Please show how to get the answers above with steps shown. One end of a uniform
ID: 1618141 • Letter: P
Question
Please show how to get the answers above with steps shown. One end of a uniform bar that has mass 5.00 kg and length 6.00 m is attached to a vertical wall by a friction less hinge. The bar is held in a horizontal position by a wire that runs from the end of the bar to the wall. The angle between the wire and the bar is 36.9 degree. A 3.00 kg object is suspended from the bar, 4.00 m from the hinge. a) What is the tension T in the wire? b) What is the magnitude of the resultant force exerted on the bar by the hinge?Explanation / Answer
Since the bar is in equilibrium, the net force acting on the bar is 0 and also, the net torque around the Hinge should also be 0 .
Net Torque, = 0
Torque due the mass of the bar = Mass of the bar* Acceleration due to gravity * Length of the Center of mass from the Hinge = 5g kg * 3m = 15g kg.m in Clockwise direction.
Torque due the mass hanging on the bar = Mass of the hanging mass *Acceleration due to gravity * Length of the mass from the Hinge = 3g kg * 4m = 12g kg.m in Clockwise direction.
Torque due to Tension,T of the wire = T sin 36.90 * 6m in the Anti-clockwise direction.
Since Net Torque, = 0 , T sin 36.90 * 6m = (15 + 12) kg.m
-> T * 0.6* 6 = 27g
a) ->Taking g = 9.8m/s2, T = 73.5N
b) T sin 36.90 = 73.5 * 0.6 = 44.1 N ( Acting upwards)
Net force downwards = (5+3)g = 78.4N ( Acting Downwards)
Net Vertical Force on the Hinge = 34.3N (acting Downwards)
Net Horizontal Force on the Hinge = T*cos 36.90 = 73.5*0.8 = 58.8N ( acting in the direction of positive x-axis)
Net resultant Force on the Hinge , Fhinge = ( 34.32+ 58.82) = 68.07N
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