You have been hired to design a family-friendly see-saw. Your design will featur

Question

You have been hired to design a family-friendly see-saw. Your design will feature a uniform board (mass M, length L) that can be moved so that the pivot is a distance d from the center of the board. This will allow riders to achieve static equilibrium even if they are of different mass, as most people are. You have decided that each rider will be positioned so that his/her center of mass will be a distance xoffset from the end of the board when seated as shown. You have selected a child of mass m (shown on the right), and an adult of mass n times the mass of the child (shown on the left) to test out your prototype.

Part (a) Derive an expression for the torque applied by the adult rider (on the left) in terms of given quantities and variables available in the palette. Assume counterclockwise is positive.

Part (b) Derive an expression for the torque applied by the child rider (on the right) in terms of given quantities and variables available in the palette. Assume counterclockwise is positive.

Part (c) Derive an expression for the torque applied by the board in terms of given quantities and variables available in the palette.

Part (d) Determine the distance d in terms of n, g, and the masses and lengths in the problem.

Part (e) Determine the magnitude of the force exerted on the pivot point by the see-saw while in use in terms of given quantities and variables available in the palette.

Explanation / Answer

As per question the adult is on the left side of the pivot point, the adult’s weight will cause the board to rotate counter clockwise. When the pivot point is moved to the left of the center of the board, the weight of the board will be on the right side of the pivot point. This means board’s weight will cause the board to rotate clockwise. Since the child is on the right side of the pivot point, the child’s weight will cause the board to rotate clockwise. For the board to be balanced the counter clockwise torque must be equal to the clockwise torque.

Let’s determine the weights.

For the board, weight = M * g = Mg

For the child, weight = m * g = mg

For the adult, weight = nm * g = nmg

To determine the torque, we must determine the distance from the pivot to each person. To determine this distance we need to the distance each person is from the center of the board. Since the board is L meters long, each half of the board is L/2 meters long. Since each person is x meter from the board, each person is (L/2 - x) meters from the center of the board.
Let K = (L/2 - x)

(a) ...........The distance from the adult to the pivot point is (K - d).
Torque = nmg(K – d) N-m

(b).............The distance from the child to the pivot point is (K + d).
Torque = mg * (K + d) N-m

(c).............The distance from the center of the board to the pivot point is d.
Torque = Mg * d N-m

Total clockwise torque = Mg * d + mg * (K + d)

Set the torques equal to each other and solve for d.

nmg(K – d) = Mg * d + mg * (K + d)
d = K * [ (n + 1) * m ] / (M + 2m)

d = (L/2 - x) [(n +1)*m] / (M + 2m) metres

The magnitude of the force exerted on the pivot point by the see-saw while in use is equal to the sum of the weights.

F = Mg + mg + nmg Newtons

By putting the values in above equations , we can determine the answers.

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