# Max reaction force - ferris wheel

1. Dec 11, 2016

### krbs

1. The problem statement, all variables and given/known data
What is the maximum reaction force exerted by a seat on the passenger of a ferris wheel? There's a diagram showing the seat at 6 o'clock, 12, 3, and 4:30... on a 12 hour clock.

2. Relevant equations
N/A

3. The attempt at a solution
I guess the reaction force is the normal force exerted by the chair. It's not the top position because the chair is also falling. Maybe it's highest at the 3 o clock position because the bottom of the chair is counteracting gravity and the back of the chair is providing the centripetal force.

(This is a textbook question, not homework.)

2. Dec 11, 2016

The ferris wheel seats can be assumed to all be at some radius r and the ferris wheel can be assumed to be rotating at a constant rate so that each seat is moving at velocity v tangent to the wheel. Did you know that the acceleration that occurs in doing circular motion is $a=v^2/r$ and points to the center of the circle? Meanwhile, gravity always has a downward force on a passenger of mass $m$ equal to $F_g=mg$ regardless of any motion that occurs. The only other force on the passenger is from the seat. What does Newton's 2nd law tell you about the vector sum of all the forces on an object? From this, you should be able to come up with the solution.

3. Dec 11, 2016

### krbs

Yeah, I'm trying to use all this info, but I cant come up with a solution =
If velocity, radius, and mass are constant then acceleration and centripetal force must be as well?

Would it be the 4:30 position because the net force is pointing at an angle to the centre of the circle, gravity's pointing downward, so the final vector has to be between the gravity and the net force. It's the longest vector?

Unless... at the bottom of the loop, is the Normal force greater than mg? Then the reactive force would be highest there...

4. Dec 11, 2016

The vector sum of the gravitational force plus the force from the seat must equal $m \vec{a}$. $\\$ i.e. $\vec{F}_g +\vec{F}_{seat}=m \vec{a}$. $\\$ $\vec{a}$ is constant in amplitude but always points to the center of the ferris wheel. Solve the equation for $\vec{F}_{seat}$ and try to determine the location for which $\vec{F}_{seat}$ is largest. It may help to use a diagram with arrows for $m \vec{a}$ and $\vec{F}_g$. $\\$ And here is a simple hint: use an upward arrow for the quantity(vector) $- \vec{F}_g$.

5. Dec 11, 2016

### krbs

I just dont get it.

At the top: Fnet = mg - N; N = mg - Fnet
Bottom: Fnet = N - mg; N = Fnet + mg
3 o' clock: Fnetx = Nsin(theta); Fnety = mg - Ncos(theta) ???
4:30: same as above, different angles??

There are no given values. So how do I determine the answer just from this?

6. Dec 11, 2016

If you solve for $\vec{F}_{seat}$, you get $\vec{F}_{seat}=-\vec{F}_g+m \vec{a}$. The vector $-\vec{F}_g$ points upward. Only at the bottom does $m \vec{a}$ also point upward. Alternatively, you could solve for the amplitude of $\vec{F}_{seat}$ by using the law of cosines(and finding the angle for which you get a maximum), but just drawing a few vector arrows should give you the answer.

7. Dec 11, 2016

### krbs

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8. Dec 12, 2016

If the ferris wheel is not moving, there is no centripetal acceleration so $m \vec{a}=0$. That makes $\vec{F}_{seat}=-\vec{F}_g$ at all locations. You should be able to add the vector $m \vec{a}$ to $-\vec{F}_g$ to get the result for $\vec{F}_{seat}$. All it requires is to add a couple of arrows together head to tail and connecting them.

9. Dec 12, 2016

### krbs

Hi, the question says the person is riding the Ferris wheel so I assume Fnet is not zero.

It would appear that Fseat is biggest at 4:30. Do I finally have a solution??

10. Dec 12, 2016

You need to make the $\vec{F}_g =mg$ and $\vec{F}_{net}=m \vec{a}$ vector arrows the same length in each diagram. Normally, the $\vec{F}_g$ arrow will be considerably longer than the $\vec{F}_{net}$ arrow, because a ferris wheel is not a ride that has strong acceleration (g) forces. As the $m \vec{a}$ changes direction, if you measure the length of the resultant vector $\vec{F}_{seat}$ carefully, you should be able to get the correct answer. The law of cosines will get you the precise answer perhaps more readily, because the cosine function result doesn't change much from the 4:30 or 7:30 position to the answer that you get at the bottom. The law of cosines reads $c^2=a^2+b^2-2ab cos(\theta')$ where $\theta'=180-\theta$. (Don't know if you have seen law of cosines in a trigonometry class). In any case $c=F_{seat}$ peaks at $\theta=0$ which is at the bottom.

11. Dec 12, 2016

### krbs

Hi again, I looked at my diagrams and noticed a lot of the mistakes >.> Fnet has constant magnitude so the vector should always be the same length. Same with gravity.

Here's my thinking:
At the top of the circle, N is smaller than mg, if not zero.
At the right, mg doesnt even contribute to the Fc because it's perpindicular to the centripetal force. So N = Fnet.
At 4:30, mg affects Fc more so N has to be a little larger to compensate. But gravity is still not counteracting Fc as much as it could...
At the bottom of the loop, though, N has to entirely overcome gravity to make Fc point centripetally. N > mg and Fc combined. That means the force from the seat must be biggest there.

For the law of cosines, would I need actual values* to solve?

12. Dec 12, 2016

Very Good !!! For the law of cosines, $A=mg$ and $B=ma$ (where $a=v^2/r$). $C^2=A^2+B^2-2AB \cos(\theta')$. As the ferris wheel turns, the side $B$ changes direction and $\theta'$ takes on different values. $C=F_{seat}$ and $C^2$ is maximum for $\theta'=180 \, degrees$ , since $cos(180)=-1$. This makes $C^2=A^2+B^2+2AB =(A+B)^2$ so that $C=A+B$ which is the result from your diagrams. I don't know how to draw diagrams on this website, but the $\theta'=180$ is the angle in the vector triangle when the passenger is at the bottom of the ferris wheel.