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 = 85.00kg and length L = 5.200m is supported by two vertical massless strings. String A is attached at a distance d = 1.500m 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 = 5.000m 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.

1) Find TA, the tension in string A.

Explanation / Answer

Its given that,

mass m1 = 85.00kg and length L = 5.2m

String A is attached at a distance d = 1.5m

m2 = 3500kg is supported by the crane at a distance x = 5.m

g = 9.807m/s2

1)Tension in string A, T(A) has to be determind.

The sum of torque about string b should be zero, so we can write:

T(A) d - m1 g L / 2 - m2 g x = 0

this will give us:

T(A) =  (m1 g L / 2 + m2 g x ) / d = g (m1 L/2 + m2 x ) / d

T(A) = 9.807 ( 85 x 5.2/2 + 3500 x 5 ) / 1.5 = 115859.898 Newtons

Hence T(A) = 115860 Newtons

Part (2) T(B) has to be calculated.

For T(B), we can write:

T(A) - T(B) - (m1+m2) g = 0

This will give us:

T(B) = T(A) - (m1+m2)g

T(B) = 115860 - ( 80 + 3500 ) 9.807 = 115860 - 35109.06

T(B) = 80751 Newtons

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