A school yard teeter totter with a total length of 4.0 m and a mass of 36 kg is

Question

A school yard teeter totter with a total length of 4.0 m and a mass of 36 kg is pivoted at itscenter. A 17 kg child sits on one endof the teeter totter. (a) Where should a parent push with a force of218 N in order to hold the teetertotter level?
1 m (from the pivot)
(b) Where should the parent push with a force of 341 N to keep the teeter totter level?
2 m (from the pivot)
(c) How would your answers to parts (a) and (b) change if the massof the teeter totter were doubled? Explain.
3 (a) Where should a parent push with a force of218 N in order to hold the teetertotter level?
1 m (from the pivot)
(b) Where should the parent push with a force of 341 N to keep the teeter totter level?
2 m (from the pivot)
(c) How would your answers to parts (a) and (b) change if the massof the teeter totter were doubled? Explain.
3

Explanation / Answer

a) =rxF (cross product simplified to rf, sincesin90=1) since the teeter totter is pivoting on its center, half itsmass and length are on each side. torque on the left side of the teeter totter: torque fromkid+torque from teeter totter =(5.1/2)(20*9.8 + [36/2]*9.8)=949.62Nm for the totter to be balanced, the torque on the right sidehas to equal to the left side. torque on right=torque from totter+torque from parent, whichhas to equal 949.62Nm (5.1/2)([36/2]*9.8+252*r)=949.62 solving for r gives 1.98m, which is the distance from thepivot point the parent must apply 252N.
b) The procedure for 329N is thesame: (5.1/2)([36/2]*9.8+329*r)=949.62 Solving for r gives 1.52m, which is the distance from thepivot the parent must apply the force.
c) The answers would be the same because the teeter totter isbalanced at its center. Because of this, the torque due to theteeter totter from both sides cancels out (ie the net torque iszero).
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