# Force of Gravity on a Man on a Ferris Wheel

• cheerspens
In summary, a man is sitting on a Ferris wheel with a mass of 50 kg. The Ferris wheel completes a single revolution every 20 seconds. The man is subject to the force of gravity which is 490 N. The attempted solution found that the chair would be pushed with a force of 473740.71 m/s2 if the period of the wheel was 0.05 rev/sec.
cheerspens

## Homework Statement

A man sitting on the edge of his seat on a Ferris Wheel has a mass of 50.0 kg. The Ferris Wheel has a radius r=30m and the ferris wheel completes a single revolution every 20 seconds. Find the force between the man and the chair.

## Homework Equations

FNET=ma
Fg=mg

$$\tau$$=2$$\pi$$r/v
a=v2/r

## The Attempt at a Solution

I found the force of gravity on the man to be 490N. I think the period is 0.05rev/sec so I set up the $$\tau$$ equation to be 0.05=2$$\pi$$(30)/v. I solved for V however and get 3769.91 m/s. It seems like too big of a number. This then gives me a very large acceleration of 473740.71m/s2.
I need to solve for a to plug into my FNET equation in order to find the force on the chair on man.
What am I doing wrong to get these large numbers?

The period is the time taken for one revolution, so T = ?

then you can use ω=2π/T

Your main error lies in the periodic time.

But I don't have a velocity so how can I solve for T? I thought I had to solve for velocity first?

Force and Circular Motion??

## Homework Statement

A man sitting on the edge of his seat on a Ferris Wheel has a mass of 50.0 kg. The Ferris Wheel has a radius r=30m and the ferris wheel completes a single revolution every 20 seconds. Find the force between the man and the chair.

## Homework Equations

FNET=ma
Fg=mg

$$\tau$$=2$$\pi$$r/v
a=v2/r

## The Attempt at a Solution

I set up FNET to be FGP-FCP=ma. I found the force of gravity on the man to be 490N and we know the person's mass so the equation is now:
490-FCP=(50)a
So in order to solve for the force between the chair and the person (FCP) you have to solve for the acceleration (a).
I think the period is 0.05rev/sec so I set up the $$\tau$$ equation to be 0.05=2$$\pi$$(30)/v. I solved for V however and get 3769.91 m/s. It seems like too big of a number. This then gives me a very large acceleration of 473740.71m/s2.
I need to solve for a to plug into my FNET equation in order to find the force on the chair on man.
What am I doing wrong to get these large numbers?

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Last edited:

Try finding the speed directly. How much distance does he cover in 20 seconds?

As for the force between the chair and the man, it looks like not enough information is given. That force depends on where the man is. For example at the "12 o' clock" position the force is different from the "6 o' clock" position.

Would it me 0.05rev/sec? And that is velocity?
(I also just attached the diagram)

You have posted duplicate threads. I will refrain from posting here until the admins merge the threads.

It may help to think (or read in your textbook) about if there is a simple relationship between the centripetal acceleration, angular speed and radius for a steady circular motion. This should give you the [STRIKE]horizontal [/STRIKE]centripetal force acting on the man which, when vectorially added to the vertical gravity force acting on the man should give you the combined force acting on the man.

Edit: I didn't know a Ferris wheel is a "vertical" wheel until a diagram was added (english being my 2nd language and all) and for that wheel the centripetal acceleration is of course not horizontal in general.

Last edited:
cheerspens said:
But I don't have a velocity so how can I solve for T? I thought I had to solve for velocity first?

You can use T to get v.

One revolution takes 20 seconds, so the period is what ?

0.05rev/sec??

cheerspens said:
0.05rev/sec??

Periodic time is the time taken for one revolution. So T is ? (they gave it to you)

Then you can find ω.

rock.freak667 said:
Periodic time is the time taken for one revolution. So T is ? (they gave it to you)

Then you can find ω.

Oh that's what I forgot and what had me confused. Well that solves everything. Thank you!

## What is the force of gravity on a man on a Ferris Wheel?

The force of gravity on a man on a Ferris Wheel is the same as the force of gravity on any object on the surface of the Earth. It is equal to the product of the mass of the man and the acceleration due to gravity (9.8 meters per second squared).

## Does the force of gravity change as the man moves on the Ferris Wheel?

Yes, the force of gravity does change as the man moves on the Ferris Wheel. As the man moves higher, the force of gravity decreases because he is farther away from the center of the Earth. As he moves lower, the force of gravity increases because he is closer to the center of the Earth.

## How does the radius of the Ferris Wheel affect the force of gravity on the man?

The radius of the Ferris Wheel does not affect the force of gravity on the man. The force of gravity is only dependent on the mass of the man and the distance from the center of the Earth. However, the radius of the Ferris Wheel does affect the speed at which the man moves and experiences the force of gravity.

## Is the force of gravity the only force acting on the man on the Ferris Wheel?

No, the force of gravity is not the only force acting on the man on the Ferris Wheel. The man also experiences a centripetal force, which is the force that keeps him moving in a circular motion on the Ferris Wheel. This force is provided by the structure of the Ferris Wheel and is directed towards the center of the circle.

## Can the force of gravity on the man be completely eliminated on the Ferris Wheel?

No, the force of gravity on the man cannot be completely eliminated on the Ferris Wheel. As long as the man is on the surface of the Earth, he will always experience the force of gravity. However, the magnitude of the force can be reduced by moving the man farther away from the center of the Earth, such as at the top of the Ferris Wheel.

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