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A schoolyard teeter-totter has a total length of 5.34 m and a mass of 35.9 kg. It is pivoted at its center. A 17.7 kg child sits on one end of the teeter-totter.

Where, or how far away from the pivot point, should a parent push vertically downward with a force of 184 N in order to hold the teeter-totter level?

Answer :

Main Answer:

To maintain balance on the teeter-totter, the parent should push vertically downward with a force of 184 N at a distance of approximately 2.09 meters from the pivot point.

Explanation:

In order to keep the teeter-totter level, the torque produced by the parent's force must balance the torque caused by the child's weight. The torque exerted by the parent's force (τ_p) is calculated using the formula τ_p = Fd, where F is the force exerted (184 N) and d is the distance from the pivot point.

The torque caused by the child's weight (τ_c) is calculated as τ_c = mgd, where m is the child's mass (17.7 kg), g is the acceleration due to gravity (9.81 m/s^2), and d is the distance from the pivot point.

Setting these torques equal to each other and solving for d gives us the distance from the pivot point where the parent should push to maintain balance.

Thus, d = (m_c/m_p) * L, where m_c is the child's mass, m_p is the parent's force, and L is the length of the teeter-totter. Plugging in the values, we get d ≈ (17.7 kg / 184 N) * 5.34 m ≈ 2.09 m.

Therefore, the parent should push downward at a distance of approximately 2.09 meters from the pivot point to hold the teeter-totter level.

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