# Calculating Normal Force for Student on Ferris Wheel

• Naeem
In summary, a chemistry student with a mass of 75 kg is riding a steadily moving Ferris wheel and experiences a normal force of 556 N at the top of the wheel. If the wheel's velocity is doubled, the normal force would quadruple in magnitude and could potentially be negative. It is important to carefully consider all units and equations when solving physics problems, and to avoid making contradictory statements.
Naeem
Q. A chemistry student with a mass of 75 kg is riding a steadily moving Ferris wheel. When she is at the top of the Ferris wheel, the normal force from the seat on to her body has a magnitude of 556 N.

Q. What would the normal force be on the student at the top of the wheel if the wheel's velocity were doubled

If the wheels velocity were doubled mv^2 / R increases by a factor of four, i.e

Weight of the student = 75 kg * 9.81 = 735.75 N - Normal Force ( 556 N )

= 179.75 N * 4 ( coz, it increases by a factor of 4 )

= 719 N

There fore Normal force = 719 N - 735.75 N = 16.75 N , is this correct!

Naeem said:
Q. A chemistry student with a mass of 75 kg is riding a steadily moving Ferris wheel. When she is at the top of the Ferris wheel, the normal force from the seat on to her body has a magnitude of 556 N.

Q. What would the normal force be on the student at the top of the wheel if the wheel's velocity were doubled

If the wheels velocity were doubled mv^2 / R increases by a factor of four, i.e

Weight of the student = 75 kg * 9.81 = 735.75 N - Normal Force ( 556 N )

= 179.75 N * 4 ( coz, it increases by a factor of 4 )

= 719 N

There fore Normal force = 719 N - 735.75 N = 16.75 N , is this correct!

Your numerical answer is probably correct, but your "stream of consciousness" approach to solving problems is going to get you in trouble when solving more compex problems. Do yourself a big favor and resolve to never put an equal sign between two things that are not equal. As a case in point

719 N - 735.75 N = 16.75 N

Well, no. This statement is false. In truth

719 N - 735.75 N = -16.75 N

A negative normal force is in the realm of possibility in this problem, as long as there is a bar or belt to hold in the passengers. How do you know the answer is not negative?

Also,

Weight of the student = 75 kg * 9.81 = 735.75 N - Normal Force ( 556 N )

You just wrote a contradiction. Ignoring for the moment the units that you left out, you said

W = 75 * 9.81 = 179.75

Since when was 75*9.81 = 179.75?

Try rewriting your solution in such a way that every statement you make is true. If you say A = B, then make sure A in fact does equal B.

Well, yes I understand my mistake ( a serious one maybe ), but

'OlderDan' you need to tone down, your disagreement, and try to be softer at times if not always. Well can't complain, since this is a free forum, and with 'Older ppl' I can understand that, it happens with them.

Tone down? You're kidding me right? Has it occurred to you that nobody else responded to your post earlier in the evening because nobody else took the trouble to try to figure out what you were doing? My reply to you was made for your benefit- not mine. It was directed specifically to what you wrote, without making any judgements about your personality or demeanor. Had I wanted to be insulting, believe me I could have done a much better job.

You don't know anything about me, or how I relate to people as a person or as an instructor. Until you do, I suggest you limit your comments to the physics.

Yes, I would like very much to limit my comments to the physics itself. And yes, thank you very much for looking through the whole problem.

I certainly would not be interested in how you would relate to, but please certainly don't give a negative impression.

I know that this is a forum, and ppl like you are doing a gr8 job, by sparing their time here, trying to help others but at the same time, go off in a tangential direction , deviate from the actual stuff,and pass of nasty comments along with it. For instance
"stream of conciousness" [ a very diginified philosophical taunt ]

Well, by writing all this, I am just trying to make you aware of these little, so called,

'mood disturbers' .

I don't like to pursue this any further. Thanks a lot!

## 1. How do you calculate the normal force for a student on a Ferris wheel?

The normal force for a student on a Ferris wheel can be calculated by using the formula: N = mg + mv²/r, where N is the normal force, m is the mass of the student, g is the acceleration due to gravity, v is the velocity of the Ferris wheel, and r is the radius of the Ferris wheel.

## 2. What is the significance of the normal force on a student riding a Ferris wheel?

The normal force is the force exerted by the Ferris wheel on the student in the direction perpendicular to the surface of the Ferris wheel. It is important because it counteracts the force of gravity and prevents the student from falling off the Ferris wheel.

## 3. How does the normal force change as the Ferris wheel rotates?

The normal force changes as the Ferris wheel rotates based on the position of the student. When the student is at the top of the Ferris wheel, the normal force is equal to the weight of the student. As the student moves downwards, the normal force decreases, reaching zero at the bottom of the Ferris wheel, and then increases again as the student moves upwards.

## 4. Can the normal force ever be greater than the weight of the student?

Yes, the normal force can be greater than the weight of the student if the Ferris wheel is moving at a high enough velocity or has a small enough radius. This is because the centripetal force (mv²/r) adds to the weight of the student, resulting in a higher normal force.

## 5. How does the normal force affect the ride experience for a student on a Ferris wheel?

The normal force affects the ride experience by providing the necessary centripetal force to keep the student moving in a circular motion without falling off. It also varies throughout the ride, which can cause changes in the perceived weight of the student and contribute to the overall thrill of the ride.

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