At the local playground, a 21-kg child sits on the right end of a horizontal tee

Question

At the local playground, a 21-kg child sits on the right end of a horizontal teeter-totter, 1.8 m from the pivot point. On the left side of the pivot an adult pushes straight down on the teeter-totter with a force of 151 N.

Part A In which direction does the teeter-totter rotate if the adult applies the force at a distance of 3.0 mfrom the pivot?

clockwise or counterclockwise

Part B In which direction does the teeter-totter rotate if the adult applies the force at a distance of 2.5 mfrom the pivot?

clockwise or counterclockwise

Part C In which direction does the teeter-totter rotate if the adult applies the force at a distance of 2.0 mfrom the pivot?

clockwise or counterclockwise

Explanation / Answer

Given,

m = 21 kg ; d = 1.8 m ; F = 151 N

A)d' = 3 m

W = mg = 21 x 9.81 = 206.01 N

Torque due to child's weight,

Tau-w = -206.01 x 1.8 = -370.82 N-m (Its downward, CW torque)

that due to applied force:

Tau-F = 151 x 3 = 453 N-m CCw torque

Net Torque = Tau-w + Tau-F = -370.82 + 453 = +82.18

Its positive implies, rotation will be CCW that is Counter clockwise

B)d = 2.5 m

Now, Tau-w remains same ; Tau-w = -370.82 N

Tau-F = 2.5 x 151 = 377.5

Net torque = -370.82 + 377.5 = +6.68 N-m

Since its positive, it will now rotate in CCW that is counter clockwise direction.

3)d = 2 m

Tau-F = 2 x 151 = 302 Nm

Net torque

Tau-net = -370.82 + 302 = -68.82 N-m

This time, its negative, so it will rotate in CW that is clockwise direction.

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