A Ferry's wheel

The people do not need to hang on at all when the force is DOWNWARD. If gravity suddenly increased 10% in a stationary ferris wheel they still would not have to hang on at all. The point of maximum SIDEWAYS force is when they need to hang on the most. Since the force is toward the centre, wouldn't this be at point halfway up or down?

The problem is badly stated, I also view the force of gravity irrelevant to the need to hold a railing, the seat bottom does that.

The problem is poorly worded!

For the answer to be correct, the wording should be changed from "do they need to exert the greatest force" to "is the total force on the occupant the greatest". The occupant doesn't need to exert the force at the bottom to remain seated in the same position.

For the problem as stated, the correct answer is M. For positions B and T the occupants do not need to exert any force to remain seated in the same position. For a normal person in a normal seated position, the downward force of the center of mass is on the same axis as the upward force from the seat. At position M, assuming sufficient friction from the seat, the occupant will need to keep her center of mass (which should be above the seat) from pitching outward by exerting force: either grabbing a bar, or using core strength.

The situation is far more complicated. There are at least four forces acting on the people:

• the force of gravity, $F_g = mg\ \downarrow$

• the normal force from their seats, $F_N\ \uparrow$

• the friction force from their seats, $F_f \leq \mu F_N\ \leftrightarrow$

• the force from the railing, $F_r$

In order to remain in their seat in the same position, the total force must be the centripetal force, $F_{cp} = m\omega^2 r$ (inward).

Unless the wheel turns at a ridiculously fast rate, $F_{cp} < mg$ and we may assume that the normal force of the seat is able to make the vertical forces work out. That is, $F_N = -m(\omega^2 r \sin\theta + g).$ This normal force is indeed maximal at the bottom of the ride.

For the horizontal equilibrium, we need a force to provie $F_{cp,x} = -m\omega^2 r\cos\theta$ . If the seat is perfectly slippery, the horizontal force must come from holding on to the railing, and will be maximal with the wheel in the middle position.

More realistically, the horizontal force is provided, at least partially, by friction from the seat $F_f$ . If the wheel is slow enough and the seat is rough enough, the passengers need not hold on at all. Note that, since $F_f \leq \mu F_N$ , passengers can increase their friction force by pushing up on the railing, so that $F_N$ increases. This makes it more challenging to determine at what point exactly the force is maximal.

Surely the seat would provide most support at the bottom position, and not the railing.

The question could be rewritten to say a small scale is hung from the railing with a 1kg weight suspended from it. At which point does the scale indicate the highest weight?

Centrifugal force is a misnomer. Inertia is the other cause of force.

Why add these two accelerations? Isn't the acceleration of circular motion is the net acceleration of the people in the cabin?

The people sitting on the seat in the cabin would need to exert least force on the railing at point B as the vertical force is maximized already pushing them to the seat. At point M is where they should hold tightly as the vector sum of the gravity and the centrifugal forces will push to the right.

The cabins in a Ferris wheel behave like a pendulum. Therefore the passengers should only feel an acceleration towards the bottom of the support, never a lateral force.

It clearly states at which point will they need to exert the greatest force not at which point is the rider experiencing the greatest force.

The wording of the problem is just wrong. "...do they need to exert...". It should say "...do they exert...".

since they are seated they only need to hold on to avoid sliding sideways or lifting off (at the top - unlikely)

Centrifugal force is not a force it and doesn’t exist in the context of the real world. It is simply the result of other forces. Also the centripetal force is acting on the object and that is what the question references. The only point of contact are the normal and friction.

I think the only force that counts is gravity. The other consideration is which direction the wheel is turning. It's not stated but I'm assuming it's moving clockwise. The speed is constant so the centripetal force is constant. To maintain one's seat the real effort is only needed between T and B (so let's call it M). Just think of the flutter in your stomach as you start to descend, that's the point where gravity gives up a bit of it's pull since you're going in the same direction as the force of gravity. So to maintain your seat exactly as going in the up direction you have to hold on a bit tighter at all points to cancel out that bit of drop in the force of gravity. All in all I think it's a poorly stated problem. Granted this isn't an elegantly worded solution but I didn't think it needed to be.

"There are two forces acting on the people, the force of gravity and centrifugal force.

The two have the same direction at the bottom of the ride"

Well, "centrifugal force" isn't a thing. There is "centripetal force" which is the net force on any object traveling in a circle of constant radius, as the rider is here. The direction of centripetal force is always toward the center. Everywhere in the circle gravity points downward. At the bottom of the circle to have a net force pointing up the total force acting on rider by the wheel must be enough to counter-act the force of gravity. (At the top of the circle, gravity itself is providing the centripetal force, so the wheel need not pull the rider down.)

I agree that the problem is poorly worded. At the bottom of the ride, the people just need to sit still to maintain their position. Same at the top. Only at M do they need to counter the centripital force (not centrifugal!) to maintain their seated upright position, by pushing away from the center of the ferris wheel (but not very hard, as the centripital force of most ferris wheels is barely noticeable.)

Disagree. I think maximum "throwing" force will be where the tangent vector is moving away from the cab: perpendicular to gravity.

Maximum force on the riders will be where the tangent vector is vertical up (adding to gravity). But I can't tell what your question is, or which way there ferris wheel is turning.

At point B the seat exerts and equal and opposite force on the body to gravity and the centrifugal force. So point M I guess.

There is no such force as centrifugal force, the centripetal force (Fcp) (towards the centre) is provided by a combination of the 2 forces on the person - weight (mg) and the force exerted by the cabin (Fcb). At the bottom Fcp = Fcb - mg. Both are vertical so no effort needed At the top Fcp = Fcb + mg. in practice mg is more than enough to provide Fp At M mg cannot contribute at all to Fcp

I think that the answer is wrong. The point at which the total force is maximised is at the bottom. However, at this point the combined centrifugal forced and gravity act in the same direction and the seat offers a force equal in magnitude but acting directly upward - hence the people do not experience any sideways force which they would have to counter in order to remain in their same seated position. What they experience is the maximum force on their bodies. However they would experience the maximum of such force in the MIDDLE positions - which was my answer.

I agree problem with problem is wording of problem.

I disagree. At the bottom of the Ferris wheel, the passenger does not have to hold on tightly to anything. They can actually let go and let gravity and inertia keep them in the same seat. The question is worded poorly.

The problem is still badly worded because of the two common meanings of the word 'exert'. The first meaning is passive as in 'the moon exerts a force on the earth' - the question uses this sense. But the second meaning is active 'the weightlifter had to really exert himself to break the world record'. Since the question talks about people 'exerting' a force, the second sense of the word is more natural. This gives a different answer.

wording the problem and answer given as right do not match. Michael Palm has explained it .

The big problem here is that the given answer is incorrect.

The wording is confusing and do not ask what is intended for