Question

A Bar Suspended by Two Vertical Strings

The figure (Figure 1) shows a model of a crane that may be mounted on a truck.A rigid uniform horizontal bar of massm1 = 80.00kg and length L = 5.100m is supported by two vertical massless strings. String A is attached at a distance d = 1.200m 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 massm2 = 3500kg is supported by the crane at a distancex = 4.900m 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. Express your answer in newtons using four significant figures.

Find TB, the magnitude of the tension in string B. Express your answer in newtons using four significant figures.

Answer 1

Sum of torque about string B

(T_A * d) - (m₁ * g * L / 2) - (m₂ * g * x) = 0

T_A = g * (m₁ * L/2 + m₂ * x) / d

= 9.81 * (80 * 5.1 / 2 + 3500 * 4.9 ) / 1 .2

T_A = 141724 N = 1.42 * 10⁵ N

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For Tb, sum of forces in the y

141724 - T_B - g * (3500 + 80) = 0

T_B= 141724 - 9.81 * 3580

T_B = 106640 N

Answer 2

[REMOVED]