A uniform wooden plank 5m long and weighing 40kg is resting on two sup

1.	A uniform wooden plank 5m long and weighing 40kg is resting on two supports 0.5 m from each end. Abody weighing 45 kg stands 1.5 m from one end ofthe wooden plank. Find the reactions at the supports.
Answer» For the plank, weight = 40 * 9.8 = 392 NFor the boy, weight = 45 * 9.8 = 441 NTotal weight = 392 + 441 = 883 N Let F1 be the force on the left support and F2 be the force at the right support. Let’s assume the boy is 1.5 meters from the left end of the beam. The sum of these two forces is equal to the total weight. F1 + F2 = 883 N This is a torque problem. Let the pivot point be at the left the support. The weight of the plank and boy will produce clockwise torques. The upward force at the right support will produce counter clockwise torque. The weight of the plank is at its center. 2.5 – 0.5 = 2 mThis is distance from the left support to the center of the plank. Clockwise torque = 392 * 2 = 784 N * m1.5 – 0.5 = 1 mThis is distance from the left support to the boy. Clockwise torque = 441 N * m Total clockwise torque = 784 + 441 = 1225 N * m 4.5 – 0.5 = 4 mThis is distance from the left support to the right support. Counter clockwise torque = F2 * 4 F2 * 4 = 1225F2 = 1225 ÷ 4 = 306.25 N F1 = 883 – 306.25 = 576.75 N

A uniform wooden plank 5m long and weighing 40kg is resting on two supports 0.5 m from each end. Abody weighing 45 kg stands 1.5 m from one end ofthe wooden plank. Find the reactions at the supports.

Answer»

For the plank, weight = 40 * 9.8 = 392 NFor the boy, weight = 45 * 9.8 = 441 NTotal weight = 392 + 441 = 883 N

Let F1 be the force on the left support and F2 be the force at the right support. Let’s assume the boy is 1.5 meters from the left end of the beam. The sum of these two forces is equal to the total weight.

F1 + F2 = 883 N

This is a torque problem. Let the pivot point be at the left the support. The weight of the plank and boy will produce clockwise torques. The upward force at the right support will produce counter clockwise torque. The weight of the plank is at its center.

2.5 – 0.5 = 2 mThis is distance from the left support to the center of the plank.

Clockwise torque = 392 * 2 = 784 N * m1.5 – 0.5 = 1 mThis is distance from the left support to the boy.

Clockwise torque = 441 N * m

Total clockwise torque = 784 + 441 = 1225 N * m

4.5 – 0.5 = 4 mThis is distance from the left support to the right support.

Counter clockwise torque = F2 * 4

F2 * 4 = 1225F2 = 1225 ÷ 4 = 306.25 N

F1 = 883 – 306.25 = 576.75 N

A uniform wooden plank 5m long and weighing 40kg is resting on two supports 0.5 m from each end. Abody weighing 45 kg stands 1.5 m from one end ofthe wooden plank. Find the reactions at the supports.

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