Static Friction carnival ride Problem

Sorry!In summary, the conversation discusses a classic carnival ride where patrons stand against the wall in a cylindrical room and the floor drops from the bottom once the room spins fast enough. The engineers making the ride want to ensure that the normal force does not exceed 1.7 times each person's weight, and therefore adjust the frequency of revolution accordingly. The conversation then delves into solving for the minimum coefficient of friction needed to achieve this. The final conclusion is that the normal force is perpendicular to the surface exerting it and the direction in this case is from the floor.
  • #1
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(us) I used the formula Fs=(us)*N. Substituting gives us (9.8)(M)=(us)(1.7)(M). Solving for us gave me (9.8)(M)/(1.7)(M)=(us). The mass cancels itself out leaving me with (us)=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.
 
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  • #2
Jack G said:
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.)
 
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Likes Sponson
  • #3
I would start from here: what is the normal force of the problem? What is its direction?
 
  • #4
A normal force is perpendicular to the surface that exerts it. What is that direction in this case?
 
  • #5
Bestfrog said:
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.
kuruman said:
A normal force is perpendicular to the surface that exerts it. What is that direction in this case?
Is that in reply to Bestfrog?
 
  • #6
haruspex said:
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.
 
  • #7
Bestfrog said:
but he seems to think that the normal force is exerted from the floor.
What led you to that conclusion?
 
  • #8
haruspex said:
Is that in reply to Bestfrog?
Yes, I (carelessly) thought I was replying to OP.
 

FAQ: Static Friction carnival ride Problem

What is static friction?

Static friction is a type of force that acts between two objects that are in contact with each other, preventing them from moving relative to each other. It occurs when there is no movement between the two objects and is caused by the microscopic irregularities on their surfaces.

How does static friction relate to carnival rides?

Carnival rides often use static friction to keep riders safely in their seats and prevent them from slipping or sliding. For example, on a roller coaster, the static friction between the wheels and the track keeps the train from derailing as it goes around curves and drops.

What factors affect the amount of static friction on a carnival ride?

The amount of static friction on a carnival ride is affected by the weight of the riders, the speed and direction of the ride, and the type and condition of the surfaces in contact. Rougher surfaces and heavier riders will result in a greater amount of static friction.

Is static friction always a desirable force on a carnival ride?

No, in some cases, too much static friction can be dangerous and cause a ride to malfunction. For example, if the wheels of a ride become too hot due to excessive friction, they may lose their grip on the track and cause the ride to derail.

How do engineers account for static friction when designing carnival rides?

Engineers use mathematical equations and computer simulations to calculate the amount of static friction that will occur on a ride and ensure that it is within safe limits for the intended use. They also use materials and surface treatments that minimize friction and lubricants to reduce it further.

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