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Inclined plane/Static Friction Problem

  • Thread starter newport
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  • #1
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1. A 250-kg crate rests on a surface that is inclined above the horizontal at an angle of 21.8°. A horizontal force (magnitude = 515 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?



2. Ffsmgcosθ
Fapplied=Fparallelcosθ




3. I made Fparallel equal to Ff. When I solved it I got a coefficient of static friction of .244, but apparently it's not correct.
 

Answers and Replies

  • #2
Doc Al
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Ffsmgcosθ
Careful here. The friction force equals μN, but N ≠ mgcosθ. The horizontal force changes the normal force, since it has a component perpendicular to the surface.

Figure out the normal force, then you can redo the problem.
 
  • #3
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I got practically the same answer.
 
  • #4
Doc Al
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How did you solve for the normal force?
 
  • #5
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I found the y component of the applied force and then added that to the normal force of the crate on the plane.
 
  • #6
Doc Al
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I found the y component of the applied force and then added that to the normal force of the crate on the plane.
I don't quite understand. Do you mean: N = Fsinθ + mgcosθ? If so, that's not quite correct.
 
  • #7
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I used [itex]\frac{tan21.8}{515}[/itex] to find it, not Fsinθ.
 
  • #8
Doc Al
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I used [itex]\frac{tan21.8}{515}[/itex] to find it, not Fsinθ.
I don't understand what that represents.

To find the normal force, consider all forces acting on the crate perpendicular to the surface:
- y component of 515 N force (up)
- Normal force (up)
- y component of weight (down)

They must add to 0. That lets you solve for the normal force.
 

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