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Static and Kinetic friction question

  1. Sep 23, 2012 #1
    1. The problem statement, all variables and given/known data
    A crate is being pushed up a hill with friction. Given: m1 = 131 kg, θ = 23°, [itex]\mu[/itex]s = 0.15, and [itex]\mu[/itex]k = 0.08. If the crate isn't being pushed anymore, what will be the magnitude of the acceleration of the crate sliding downhill?

    What force must be exerted to move the crate uphill at constant speed?

    What force must be exerted to move the crate downhill at constant speed?
    2. Relevant equations
    [itex]\Sigma[/itex]Fx = m*a

    [itex]\Sigma[/itex]Fy = m*a

    3. The attempt at a solution

    I drew a free body diagram of the crate, m1. The crate is diagonal since it's on an incline, so I made my coordinate system correlate with that. the only diagonal I had was Mg, so I found the x and y components of it. So Mxg = MgSinθ and Myg = MgCosθ.
    Overall the forces acting upon the box were Mg, N, and f (friction). So for [itex]\Sigma[/itex]Fx = M*a:
    -MgSinθ - f = m*ax
    -MgSinθ - [itex]\mu[/itex]s*N = M*ax

    [itex]\frac{-MgSinθ - ms*N}{M}[/itex]=ax

    For [itex]\Sigma[/itex]Fy = M*a:
    N-MgCosθ=M*ay (there's no vertical a, so the right side is 0)
    N = MgCosθ

    Since I'm trying to find horizontal ax, I'll sub in MgCosθ in for N.

    Now I have [itex]\frac{-MgSinθ - ms*MgCosθ}{M}[/itex]=ax
    I can factor out the M...

    -gSinθ - [itex]\mu[/itex]s*gCosθ =ax (the M's cancel on top/bottom)
    Bring out a g and -1 from both...

    -g(Sinθ + [itex]\mu[/itex]s*Cosθ) =ax

    Plug in and get my answer for part 1? The real question I guess is if my math was right and if I'm using [itex]\mu[/itex]k instead of [itex]\mu[/itex]s. I have no idea how to start the other 2 parts.
     
  2. jcsd
  3. Sep 23, 2012 #2

    BruceW

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    Homework Helper

    Yes, you've got it correct. But you must be very careful with the friction. It will change direction depending on whether the crate is moving uphill or downhill. Your current answer assumes that the crate is moving uphill, since the friction is in the same direction as gravity (downhill). But if the crate is moving downhill, then of course the sign of the friction will become opposite of gravity. Why do you have a problem with the next two parts? Think about what constant speed means. What must be the total force on the object in this case?
     
  4. Sep 24, 2012 #3
    Oh wow, I didn't even notice that! Alright, so friction is moving against the crate moving downhill, which means that it's not -f, but +f.
    As for the next two parts, wouldn't a constant speed mean that a = 0? Specifically ax. So if that's right The F for uphill would be F = MgSinθ + [itex]\mu[/itex]s*N?
     
  5. Sep 24, 2012 #4

    BruceW

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    Homework Helper

    Yep, you got it. When the crate is sliding uphill, the friction and gravity are both acting downhill, while the applied force must be uphill. So you are correct that friction and gravity are both in the same direction in this case.

    One other thing is the coefficient of friction. You have taken mus to be the coefficient of sliding friction, but I think it might be the other way around. You can work it out logically, by looking at the values of the two coefficients of friction.
     
  6. Sep 24, 2012 #5
    Yeah I figured that [itex]\mu[/itex]k was the right one to use. I went through with it and found the right answer, Just that and a couple sign errors got me. Also for the last two parts, they're literally just asking the same thing. When, ax = 0, what is F, while part one is saying if F = 0, find ax. I could practically do it all in one sweep. Anyways thank you for the help! Got me set on the right track for sure :)
     
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