The figure ( Figure 1 ) shows a model of a crane that may be mounted on a truck.
ID: 1288786 • Letter: T
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 = 100.0kg and length L = 5.500m is supported by two vertical massless strings. String A is attached at a distance d = 1.900m 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 = 5.300m from the left end of the bar.
Throughout this problem, positive torque is counterclockwise and use 9.807m/s2 for the magnitude of the acceleration due to gravity.
Find TA, the tension in string A.
Find TB, the magnitude of the tension in string B.
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 = 100.0kg and length L = 5.500m is supported by two vertical massless strings. String A is attached at a distance d = 1.900m 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 = 5.300m from the left end of the bar. Throughout this problem, positive torque is counterclockwise and use 9.807m/s2 for the magnitude of the acceleration due to gravity. Find TA, the tension in string A. Find TB, the magnitude of the tension in string B.Explanation / Answer
with respect to point B
Ta*d = m1*g*(L/2) + m2*g*x
Ta*1.9 = (100*9.807*2.75)+(3000*9.807*5.3)
Ta = 83488.53 N
with respect to point A
Tb*d = m1*g*((L/2)-d) + m2*g*(x-d)
Tb*1.9 = (100*9.807*(2.75-1.9)) + (3000*9.807*(5.3-1.9))
Tb = 53086.83 N
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