Question

The figure (Figure 1) shows a model of a crane that may be mounted on a truck.A rigid uniform horizontal bar of mass = 100.0 and length = 5.800 is supported by two vertical massless strings. String A is attached at a distance = 1.600 from the left end of the bar and is connected to the top plate. String B is attached to the left end of the bar and is connected to the floor. An object of mass = 2000 is supported by the crane at a distance = 5.600 from the left end of the bar.

Answer 1

m1 = 95 kg
L= 5.2m
d = 2.0m
m2 = 3500kg
x = 5.00m
g = 9.807 m/s/s


TA = [g(m1 * L/2 + m2 * X)] / d

TB = TA - (g * (m2 + m1))

just plug in your values!

Answer 2

Net torque is zero.
0 = (T_b)(d) - (m₁)(g)(L/2 - d) - (m₂)(g)(x - d)
T_b = g[(m₁)(L/2 - d) + (m₂)(x - d)] / d
Tb = (9.81 m/s²)[(100 kg)(5.8/2 - 1.6 m) + (2000 kg)(5.6 - 1.6m)] / (1.6 m)
T_b =49847 N
Net force is zero.
0 = T_a - (m₁)(g) - (m₂)(g) - T_b
T_a = T_b + g(m₁ + m₂)
T_a =49847 + (9.81 )(100+ 2000 )
T_a = 70488N or 70.488KN

Answer 3

figure