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Friction in an inclined right-angle trough

  1. Jun 15, 2004 #1
    The following problem is from my physics textbook:

    A crate slides down an inclined right-angled trough. The coefficient of kinetic friction between the crate and the trough is [itex]\mu_k[/itex]. What is the acceleration of the crate in terms of [itex]\mu_k[/itex], [itex]\theta[/itex], and g?

    The problem was accompanied by a drawing which I have tried to reproduce and included as an attachment.

    The answer I come up with when I try this is [itex]g(sin \theta - \mu_k cos \theta)[/itex], which differs from the book's answer in that the book's answer has a factor of √2 sitting in front of the cosine.

    I think I understand how to do problems involving friction and inclined planes in general, but I do not understand why it matters that the crate is in a right-angle trough, or in a trough at all for that matter. I recognize that a factor of √2 would result from an isosceles right triangle, or from a 45-degree angle, which you could certainly form out of the many right angles present in the problem, but I don't see how any of them would affect the final answer. I have tried to calculate the force the crate exerts on each side of the trough, and then take the vector sum of those two forces to find the net normal force acting upon the crate. When I did this, I found the result to be equal in magnitude to the gravitational force acting orthogonal to the trough - in other words, right back where I started. Perhaps I made a mistake here?

    How can I correctly solve this problem?
     

    Attached Files:

  2. jcsd
  3. Jun 15, 2004 #2

    AKG

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    Okay, you know that the net normal force will simply be perpendicular to the "plane" (imagine the trough was a plane), because the left and right components of the normal forces from the right and left parts of the trought, respectively, will cancel out. If you look a the trough from the side, you can pretty much treat it like a plane (for now), so you have a normal force, and a gravitational force (worry about the friction later). Resolve the gravitational force into perpendicular to the plane and parallel to the plane components. The perpendicular component has magnitude [itex]mg \cos \theta[/itex]; so must the net force. This tells you that each individual normal force (the two of them) have a perpendicular-to-the-plane component of [itex](mg \cos \theta )/2[/itex], and some left or right component. Because of the shape of the trought, being a right angled trough at a 45 degree angle to the ground, you know that the normal forces created by both troughs must be at a 45 degree angle to the ground (because they are at an angle perpendicular to the individual left and right surfaces of the trough). So the normal force from an individual half of the trough is:

    [tex]\frac{(mg \cos \theta )/2}{\sin 45 ^{\circ}} = \frac{mg \cos \theta}{\sqrt{2}}[/tex]

    The force of friction is of course the sum of the frictions caused by each plank, which is in turn:

    [tex]2 \times \mu _k \times \frac{mg \cos \theta}{\sqrt{2}} = \sqrt{2} mg \cos \theta[/tex]

    There's your frictional force. The force down the trough due to gravity is obviously [itex]mg \sin \theta[/itex]. Since acceleration is net force divided by mass, you get acceleartion to be:

    [tex]g(\sin \theta - \sqrt{2} \mu _k \cos \theta )[/tex]
     
    Last edited: Jun 15, 2004
  4. Jun 16, 2004 #3
    Thanks a lot, AKG!
     
  5. Oct 11, 2011 #4
    I do not understand this problem, and since this is my first 3D problem, I can't figure out how to draw a FBD either, so I can't figure out how many forces acting on the crate. It'd be nice if someone could help me out. Thanks
     
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