Coefficient of static friction problem

In summary: These forces are acting on the riders and are not part of the forces acting on the ride. The force acting on the ride is the tension force. In summary, the force that points upward, preventing the riders from falling downward, is the tension force.
  • #1
Seung Lee
6
0

Homework Statement


I got this problem:
A wedge of mass m = 36.1 kg is held in place on a fixed plane that is inclined by an angle θ = 21.3° with respect to the horizontal. A force F = 302.3 N in the horizontal direction pushes on the wedge, as shown in Figure 4.25a. The coefficient of kinetic friction between the wedge and the plane is μk = 0.159. Assume that the coefficient of static friction is low enough that the net force will move the wedge.

http://semmedia.mhhe.com/physics/bau...dge/index.html
basically what's there except friction would be the other way around (since its accelerating up the slide.

Homework Equations


FBD
Sum of all forces = ma

The Attempt at a Solution


I did a free body diagram and got:
Summation of forces in y:
Normal force = mg sin theta + Appliedforce sin theta

Then for the forces in x:
mgcos + friction - Aplliedforce sin = 36.1a

I did solve for everything but strangely the acceleration gives me acceleration = 3.2

Can any1 help me if I am doing something wrong? Like getting cos and sin mixed?
 
Physics news on Phys.org
  • #2
Check "Normal force = mg sin theta + Appliedforce sin theta"
mg is in a different direction than F, so how can it have the same coëfficiënt ?
also check the angle. Why do you state 21.3 and the text 22 degrees ?

What is so trange about acceleration 3.2 ?
You sure the picture and your exercise match ? Why do you want to calculate the acceleration if the exercise says the task is to draw an FBD ?

Oh, and: welcome to PF :)
 
  • #3
Oh! Ok, I think fixing the cos and sin around helped me out
it was mg cost + Forceappliedsin = Normal force (sum of forces in y)
and mg sin + friction - Forceappliedcos = ma

I just have another question. The acceleration here would be negative right? My teacher had told me that itw as positive but I wasn't sure why but I'm also pretty sure it's actually negative.

Thank you for the welcoming too :D

P.S.:
Just a fast MC question I had a doubt with:
The Tornado is a carnival ride that consists of a hollow vertical cylinder that rotates rapidly about its vertical axis. As the Tornado rotates, the riders are pressed against the inside wall of the cylinder by the rotation, and the floor of the cylinder drops away. The force that points upward, preventing the riders from falling downward, is
tension, friction, gravity or normal force.

I'm pretty sure its friction due to the fact that normal would act perpendicular (to the center of the cylinder) and gravity would point downward and well tension would not exist.
Am i right?
 
  • #4
Seung Lee said:
Oh! Ok, I think fixing the cos and sin around helped me out
it was mg cost + Forceappliedsin = Normal force (sum of forces in y)
and mg sin + friction - Forceappliedcos = ma

I just have another question. The acceleration here would be negative right? My teacher had told me that itw as positive but I wasn't sure why but I'm also pretty sure it's actually negative.

Thank you for the welcoming too :D

P.S.:
Just a fast MC question I had a doubt with:
The Tornado is a carnival ride that consists of a hollow vertical cylinder that rotates rapidly about its vertical axis. As the Tornado rotates, the riders are pressed against the inside wall of the cylinder by the rotation, and the floor of the cylinder drops away. The force that points upward, preventing the riders from falling downward, is
tension, friction, gravity or normal force.

I'm pretty sure its friction due to the fact that normal would act perpendicular (to the center of the cylinder) and gravity would point downward and well tension would not exist.
Am i right?
You are right about your last comment. Note that the normal force is acting as the centripetal force. And the friction force is acting upward.
 
  • #5


I would first commend you for attempting to solve the problem and using the correct equations. However, it seems like you may have made a mistake in your free body diagram. The normal force should be perpendicular to the plane, not inclined at an angle. It should be equal to the weight of the wedge, mg, and point directly upwards. Additionally, the applied force should be parallel to the plane, and the friction force should be opposite in direction, pointing downwards.

With that in mind, your equations should be:

Sum of forces in y: Normal force - mgcosθ = 0

Sum of forces in x: Applied force - friction force - mgsinθ = ma

Since the wedge is accelerating, the friction force should be equal to the coefficient of kinetic friction (μk) multiplied by the normal force (which is equal to mgcosθ). So your final equation should be:

F - μk(mgcosθ) - mgsinθ = ma

Solving for a, you should get an acceleration of approximately 1.9 m/s^2.

I would also suggest double checking your calculations to make sure you are using the correct values for the mass, angle, and coefficient of kinetic friction. And remember to always check your free body diagram to make sure all forces are correctly labeled and accounted for. Good luck!
 

1. What is the coefficient of static friction?

The coefficient of static friction is a measure of the force required to keep an object at rest on a surface. It is represented by the symbol μs and is a dimensionless number that is specific to a particular material and surface.

2. How is the coefficient of static friction calculated?

The coefficient of static friction can be calculated by dividing the maximum force of static friction by the normal force acting on an object. This can be represented by the equation μs = Fmax/N, where Fmax is the maximum force of static friction and N is the normal force.

3. What factors affect the coefficient of static friction?

The coefficient of static friction is affected by the materials of the object and surface in contact, the roughness of the surface, and the amount of force applied. It is also dependent on external factors such as temperature and humidity.

4. How is the coefficient of static friction different from the coefficient of kinetic friction?

The coefficient of static friction applies to objects at rest, while the coefficient of kinetic friction applies to objects in motion. The coefficient of kinetic friction is typically lower than the coefficient of static friction, as it takes less force to keep an object in motion than it does to overcome its initial state of rest.

5. How does the coefficient of static friction relate to the laws of motion?

The coefficient of static friction is directly related to Newton's First and Third Laws of Motion. It explains that an object at rest will remain at rest unless acted upon by an external force, and that for every action, there is an equal and opposite reaction. In the case of static friction, the force required to keep an object at rest is equal and opposite to the force applied to it.

Similar threads

  • Introductory Physics Homework Help
Replies
11
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
11
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
6K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
3K
  • Introductory Physics Homework Help
Replies
13
Views
821
  • Introductory Physics Homework Help
Replies
5
Views
5K
  • Introductory Physics Homework Help
Replies
28
Views
5K
Back
Top