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

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Homework Help Overview

The problem involves analyzing the forces acting on a rider in a Ferris wheel, specifically focusing on the normal force at various points in the ride, including the midpoint. The subject area includes concepts of centripetal acceleration and forces in circular motion.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to derive the normal force using a force balance approach, while some participants question the use of sine in the calculations and the direction of forces involved.

Discussion Status

Participants are exploring different interpretations of the forces acting on the rider, particularly the relationship between gravitational force and centripetal acceleration. There is an ongoing discussion about the correctness of the original poster's reasoning compared to the solution manual, with no explicit consensus reached.

Contextual Notes

There is mention of a solution manual that provides a different answer, which has led to confusion and questioning of assumptions regarding the direction of forces. The original poster has expressed uncertainty about their approach and seeks clarification on the discrepancies.

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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  • Ferris Wheel.jpg
    Ferris Wheel.jpg
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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
 

Attachments

  • figure for liquidxlax.jpg
    figure for liquidxlax.jpg
    8.4 KB · Views: 759
For part (d), isn't θ = 0?
 

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