The figure (Figure 1) shows a model of a crane that may be mounted on a truck.A

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 m1 = 85.00 kg and length L = 5.800 mis supported by two vertical massless strings. String A is attached at a distance d = 1.500 m 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 m2 = 3000kg is supported by the crane at a distance x = 5.600 m from the left end of the bar.

Throughout this problem, positive torque is counterclockwise and use 9.807 m/s2 for the magnitude of the acceleration due to gravity.

Part A) Find TA, the tension in string A.

Express your answer in newtons using four significant figures.

Part B) Find TB

, the magnitude of the tension in string B.

Express your answer in newtons using four significant figures.

Explanation / Answer

Mass of horizontal bar, m1 = 85.0 Kg
Length of horizontal bar, L1 = 5.8 m
d = 1.5 m
x = 5.6 m
m2 = 3000 Kg

(a)
As the Bar is at Equilirbrium, Sum of Net Force acting on Bat at any point = 0
Calculating Sum torque about string B -
m2 * x + m1 * L/2 - Ta * d = 0
3000 * 9.8 * 5.6 + 85.0 * 9.8 * 5.8/2 = Ta * 1.5
Solvong for Ta
Ta = 111370 N
Tension in the String A, Ta = 111370 N

(b)
Sum Of Vertical Forecs = 0
Tb = Ta - m1*g - m2*g
Tb = 111370 - (85 + 3000)*g
Tb = 81137 N
Tension in the String B, Tb = 81137 N

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