A school yard teeter-totter with a total length of 5.34 m and a mass of 35.9 kg

Question

A school yard teeter-totter with a total length of 5.34 m and a mass of 35.9 kg is pivoted at its center. A 17.7-kg child sits on one end of the teeter-totter. Where (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?

b) Where should the parent push with a force of 284 N?

c)Where should the parent push with a force of 184 N if the mass of the teeter-totter were doubled?

d)Where should the parent push with a force of 284 N if the mass of the teeter-totter were doubled?

Explanation / Answer

The mass of the teeter-totter M = 35.9kg

the length of the teeter-totter L = 5.34m

the mass of the child m = 17.7kg

the force applied by parent F = 184N

Now from moments forces, about center

     (17.7kg)(9.8) (2.67) = (184)x

then x = 463.14/184

           = 2.52m from pivot

(b) If the force applied by parent F = 284N

then again from principle of moments of force

        (17.7)(9.8)(2.67) = (284)x

then x = 1.63 m from pivot

(c) When the teeter -totter mass is doubled,

the parent has to applied the force 184N at 2.52m from pivot

there is no change weather the mass of the totter changed.

(d) Whenthe mass doubled, the point is x = 1.63 m only

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