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Two children sit on a seesaw that is in rotational equilibrium. The first child has weight W and sits at distance d from the pivot. If the second child sits at a distance of 7*d from the pivot, what must be the weight of the second child

Sagot :

jay40891

Answer:

W/7

Explanation:

By principle of moments,

Sum of clockwise moment = sum of anticlockwise moment

Weight × 7d = W × d

Weight = W/7

mickymike92

Since the two children are in rotational equilibrium, the weight of the second child is W/7.

How can the weight of the second child be determined?

The weight of the second child can be determined from the principle of moments.

The principle of moments states that for a body in equilibrium, the sum of the clockwise moments and anticlockwise moments about a point is zero.

Let the weight of the second child be X

From the principle of moments:

W × d = 7×d × X

X = W/7

Therefore, the weight of the second child is W/7.

Learn more about principle of moments at: https://brainly.com/question/20298772

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