What is the Normal Force on a Rider in a Ferris Wheel at the Midpoint?

AI Thread Summary
The discussion revolves around calculating the normal force on a rider in a Ferris wheel at the midpoint of the ride. The original problem involves determining centripetal acceleration and the forces acting on a 40.0-kg rider at various points, including the midpoint. The user expresses confusion regarding their calculations for the normal force, specifically questioning the use of sine in the solution manual's formula. The solution manual suggests using the Pythagorean theorem to combine gravitational and centripetal forces, while the user believes their approach is valid but misinterpreted. Ultimately, the key point is the distinction between the directions of gravitational and centripetal forces and how they combine to affect the normal force.
vadiraja
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Homework Statement


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}<br /> \sum {{F_r} = m{a_r} = (\ddot r} - r{\left( {\dot \theta } \right)^2}) = \frac{{ - m{v^2}}}{r} = - mg\sin (\theta ) + n\\<br /> n = mg\sin (\theta ) - \frac{{m{v^2}}}{r}\\<br /> \theta = 0,n = \frac{{ - m{v^2}}}{r}<br /> \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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because the direction of the gravity and the acceleration are not in the same direction.
 
I don't understand the sin(θ) in your answer.
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.
 
Liquidxlax said:
because the direction of the gravity and the acceleration are not in the same direction.

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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For part (d), isn't θ = 0?
 
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