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Friction Problem of a Crate Sliding Down a Ramp

  1. Jul 3, 2017 #1
    1. The problem statement, all variables and given/known data
    A steel washer is suspended inside an empty shipping crate from a light string attached to the top of the crate. The crate slides down a long ramp that is inclined at an angle of 38 ∘ above the horizontal. The crate has mass 239 kg . You are sitting inside the crate (with a flashlight); your mass is 59 kg . As the crate is sliding down the ramp, you find the washer is at rest with respect to the crate when the string makes an angle of 71 ∘ with the top of the crate.

    What is the coefficient of kinetic friction between the ramp and the crate?

    Note: u=coefficient of friction

    2. Relevant equations
    Newton's Second Law: F=ma
    Kinetic Friction: Fk=uN
    3. The attempt at a solution

    I have defined my xy-coordinate system with the x-axis along the ramp (negative going up the ramp) and the y-axis perpendicular to the ramp.

    Forces on the crate: Fx= Mgsin38 - uMgcos38=ma. (Eq. 1)
    There is no net force in the y-direction on the crate.

    Gravity is acting in the positive x-direction (for how I defined the xy-coordinates), and friction acts opposite to the direction of motion, so it is negative. So Fx= (x-component of gravity)-(friction force). Friction force is equal to uN, and the Normal force is equal to the y-component of gravity (which is Mgcos38).

    The washer is the trickier part to me. It seems that I need to use the washer to find the acceleration. Here are the equations I have for the forces on the washer.

    Note: T=tension of rope on washer. Also, little m denotes mass of washer, whereas big M denotes mass of crate.

    Forces on the washer: Fx= Tcos71+mgsin38. (Eq. 2)

    Both gravity and the tension act on the string in the positive x-direction (I think).

    Since m and T are not given, they somehow need to be eliminated, so I defined a second equation.

    Forces in the y-direction on washer: Fy= Tsin71 - mgcos38=0. (Eq. 3)

    Since the washer is not moving in the y-direction, the sum of the forces is 0.

    I then solved Eq. 3 for T, getting:

    T= [mgcos(38)] / [sin(71)]

    I plugged this into Eq. 2, getting:

    Fx= [mcos(38)cos(71)] / [sin(71)] - mgsin(38)=ma.

    Since every term has m, they can all cancel out, leaving:

    Fx= [cos(38)cos(71)] / [sin(71)] + gsin(38) = a = 6.3 m/s^2

    Now that acceleration is known, there is only one unknown in Eq. 1, the friction coefficient.

    Mgsin38-uMgcos38= Ma (This is just Eq. 1 restated)

    Solving for u:

    Since M is in every term, I can cancel out M: gsin38-ugcos38=a.

    Rearranging the equation to isolate u: gsin38-a=ugcos38

    Divide both sides by gcos38: tan38- [a] / [cos38] =u.

    u= -0.35. I figure the negative is since it's just in the opposite direction.
     
  2. jcsd
  3. Jul 3, 2017 #2
    If you picture the situation, does the string hang in the downhill-half of the box, or the uphill-half of the box?
    So should the x-component of tension be positive or negative (according to your coordinates)?

    I am speaking in particular about this equation:
     
  4. Jul 3, 2017 #3
    Turns out my significant digits weren't correct and that was the issue. The correct answer was 0.345. For anyone with a similar problem, I think my method was correct.
     
  5. Jul 3, 2017 #4
    But you said the x-acceleration is 6.3 m/s/s right? What if there is no friction, then the acceleration would be g*sin(38deg)≈6.0 m/s/s ....... So something is clearly not right if you calculated acceleration faster than the frictionless case!
     
  6. Jul 3, 2017 #5

    haruspex

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    No, it's negative because you have the string in the upper half of the crate instead of in the lower half, as @Hiero pointed out.
     
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