Static Friction carnival ride Problem

[Jack G]

Homework Statement

In a classic carnival ride, patrons stand against the wall in a cylindrically shaped room. Once the room gets spinning fast enough, the floor drops from the bottom of the room! Friction between the walls of the room and the people on the ride make them the “stick” to the wall so they do not slide down. In one ride, the radius of the cylindrical room is R = 7 m and the room spins with a frequency of 20.8 revolutions per minute.

Homework Equations

To be safe, the engineers making the ride want to be sure the normal force does not exceed 1.7 times each persons weight - and therefore adjust the frequency of revolution accordingly. What is the minimum coefficient of friction now needed?

The Attempt at a Solution

Since the normal force(Fn) is a force (F=M*A) and can not exceed 1.7 times the mass of the object I used the inequality Fn<(1.7)(M). The force of static friction should be equal to the force of gravity in this case as it is holding someone steady so it should be Fg=Fs=(9.8)(M). To solve for the static friction coefficient(u_s) I used the formula Fs=(u_s)*N. Substituting gives us (9.8)(M)=(u_s)(1.7)(M). Solving for u_s gave me (9.8)(M)/(1.7)(M)=(u_s). The mass cancels itself out leaving me with (u_s)=9.8/1.7=5.7647 but this is wrong I guess. I am not really sure why though, any help or a push in the right direction would be appreciated.

 

[Doc Al]

Since the normal force(Fn) is a force (F=M*A) and can not exceed 1.7 times the mass of the object I used the inequality Fn<(1.7)(M).

Not mass, but weight. (Correct that and you'll be fine.)

 

[Bestfrog]

I would start from here: what is the normal force of the problem? What is its direction?

 

[kuruman]

A normal force is perpendicular to the surface that exerts it. What is that direction in this case?

 

[haruspex]

I would start from here: what is the normal force of the problem? What is its direction?

Jack G does not seem to have a problem with that. Doc Al made the useful response.

A normal force is perpendicular to the surface that exerts it. What is that direction in this case?

Is that in reply to Bestfrog?

 

[Bestfrog]

Jack G does not seem to have a problem with that.

I hope not to take him in the wrong way, but he seems to think that the normal force is exerted from the floor.

 

[Doc Al]

but he seems to think that the normal force is exerted from the floor.

What led you to that conclusion?

 

[kuruman]

Is that in reply to Bestfrog?

Yes, I (carelessly) thought I was replying to OP.