The figure ( Figure 1 ) shows a model of a crane that may be mounted on a truck.
ID: 1292304 • 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 = 80.00kg and length L = 5.300m 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 = 3500kg is supported by the crane at a distance x = 5.100m 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.
Part A
Find TA, the tension in string A.
Part B
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 = 80.00kg and length L = 5.300m 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 = 3500kg is supported by the crane at a distance x = 5.100m 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. Part A Find TA, the tension in string A. Part B Find TB, the magnitude of the tension in string B.Explanation / Answer
Sum torque about string B
Ta*d-m1*g*L/2-m2*g*x=0
solve for Ta
Ta=g*(m1*L/2+m2*x)/d
plug in the numbers
Ta=9.81*(80*5.3/2+3500*5.1)/1
177188.22 N
for Tb, sum forces in the y
177188.22-Tb-g*3580=0
solve for Tb
Tb=177188.22-9.81*3580
142068.42 N
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