Static friction?

  • Thread starter samantha.
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  • #1
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Homework Statement


A 224-kg crate rests on a surface that is inclined above the horizontal at an angle of 20.1°. A horizontal force (magnitude = 543 N and parallel to the ground, not the incline) is required to start the crate moving down the incline. What is the coefficient of static friction between the crate and the incline?

m=224kg
Force applied = 573


Homework Equations


F=ma
Force of Friction=(co-eff static friction)*(Normal force)


The Attempt at a Solution



So I calculated the normal force to be 2063.6N using N=m*g*cos(theta).

Then, I assumed that 573 was the limit for static friction. So the forces would have to equal up to 0 for there to be no acceleration.

Fapp - Ff = 0
573N - 2063N*(coeff. static) = 0
2063N*(coeff. static)/2063N=-573N/2063N
coeff. static = -0.278

I tried both this value and the absolute value and about 7 other values. I've only got three tries left and I have no clue what to do! Any help would be so great!! Thanks in advance.
 

Answers and Replies

  • #2
Doc Al
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So I calculated the normal force to be 2063.6N using N=m*g*cos(theta).
You need to consider the applied force since it will affect the normal force. Hint: To find the normal force, analyze all force components perpendicular to the surface. They must add to zero.
 
  • #3
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Thanks for your quick reply.
I'm not really sure how to incorporate the applied force into the normal force since it's purely a horizontal force..

Sorry if I'm a bit slow. I'm a chemist.
 
  • #4
993
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573N - 2063N*(coeff. static) = 0
2063 x coef of static friction is along the plane but 573 is along the horizontal and not along the inclined plane.

One has to find the component of the weight and of the applied force ALONG the inclined plane.
 
  • #5
Doc Al
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45,029
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I'm not really sure how to incorporate the applied force into the normal force since it's purely a horizontal force..
The applied force is horizontal, but the incline is not. Find the components of the applied horizontal force parallel and perpendicular to the incline surface.
 
  • #6
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Thanks!

To calculate the component of the applied force that was parallel to the incline I did this;

Fapp(x)=573*cos(20.1)
= 538.1N

Applying this to the equation I used previously,
2063.6N*U-538.1N = 0
0.261.

I confidently entered this answer and I was once again rejected. :(
 
  • #7
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Woops. I used the wrong value. But doing so, I fixed it to the correct one but I was still wrong.
 
  • #8
Doc Al
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45,029
1,325
To calculate the component of the applied force that was parallel to the incline I did this;

Fapp(x)=573*cos(20.1)
= 538.1N
Looks good.

Applying this to the equation I used previously,
2063.6N*U-538.1N = 0
That equation is incorrect: You're still using the wrong value for the normal force and you are ignoring the component of the weight acting parallel to the incline.
 
  • #9
15
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Ohhh you're very right. Thanks so much again!
 

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