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 = 9 kg, length L = 11 m) 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 = 23 cm from the end of the board when seated as shown. You have selected a child of mass m = 29 kg (shown on the right), and an adult of mass n =3 times the mass of the child (shown on the left) to test out your prototype.

Determine the distance d in m.

Explanation / Answer

For the adult, weight = 3*29 * 9.8 = 852.6 N
For the board, weight = 9 * 9.8 = 88.2 N
For the child, weight = 29 * 9.8 = 284.2 N

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 11 meters long, each half of the board is 5.5 meters long. Since each person is 0.23 meter from the board, each person is 5.27 meters from the center of the board.

(5.5 – 0.23 = 5.27)


The distance from the adult to the pivot point is 5.27 – d.

Torque =852.6 * (5.27 – d) = 4493– 852.6 * d


The distance from the center of the board to the pivot point is d.
Torque =88.2 * d

The distance from the child to the pivot point is 5.27 + d.

Torque = 284.2 * (5.27 + d) = 1625.6 + 284.2 * d

Total clockwise torque = 88.2 * d+ 1625.6 + 284.2 * d = 372.4* d + 1625.6

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


4493– 852.6 * d = 372.4* d + 1625.6
1225 * d = 2867.4
d = 2867.4 ÷ 1225 = 2.34 meter

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