Family-Friendly See-Saw Physics Design – Full Solution

Question:

You have been hired to design a family-friendly see-saw. Your design will feature a uniform board of mass M = 9.31 kg and length L = 14.02 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 masses. You have decided that each rider will be positioned so that his/her center of mass is a distance x_offset = 13.18 cm = 0.1318 m from the end of the board.

A child of mass m = 22.9 kg and an adult of mass n = 68.7 kg (3 times the child’s mass) are placed on either end. Determine:

The distance d from the center of the board to the pivot for static equilibrium.
The magnitude of the force exerted on the pivot point.

Answer:

Step 1: Distance from the Center for Each Rider

The board is symmetric, so half its length is:

L / 2 = 14.02 / 2 = 7.01 m

Subtracting the offset from the end gives:

Distance = 7.01 m – 0.1318 m = 6.8782 m

Step 2: Applying the Torque Balance (Static Equilibrium)

To maintain equilibrium, total torque about the pivot must be zero.

Torque by child (counterclockwise): τ_child = m × g × (6.8782 + d)
Torque by adult (clockwise): τ_adult = -n × g × (6.8782 – d)
Torque by board (counterclockwise): τ_board = M × g × d

Setting the sum of torques to zero:

m·g·(6.8782 + d) – n·g·(6.8782 – d) + M·g·d = 0

Cancel g and simplify:

m(6.8782 + d) – n(6.8782 – d) + M·d = 0

m·6.8782 + m·d – n·6.8782 + n·d + M·d = 0

d(m + n + M) = (n – m)·6.8782

Step 3: Substituting the Values

d = ((68.7 – 22.9) × 6.8782) / (22.9 + 68.7 + 9.31) = (45.8 × 6.8782) / 100.91 ≈ 315.002 / 100.91 ≈ 3.12 m

Result: Distance of the pivot from the center = 3.12 m

Step 4: Magnitude of the Force Exerted on the Pivot

In static equilibrium, the upward force from the pivot equals the total weight of the system:

F_pivot = (m + n + M) × g = (22.9 + 68.7 + 9.31) × 9.8 = 100.91 × 9.8 ≈ 988.92 N

Result: Force on the pivot ≈ 989 N

Conclusion:

To maintain balance in a family-friendly see-saw design where one rider is heavier than the other, the pivot must be shifted by approximately 3.12 meters from the board’s center. This ensures static equilibrium. The total force the pivot exerts upward is approximately 989 newtons, which balances the total downward weight of all components.

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