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A parent pushes a balanced frictionless playground merry-go-round. The parent exerts a force F tangent to the merry-go-round resulting in a torque of 240 N . m, the distance between the center of the merry-go-round and the point of application of the force is 1.6 m. (b) What, if any, is the magnitude of the horizontal force exerted by the merry-go-round on the man?

Respuesta :

Answer:

F = 150 N

Explanation:

Torque being applied by parent

= F  x r

where F is force applied tangentially and r is point of application of force from axis of rotation ( center )

given torque = 240 Nm

r = 1.6 m

F x r = 240

F x 1.6 = 240

F = 150 N

So a force of 150 N is applied by the parent to achieve the given torque.

This force will also be applied by the merry - go- around  on the parent as reaction force according to Newton's third law. The parent will not be pushed back by this force  because he is standing on a rough surface . Friction of surface will help him remain stationary on the ground.

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