The weight of the beam is 4.0N and acts at a point 20cm from the pivot. A 2.0N weight hangs 10cm from the pivot. An upward force U is needed to keep the beam horzontal. What is the size of U?
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Since the weight of the beam acts on the centre of mass, we can deduce from given information that the total length of the beam is 40cm.
If force U acts at the end of the beam, torque about pivot needs to be zero for the beam to remain horizontal.
Torque, T = (20).(4) + (10).(2) - (40).(U) = 0
=> 80 + 20 = 40U
=> U = 100/40 = 2.5 N
Therefore a force of 2.5N at the end of the beam is required to keep it horizontal.
If force U acts at the end of the beam, torque about pivot needs to be zero for the beam to remain horizontal.
Torque, T = (20).(4) + (10).(2) - (40).(U) = 0
=> 80 + 20 = 40U
=> U = 100/40 = 2.5 N
Therefore a force of 2.5N at the end of the beam is required to keep it horizontal.
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