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Finding magnitude of tension that causes object to slip

  1. Oct 24, 2016 #1
    1. The problem statement, all variables and given/known data
    A rope attached to a 19.0 kg wood sled pulls the sled up a 19.0° snow-covered hill. A 14.0 kg wood box rides on top of the sled.

    EDIT: the actual question to be answered, oops.

    What's the magnitude of the tension that will cause the box to slide?

    Mass of box (mb) = 14.0kg
    Mass of sled and box (ms) = 33.0kg
    Coefficient of static friction for wood on wood (μs) = 0.5
    Coefficient of kinetic friction for wood on snow (μk) = 0.06
    Angle of slope (θ) = 19°
    2. Relevant equations
    Fnet = m*a
    Fs = μsn
    Fk = μkn

    3. The attempt at a solution

    I've drawn the free body diagrams for both just the box, and for the sled and box combination as they sit on the slope. I've drawn graphs of both and made a chart of the force's x and y components. Getting the relevant forces for each, I've composed two Fnet equations in the x direction.

    Equation for net force on the box in the x direction:

    Fnetbx = mba
    -Fgsin(19°) + fs = mba
    -mbg*sin(19) + mbs = mba
    -14(9.8)sin(19) + 14(9.8)(0.5) = 14a
    -44.668 + 68.6 = 14a
    23.932 = 14a
    acceleration = 1.71
    I found the acceleration and the answer seems reasonable given the context of the problem.

    Equation for net force on the sled+box in the x direction:

    Fnetsx = msa
    T + fs - fk - Fg1sin(19) - Fg2sin(19) = msa
    T + mbs - msk - mbgsin(19) - msgsin(19) = msa
    T + 14(9.8)(0.5) - 33(9.8)(0.06) - 14(9.8)sin(19) - 33(9.8)sin(19) = 33(1.71)
    T + 68.6 - 19.404 - 44.668 - 105.289 = 56.43
    T - 100.761 = 56.43
    T = 157.191
    Having the acceleration from the first equation, I solved for the tension on the rope. This number concerns me, 157N seems a little low for a force that is pulling that much weight up the slope.

    Having used the coefficient for static friction of wood on wood in finding this tension, I was thinking that any force greater than it would cause the box on the sled to slip. My answer of 157.2 is incorrect.

    After this I tried going back to just the box.
    If fsmaximum = μs*n
    fsmaxmimum = (0.5)(14)(-Fgcos(19))
    fsmaxmimum = (0.5)(14)(-14*9.8*cos(19))
    fsmaxmimum = 908
    This number doesn't make any sense to me and its clear now that I'm lost. :]

    Did I make a mistake somewhere along the lines, or have I not taken this problem far enough? Thank you in advance.
    Last edited: Oct 24, 2016
  2. jcsd
  3. Oct 24, 2016 #2


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    The normal force on the box is not the weight of the box. That would be the case on a horizontal surface, not on an incline.

    On edit: Welcome to PF.
  4. Oct 24, 2016 #3
    Thank you for the welcome!
    I considered this and thought I had done right when instead of multiplying the mass of the object by just gravity, as is the case when the object is on a flat surface, I multiplied the mass by the y component of its weight. I see now where I went wrong because that value still isn't parallel with the normal force. I'll attempt again with this information thank you
  5. Oct 25, 2016 #4


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    There is more. Lose the fs in this equation. If you are writing the net force on the two-mass system the force of static friction on the wood is equal and opposite to that on the sled so they add up to zero.
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