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    A student could either pull or push, at an angle of 30o from the horizontal, a 50-kg crate on a horizontal surface, where the coefficient of kinetic friction between the crate and surface is 0.20. The crate is to be moved a horizontal distance of 15 m.
    (a) Compared with pushing, pulling requires the student to do (1) less, (2) the same, or (3) more work.
    (b) Calculate the minimum work required for both pulling and pushing.
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    Quote Originally Posted by Candy
    A student could either pull or push, at an angle of 30o from the horizontal, a 50-kg crate on a horizontal surface, where the coefficient of kinetic friction between the crate and surface is 0.20. The crate is to be moved a horizontal distance of 15 m.
    (a) Compared with pushing, pulling requires the student to do (1) less, (2) the same, or (3) more work.
    (b) Calculate the minimum work required for both pulling and pushing.
    I can't do diagrams. Sorry!

    I am assuming in all of this that the applied force moves the crate at a constant velocity. The problem can't be done otherwise!

    a) The work done by a (constant) force is:
    W = \vec F \cdot \vec s = Fs \, cos \theta_{Fs}
    where the angle is the angle between the force and the displacement vectors (placed, as always, tail to tail.)

    Now, there are two ways in which pushing or pulling will affect the result. First, it will technically change the angle in the work formula. For pulling we have theta = 30 degrees, for pushing we have theta = -30 degrees. (I'm making an assumption about the statement of the problem here. If you are pulling I'm assuming you are pulling upward, if you are pushing I'm assuming you are pushing downward. I would personally ask the professor about this assumption, but as I can't do that I have to choose something.) Since cos(-theta) = cos(theta) there is no difference in the work from this source.

    Second it will change the normal force on the crate. Sketch a free body diagram for the crate. There will be four forces present: the weight, the normal force on the crate, the friction force, and the applied (push or pull) force. For reference purposes I am calling N the value of the normal force in the absence of the applied force.

    Case 1. The applied force has an upward component so according to Newton's second law in the vertical direction, the normal force is less than N. Thus the friction force, which this the coefficient of friction times the normal force, is smaller than \mu N, so the applied force is smaller than...

    Case 2. The applied force has a downward component so according to Newton's second law in the vertical direction, the normal force is more than N. Thus the friction force is larger than \mu N, so the applied force is greater than in Case 1.

    Thus F will be larger in Case 2, and thus the work done will be greater in Case 2.

    b) Pulling: Using the FBD we can see that (in the vertical direction) N = w - Fcos(theta). Thus the friction force f = (mu)(w - Fsin(theta)). Now, using Newton's second in the horizontal direction we get Fcos(theta) = f. (Recall that f is in the opposite direction to the horizontal component of F and that the acceleration of the crate is 0 if it is moving at constant velocity.) Finally w = mg Thus:
    Fcos(\theta) = \mu mg - \mu Fsin(\theta)

    F = \frac{\mu mg}{cos(\theta) + \mu sin(\theta)}

    Finally W = Fs cos(\theta) = \frac{\mu smgcos(\theta)}{cos(\theta) + \mu sin(\theta)}
    (I'll let you do the numbers on this.)

    Pushing: We have the same situation except that now the normal force is N = w + Fcos(theta), since F now has a downward component. The rest is the same except for the change in sign on Fcos(theta). So we get
    f = (mu)(w + Fsin(theta))
    Fcos(theta) = f (as before)
    So:
    F = \frac{\mu mg}{cos(\theta) - \mu sin(\theta)}
    and
    W = Fs cos(\theta) = \frac{\mu smgcos(\theta)}{cos(\theta) - \mu sin(\theta)}

    -Dan
    Last edited by topsquark; June 30th 2006 at 05:19 AM. Reason: Typo
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