# Normal force in Ferris wheel

1. Dec 6, 2011

1. The problem statement, all variables and given/known data
The original problem is
Figure P6.61 shows a Ferris wheel( I have attached the image) that rotates four times
each minute. It carries each car around a circle of diameter
18.0 m. (a) What is the centripetal acceleration of a
rider? What force does the seat exert on a 40.0-kg rider
(b) at the lowest point of the ride and (c) at the highest
point of the ride? (d) What force (magnitude and direction)
does the seat exert on a rider when the rider is
halfway between top and bottom?

I have already done a), b) and c) with no problem. I also have a solution manual to check and I know that I am correct.

For d) This is my reasoning
$$\begin{array}{l} \sum {{F_r} = m{a_r} = (\ddot r} - r{\left( {\dot \theta } \right)^2}) = \frac{{ - m{v^2}}}{r} = - mg\sin (\theta ) + n\\ n = mg\sin (\theta ) - \frac{{m{v^2}}}{r}\\ \theta = 0,n = \frac{{ - m{v^2}}}{r} \end{array}$$

however, this is not what the solution manual says. This is what the solution manual says:

$$n = m\sqrt {{g^2} + {a_r}^2}$$

I am not dealing with the numbers yet. I want to know why I am wrong and the solution manual is correct. Thankyou.

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2. Dec 6, 2011

### Liquidxlax

because the direction of the gravity and the acceleration are not in the same direction.

3. Dec 6, 2011

### Delphi51

It is easy to see the book answer. The seat has to push up with force mg to cancel weight. And it also has to push in with ma to provide the centripetal force. Combine those with the Pythagorean theorem to get its answer.

4. Dec 6, 2011

it is obviously true that the gravity and the acceleration are not acting in the same direction. However, my solution does not imply that they are acting in the same direction. I have put a figure for clarification.

The forces acting in the radial direction are
mgsinθ and n which are both point into the center

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5. Dec 6, 2011

### Delphi51

For part (d), isn't θ = 0?