Solutions to  Equilibrium Problems

Section 3.3

1. Draw a free body diagram

Horizontal:

T_h = T cos 60°
T_h  = (1.0 X 10⁴ N) cos 60°
T_h  = 5.0 X 10³ N

Vertical:

T_v = T sin 60°
T_v =  (1.0 X 10⁴ N) sin 60°
T_v =8.7 X 10³ N

2. From Free Body Diagram

F_net = T_v +T_A + T_A
F_net =  ma
F_net  = 0

T_v = -T_A - T_A
T_v = -2T_A
T_v = -2(-100.0 N) cos 70°

T_v = 68.4 N

3. a)

b) d_v  = (1.5 m) sin 1.5°
d_v  = 0.039 m
d_v  = 3.9 cm

c) F_net  = 2Tv + F_g
F_net = ma
F_net = 0
F_g = -2T_v

m = (-2Tsinq)/g
m = (-2(85 N) sin 15⁰)/(-9.8 N/kg)
m  = 0.45 kg

4. a)

tan q = 1.90 m / 0.650 m
q = 71.1°

F_net =  mg + 2F_Bv
F_net  = ma
F_net  = 0

0 =  mg  + 2F_B sin q
F_B = -mg/2sinq
F_B = 20.7 N

b) F_h  = F_B cosq
F_h  = (20.7 N) cos 71.1°
F_h  = 6.71 N

c) F_v  = F_B sin q

F_v  = (20.7 N) sin 71.1°
F_v  = 19.6 N [down] (not including the weight of the beams)

6.

F||  = mg sin q
F_f  =   µ mg cos q
F_net  = T + F_f + F||
F_net  =  ma = 0

T = -F|| + F_f
T = -mg sin q + µ mg cos q

T = -(400.0 kg)(9.8 N/kg)(sin 30° - (0.25) cos 30°)

T =1.11 X10³ N

Section 3.4

1. a)

b) t =  rF sin q
t = r m g sin q
t = (1.50 m)(45.0 kg)(9.8 N/kg) sin 40°
t = 425 N·m

2. a) t = 2.0 X 10³ N·m
r =1.5 m
q = 90°

F = ?
F = t/r F sin q

F = 1.3 X 10³ N

3.

a) V_w = 10.0 L
D_w  = 1000 kg/m³

V_w = (10.0 L)(1000 cm^3/1 L)
Vw _ 0.0100 m³

m_w  = D_w·V_w
m_w  = (1000 kg/m3)(0.0100 m3)
m_w  = 10.0 kg

F_g  = mg
F_g  = (10.0 kg)(9.8 N/kg)
F_g  = 98.0 N

b) Position B provides the greatest torque because the weight is directed at 90° to the wheel’s rotation.

c) t_A = r F sin q
t_A = (2.5 m)(10.0 kg)(9.8 N/kg) sin 45°
 t_A = 1.7 _ 102 N·m
 t_B =  r F sin q
t_B = (2.5 m)(10.0 kg)(9.8 N/kg) sin 90°
t_B = 2.4 X 10² N·m
 t_B =  t_C

t_C  = 1.7 X 10² N·m

d) A larger-radius wheel or more and larger compartments would increase the torque.

Section 3.5

1.

q = 90°

r₁ = ?
m₁ = 45.0 kg
m₂  = 20.0 kg (0.75/3.0)
m₂  = 20.0 kg (0.75/3.0)
r₂ = 0.75 m /2
r₂ = 0.375 m

m₃ = 20.0 kg - m₂
m₃  = 15.0 kg = 15.0 m

r₃ = (3.0 m  - 0.75 m) /2 = 1.12 m
t_net =  0 = t₁ + t₂ + t₃
0 = r₁ F₁sin q₁+ r₂F₂sin q₂+ r₃F₃sin q₃
r₁ =  (r₃F₃  -   r₂F₂)/F₁ = 0.332 m

2. a)

t _t-t  = r F sin q = 147 N·m

This torque applies to both sides of the teeter-totter, so the torques balance each other.

b)

t _h - t _L = 0
t _h  = t _L
r_L = (r_Hm_Hg)m_Lg =  2.63 m

c)

At maximum height:

t_H = (1.75 m)(45 kg)(9.8 N/kg) sin 75.5° = 7.5 X 10² N·m

% = 2.6%