A student of weight 678 N rides a steadily rotating Ferris wheel (the student sits upright). At the highest point, the magnitude of the normal force on the student from the seat is 586 N. (a) What is the magnitude of at the lowest point? If the wheel's speed is doubled, what is the magnitude FN at the (b) highest and (c) lowest point?

(a) Magnitude of seat force at lowest point = 678 + 92 = 770

(b) Force exerted by the seat (highest point)  = 310 N

(c) Force exerted by seat (lowest point) = 1046 N

Explanation:

At the highest point the magnitude of force by the seat = 586 N

The weight of the student = 678 N

Thus, at the highest point, the difference in this force is due to the centrifugal force acting on the boy. This can be calculated as follows:

Centrifugal Force = 678 - 586 = 92 N

(a) The magnitude of force on the student by the seat at the lowest point will have to appose both the student's weight and centrifugal force acting towards the ground. This would mean:

Magnitude of seat force at lowest point = 678 + 92 = 770 N

(b) If the wheel's speed is doubled, the centrifugal force will change accordingly. The equation of centrifugal force is given below:

We can see from this that the force is directly proportional to the square of the velocity. So if the velocity is doubled, the centrifugal force increases four times.

So at the highest point the centrifugal force will decrease the force of weight acting on the seat. The seat force would then be:

Force exerted by the seat = Weight - Centrifugal force

Force exerted by the seat = 678 - (4 * 92)

Force exerted by the seat (highest point)  = 310 N

(c) The force exerted by the seat at the lowest point will be the centrifugal force plus the weight.

This is:

Force exerted by seat = Centrifugal force + Weight

Force exerted by seat = (4 * 92) + 678

Force exerted by seat (lowest point) = 1046 N

nah.   i'm just sleepy.

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