Homework hellp. coefficent of friction.

In summary, the problem involves finding the friction force and coefficient of friction for an 18.0 kg box on a 37.0 degree incline with an acceleration of 0.270 m/s/s. Using Newton's laws and considering the normal force on the incline, the friction force can be calculated to be 101.34 N and the coefficient of friction is found to be 0.719. The principles used apply even if there was no friction, and a free body diagram can help with understanding the problem.
  • #1
briiannnaa04
12
0

Homework Statement


An 18.0 kg box is released on a 37.0 degree incline and accelerates down the incline at 0.270 m/s/s. Find the friction force impeding it's motion. How large is the coefficient of friction?


Homework Equations


Ff=uFn


The Attempt at a Solution


I have no idea how to do this.
 
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  • #2
Can you do the problem if there were no friction, that is just asking for acceleration down an inclined plane?
 
  • #3
ohh, maybe how can i do it without friction though?
 
  • #4
Well you can't but the principles used apply:

In other words the acceleration from gravity along the surface of the ramp is sin(theta) * g. It is not 9.8. So if there were no friction a=9.8 sin(37). What is this come out to be?
 
  • #5
5.9?
 
  • #6
OK. Now we are told that the acceleration is not 5.9 but a measly 0.27.

So let's write an eqn using Newtons laws.

ma=sum of all forces ? is the frictional force

18.0*0.27=18*5.9-? Let's not worry about the components that make up ? yet. Can you compute ?
 
  • #7
4.86=106.2-? so 101.34? or is there more than one other force?, Thank you for helping I'm sorry I'm really bad at physics.
 
  • #8
Should be 18(5.9-0.27)

Ok now the tricky part. This frictional force is equal to the normal force * u where u is the coefficient of friction.

On level ground normal force is just opposite the weight. But on a plane it is
weight* cos(37)

So normal force is 18*9.8 cos (37). Let's call this N.

Then we have N*u=the same ? as you calculated before. can you get u?
 
  • #9
176.4 cos(37)=140.9
N*u=101.34
140.9u=101.34
so u=.719
 
  • #10
By george I think you got it. It makes more sense with a free body diagram.
 
  • #11
Thank you so much!
 

Related to Homework hellp. coefficent of friction.

1. What is the coefficient of friction?

The coefficient of friction is a measure of the amount of resistance between two surfaces when they are in contact with each other. It is a dimensionless number that ranges from 0 to 1, with higher values indicating higher levels of friction.

2. How is the coefficient of friction calculated?

The coefficient of friction is calculated by dividing the force required to move an object by the weight of the object. This can be determined through experiments or by using known values for the materials involved.

3. What factors affect the coefficient of friction?

The coefficient of friction can be affected by a variety of factors, including the roughness of the surfaces, the materials involved, the force applied, and the temperature.

4. Why is the coefficient of friction important?

The coefficient of friction is important because it helps us understand and predict how objects will move and interact with each other. It is also essential in engineering and design, as it can influence the performance and durability of products.

5. How can the coefficient of friction be reduced?

The coefficient of friction can be reduced by using lubricants, smoothing surfaces, or changing the materials involved. Additionally, reducing the force applied or changing the angle of contact can also lower the coefficient of friction.

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