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

In summary, the Homework statement is about a Ferris wheel that rotates four times each minute. It carries each car around a circle of diameter 18.0 meters. The centripetal acceleration of a rider is -9.81 meters/second2. The seat exerts a force of -40.0 kilograms on a 40.0-kilogram rider when the rider is halfway between the top and bottom of the ride.
  • #1
vadiraja
12
0

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
[tex]\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}[/tex]

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

[tex]n = m\sqrt {{g^2} + {a_r}^2} [/tex]

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

Attachments

  • Ferris Wheel.jpg
    Ferris Wheel.jpg
    9.2 KB · Views: 633
Physics news on Phys.org
  • #2
because the direction of the gravity and the acceleration are not in the same direction.
 
  • #3
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.
 
  • #4
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: 672
  • #5
For part (d), isn't θ = 0?
 

What is normal force in a Ferris wheel?

Normal force is the force exerted by a surface on an object that is in contact with it. In the context of a Ferris wheel, it is the force that the seat exerts on the rider as they move along the circular path.

How is normal force related to gravity in a Ferris wheel?

Normal force and gravity are two equal and opposite forces that act on an object in contact with a surface. In a Ferris wheel, the normal force from the seat balances out the force of gravity on the rider, keeping them in place and preventing them from falling out.

What factors affect the normal force in a Ferris wheel?

The normal force in a Ferris wheel is affected by the speed of the wheel, the radius of the wheel, and the mass of the rider. As the speed or radius increases, the normal force also increases, while a heavier rider will experience a greater normal force.

Why is normal force important in a Ferris wheel?

Normal force is important in a Ferris wheel because it allows the rider to stay safely in their seat as they move along the circular path. Without the normal force exerted by the seat, the rider would experience a feeling of weightlessness and could potentially fall out of the ride.

Can normal force be greater than gravity in a Ferris wheel?

Yes, normal force can be greater than gravity in a Ferris wheel. This can occur when the wheel is moving at a high speed or when the rider is heavier, causing the seat to exert a greater normal force on the rider to keep them in place.

Similar threads

  • Introductory Physics Homework Help
3
Replies
95
Views
3K
  • Introductory Physics Homework Help
Replies
4
Views
498
  • Introductory Physics Homework Help
Replies
9
Views
809
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
18
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
594
  • Introductory Physics Homework Help
Replies
3
Views
126
  • Introductory Physics Homework Help
2
Replies
41
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
527
  • Introductory Physics Homework Help
Replies
7
Views
1K
Back
Top