ENGR0135 - Statics and Mechanics of Materials 1 (2161) Homework #1 Solution Set 1. The masses of the two bodies are m 1 = W 1 g = 500 lb 32.2 ft/s 2 = 15.5280 lb · s 2 ft = 15.5280 slug m 2 = W 2 g = 1500 lb 32.2 ft/s 2 = 46.5839 lb · s 2 ft = 46.5839 slug Then, the gravitational force of attraction between the two bodies is F = G m 1 m 2 r 2 = 3.439 × 10 -8 ft 3 slug · s 2 (15.5280 slug)(46.5839 slug) (5 ft) 2 =9.9505 × 10 -7 slug · ft s 2 Thus, F =9.95 × 10 -7 lb 2. The weight in SI units is W = (150 lb)(4.448 N/lb) = 667.2N Then, the mass in SI units is m = W g = (667.2 N) (9.81 m/s 2 ) = 68.0122 kg Thus, m = 68.0 kg 3. Substituting the following dimensions Re =1 , ρ = M L 3 , v = L T , L = L into the equations for the Reynolds number gives 1= (M/L 3 )(L/T )(L) μ = (L/T )(L) ν Then, solving for the dimensions of μ and ν gives μ = M LT , ν = L 2 T
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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