600 cos 30◦ sin 40◦ = 41.5570 lb

The y-component of the resultant is the sum of the y-components of the three forces:

Ry = 300 cos 40◦ sin 60◦ − 450 cos 35◦ sin 45◦ − 600 cos 30◦ cos 40◦ = −459.677 lb

The z-component of the resultant is the sum of the z-components of the three forces:

Rz = 300 sin 40◦ + 450 sin 35◦ + 600 sin 30◦ = 750.946 lb

Thus, the magnitude of the resultant is

R =√R2

x +R2y +R2

z = 881.448 lb = 881 lb

The angles between the line of action of the resultant and the positive coordinate axesare

θx = cos−1Rx

R= 87.3◦

θy = cos−1Ry

R= 121.4◦

θz = cos−1Rz

R= 31.6◦