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A 60 kg adult and a 30kg child are passengers on a rotor ride at an amusement park as shown in the diagram above. When the rotating hollow cylinder reaches a certain constant speed, v, the floor moves downward. Both passengers stay "pinned" against the wall of the rotor, as shown in the diagram below.

The magnitude of the frictional force between the adult and the wall of the spinning rotor is F. What is the magnitude of the frictional force between the child and the wall of the spinning rotor?

Respuesta :

Answer:

 fr ’= ½ F

Explanation:

For this exercise we use the translational equilibrium equation, on the axis parallel to the wall

              fr - W = 0

              fr = W

for the adult man they indicate that the friction force is equal to F

              F = M g

we write the equilibrium equation for the child

             fr ’= w’

             fr ’= m g

in the statement they tell us that the mass of the adult is 2 times the mass of the child

             M = 2m

we substitute

            fr ’= M / 2 g

            fr ’= ½ Mg

we substitute

            fr ’= ½ F

therefore the force of friction in the child is half of the friction in the adult

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