Question

A uniform wooden plank MN of length 5 m weighs 50 N. It is supported at end M at a point S,
1 m from the other end N. Determine the maximum weight W that can be placed at the end N
so that the plank does not topple.

Answer 1

Answer:

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For the plank, weight = 40 * 9.8 = 392 N

For the boy, weight = 45 * 9.8 = 441 N

Total 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 m

This is distance from the left support to the center of the plank.

Clockwise torque = 392 * 2 = 784 N * m

1.5 – 0.5 = 1 m

This 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 m

This is distance from the left support to the right support.

Counter clockwise torque = F2 * 4

F2 * 4 = 1225

F2 = 1225 ÷ 4 = 306.25 N

F1 = 883 – 306.25 = 576.75 N

Answer 2

Answer:

For the plank, weight = 40 * 9.8 = 392 N

For the boy, weight = 45 * 9.8 = 441 N

Total 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 m

This is distance from the left support to the center of the plank.

Clockwise torque = 392 * 2 = 784 N * m

1.5 – 0.5 = 1 m

This 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 m

This is distance from the left support to the right support.

Counter clockwise torque = F2 * 4

F2 * 4 = 1225

F2 = 1225 ÷ 4 = 306.25 N

F1 = 883 – 306.25 = 576.75 N