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

14 N.

Explanation:

Let the force applied by the boy pulling to the left be x.

From the question given above, the following data were obtained:

Net force (Fₙ) = 10 N toward the right

Force applied by the boy pulling to the left = x

Next, we shall determine the total force in the left direction. This can be obtained as follow:

Force in the left direction (Fₗ) = x + 8

Next, we shall determine the total force in the right direction. This can be obtained as follow:

Force in the right direction (Fᵣ) = 11 + 21 = 32 N

Finally, we shall determine the force applied by the boy pulling to the left direction (i.e the value of x) as follow:

Net force (Fₙ) = 10 N toward the right

Force in the left direction (Fₗ) = x + 8

orce in the right direction (Fᵣ) = 32 N

Fₙ = Fᵣ – Fₗ (since the net force is toward the right direction)

10 = 32 – (x + 8)

Clear bracket

10 = 32 – x – 8

10 = 32 – 8 – x

10 = 24 – x

Collect like terms

10 – 24 = – x

–14 = –x

Divide both side by –1

x = –14/–1

x = 14 N

Thus, the force applied by the boy pulling to the left direction is 14 N.