Determine Coefficient of Friction

In summary: That is all you need to remember.ChetIn summary, the coefficient of friction between the crate and the ramp can be determined by setting the net force equal to zero and solving for the coefficient of friction using the formulas for net force and the applied force. It is important to use a free body diagram and remember that the sum of forces is equal to mass times acceleration.
  • #1
killaI9BI
51
0

Homework Statement



You are pushing a 53 kg crate at a constant velocity up a ramp onto a truck. The ramp makes an angle of 22 degrees with the horizontal. If your applied force is 373 N, what is the coefficient of friction between the crate and the ramp?


Homework Equations



μ = Ff/Fn
Fn = m X g X cos θ
Fnet = Fa – mg sin θ +or- μmg cos θ

The Attempt at a Solution



Fn = 53 X 9.81 X cos 22
Fnet = 373 – 53 X 9.81 sin 22 + μ X 53 X 9.81 cos 22

The book's answer is 0.37 but I just can't get the rest of the equation. Any suggestions are appreciated.
 
Physics news on Phys.org
  • #2
Friction always opposes motion, not acceleration, if you're pushing up the ramp, which way should friction be?

You are applying 373N to move the crate at a CONSTANT VELOCITY, what does that tell you about the net force?

And to be honest, I'm not sure I know what you mean by the rest of the equation? Can you clarify please?
 
  • #3
Is the frictional force in the same direction as the applied force or in the opposite direction? If the crate is moving at constant velocity, what is the net force?

Chet
 
  • #4
BiGyElLoWhAt said:
Friction always opposes motion, not acceleration, if you're pushing up the ramp, which way should friction be?

You are applying 373N to move the crate at a CONSTANT VELOCITY, what does that tell you about the net force?

And to be honest, I'm not sure I know what you mean by the rest of the equation? Can you clarify please?

Me either - just using equations from notes that I took during the lesson.

I'm not really sure what it tells me about the net force...

the formula that I have for net force is:
Fnet = Fa – mg sin θ +or- μmg cos θ

where Fa is applied force, m is mass, g is acceleration due to gravity and μ is the coefficient of friction
 
  • #5
Chestermiller said:
Is the frictional force in the same direction as the applied force or in the opposite direction? If the crate is moving at constant velocity, what is the net force?

Chet

I'm thinking the frictional force would be in the opposite direction as the applied force.

I'm going to see if I can find another formula for net force that doesn't require the value of coefficient of friction.
 
  • #6
ohhh I see now; because it's moving at a constant velocity, the net force must be zero so the applied force equals the force of friction

right?
 
  • #7
I'm still in left field somewhere
 
  • #8
Go back to your equation and set the net force equal to zero. Then solve for the coefficient of friction, since you know everything else in the equation. In your problem, the applied force does not equal the force of friction because there is also a component of gravitational force present along the slide. Did you draw a free body diagram, or haven't you learned about that yet?

Chet
 
  • Like
Likes 1 person
  • #9
killaI9BI said:
ohhh I see now; because it's moving at a constant velocity, the net force must be zero so the applied force equals the force of friction

right?


Not quite. Look at your sum of the forces. How many forces are there? Youre right the net force is 0, but if the applied force equalled the force of friction, what would you say about this
##F_{applied}-F_{friction}=?##
?

Look at your sum of forces equation. The one you posted recently is right. I am not sure if you changed the original one, but there was a sign error in it. Youre almost to the answer, you just have to step back and look at what you have.
 
  • Like
Likes 1 person
  • #10
Chestermiller said:
Go back to your equation and set the net force equal to zero. Then solve for the coefficient of friction, since you know everything else in the equation. In your problem, the applied force does not equal the force of friction because there is also a component of gravitational force present along the slide. Did you draw a free body diagram, or haven't you learned about that yet?

Chet

that worked! I have learned about the FBDs but didn't think it'd really help. I will try them in the future.

Thank you.
 
  • #11
killaI9BI said:
that worked! I have learned about the FBDs but didn't think it'd really help. I will try them in the future.

Thank you.
The value of using FBDs cannot be overemphasized.

Chet
 
  • Like
Likes 1 person
  • #12
BiGyElLoWhAt said:
Not quite. Look at your sum of the forces. How many forces are there? Youre right the net force is 0, but if the applied force equalled the force of friction, what would you say about this
##F_{applied}-F_{friction}=?##
?

Look at your sum of forces equation. The one you posted recently is right. I am not sure if you changed the original one, but there was a sign error in it. Youre almost to the answer, you just have to step back and look at what you have.

Fapplied - Ffriction = Fnet - FgII
where FgII is the gravitational force parallel to the incline

... according to my notes :wink:
 
  • #13
Looks good man, good job.

I just want to add one thing: don't focus on your notes, sum of the forces = ma. done and done.
 

1. What is the coefficient of friction?

The coefficient of friction is a dimensionless value that represents the amount of resistance between two surfaces in contact with each other. It is a measure of the force required to overcome the frictional force between two objects.

2. How is the coefficient of friction determined?

The coefficient of friction can be determined by conducting an experiment where one object is dragged or pushed against another object with a known force. By measuring the force required to move the object and the weight of the object, the coefficient of friction can be calculated using the formula μ = F/W, where μ is the coefficient of friction, F is the force, and W is the weight.

3. What factors affect the coefficient of friction?

The coefficient of friction is affected by several factors including the roughness of the surfaces in contact, the weight of the objects, and the type of material the objects are made of. It can also be affected by the presence of lubricants or contaminants on the surfaces.

4. Why is the coefficient of friction important?

The coefficient of friction is important in many practical applications, such as designing machines and structures, determining the efficiency of mechanical systems, and preventing accidents. It is also used in physics and engineering calculations.

5. How can the coefficient of friction be reduced?

The coefficient of friction can be reduced by using lubricants between the surfaces in contact, smoothing or polishing the surfaces, or changing the materials of the objects in contact. Additionally, reducing the weight of the objects can also decrease the coefficient of friction.

Similar threads

  • Introductory Physics Homework Help
Replies
20
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
652
  • Introductory Physics Homework Help
Replies
20
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
18
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
4K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
6K
  • Introductory Physics Homework Help
Replies
16
Views
2K
Back
Top