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Work Required to Push/Pull a Crate

  1. Mar 27, 2007 #1
    1. The problem statement, all variables and given/known data
    A student could either pull or push, at an angle of 30 degrees from the horizontal, a 50 kilo crate on a horizontal surface, where the coeficient of kinetic friction between the crate and the surface is 0.2. The crate is to be moved a horizontal distance of 15m. (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.


    2. Relevant equations
    W=Fd
    F=mgcos30
    f_k=u_k(F_n)


    3. The attempt at a solution
    For (a), unless somebody can correct me and/or lead me in the right direction, I believe that the amount of work that is done to either push or pull the crate the required distance would be same. Conceptually, if you are wanting to move the crate in a certain direction, regardless of whether you decide to push or pull, assuming that the force to move the object is being applied from the same angle, the frictional force acting on the crate would still be resisting the force in the opposite direction of the movement of the crate. The work required to move the crate would be same, just that when you actually do it, you end up utilizing different muscle(s) when trying to move it either by push or pull, which can sometimes lead one to think that doing one method means exerting more work than the other. Please correct/advise me if this is the wrong mode of thinking.

    For (b):

    F=mgcos30
    F=(50kg)(9.81m/s/s)cos30
    F=424.79N

    f_k=u_k(F_n)
    f_k=(0.2)(50kg)(9.81m/s/s)
    f_k=98.1N

    F is being applied in the (+) direction, with the appropriate f_k resisting it in the (-) direction, therefore F-f_k=F_net

    F_net=424.79N-98.1N
    F_net=326.69N

    W=F_net(d)
    W=326.69N(15m.)
    W=4900.35J

    Thank you in advance for any help.
     
  2. jcsd
  3. Mar 27, 2007 #2

    Doc Al

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    Staff: Mentor

    You're thinking is incorrect. The work would be the same if the friction force was the same in both cases. Is it? Hint: Reconsider how you are calculating the normal force.
     
  4. Mar 27, 2007 #3
    Obviously not, though it may lead me to believe the old adage that pushing an object is less strenuous than pulling the object with a rope, which may potentially imply that there is possibly less work to be conducted if one were to push the object versus pulling it.

    That being said, conceptually, if you were to push the object at said 30 degrees from the horizontal, would the friction force be acting in the same direction of the movement of the object? Also is my work in the OP valid only on the assumption that the object in question is being pulled and not being pushed, which would obviously yield a different result?
     
  5. Mar 28, 2007 #4

    Doc Al

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    Staff: Mentor

    The kinetic friction always opposes the motion of the crate, so its direction is not the issue. (But the magnitude of the friction force will be different, depending on whether you are pulling up at 30 degrees versus pushing down at 30 degrees.)

    Unfortunately, your work is not valid for either situtation.

    I don't understand what you are calculating here. The applied force has to be just enough so that it's horizontal component (F cos30) overcomes the kinetic friction.

    Here you are calculating the friction force, but using the wrong value for F_n. If the applied force had no vertical component, then this would be correct: the normal force would equal the weight of the crate. But the applied force does have a vertical component which changes the normal force. (In one case the applied force increases the normal force; in the other case it decreases it.)

    Start by drawing yourself a diagram showing the forces acting on the crate in both situations. (A separate diagram for each.) Set up equations for vertical and horizontal force components. This will allow you to solve for the applied force (different in each case) that will just be enough to overcome friction. Once you have the force, you can calculate the minimum work required in each situation.
     
  6. Mar 28, 2007 #5
    Let's give it another whack:

    The vertical component of the applied force would be the equivalent of mgsin30, which in this problem would be 245.25N. Now how this affects the normal force (whether this figure is in the + or - direction) depends on whether the object is being pulled or pushed.

    Pull:
    mgsin30 is acting in the opposite of direction of the crate (mg), so would it's effect on the normal force create an increase (???) in the aggregate amount of normal force? That is:

    mgsin30=(50kg)(9.81m/s/s)sin30=245.25N (in the upwards direction)
    Adding to the established mg=(50kg)(9.81m/s/s)=490.5N
    490.5N+245.25N=745.75N

    Push:
    mgsin30 is acting in the same direction of the crate (mg), so would this create a decrease (???) in the aggregate normal force? That is:

    mgsin30=(50kg)(9.81m/s/s)sin30=245.25N (in the downwards direction)
    Adding to the established mg=(50kg)(9.81m/s/s)=490.5N
    490.5N-245.25N=245.25N

    Using the new normal forces (assuming these are correct, probably not :frown:), calculate the frictional forces assuming u_k is 0.2:

    Pull:
    f_k=u_k(F_n)
    f_k=(0.2)(745.25N)
    f_k=149.05N

    Push:
    f_k=u_k(F_n)
    f_k=(0.2)(245.25N)
    f_k=49.05N

    Am I heading in the right direction to point, or have I taken a couple more steps backward??? :grumpy:
     
  7. Mar 28, 2007 #6

    Doc Al

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    Staff: Mentor

    Let's sort this out. If the applied force is F, then the magnitude of its vertical component would be Fsin30. For some reason, you seem to think that the weight of the crate (mg) is the applied force--no reason to think so! The applied force is unknown--you have to solve for it.
    This is good.

    Just the opposite. Without the upward applied force, the floor must exert a normal force equal to the full weight, F_n = mg. But since the applied force pulls up it supports some of the crate's weight, thus the floor does not need to push up so hard, so F_n is less that mg.

    Mathematically: Let's analyze the vertical forces on the crate:
    (1) The weight (mg) acting down: -mg
    (2) The applied force acting up: Fsin30
    (3) The normal force acting up: F_n

    They must add to zero, so: Fsin30 -mg + F_n = 0
    Which gives us: F_n = mg - Fsin30

    For the pushing down case, the analysis is similar except that the vertical component of the applied force is downward (- Fsin30) which adds to the normal force.

    All this is just to find the normal force and thus the friction force (in terms of the unknown applied force F). You still need to set the horizontal component of the applied force (Fcos30) equal to the friction in order to solve for F. Then you can--finally--calculate the value of the work required to overcome friction.

    Time for another whack! :wink:
     
  8. Mar 28, 2007 #7
    I'm guesstimating that the amount of the applied force F will most likely be somewhere in the range of 125-175 newtons. I'll have to keep hacking at it and see what I end up with in the end. :wink:
     
  9. Mar 28, 2007 #8
    Using calculator work, I deduced that the applied force (F) for the pulling scenario is ~101N, now how to get the correct equation to get to that is rather confusing to me, as I would like to show some work for this, but this is how I got it (somehow):

    If F_n=mg-Fsin30, then F_n=490.5N-(0.5)F [since sin30=0.5]
    As I said, I deduced 101N to be nearest to the actual answer by trial and error, so if the math is right:

    F_n=490.5N-(101N)sin30
    F_n=490.5N-55.5N
    F_n=435N

    f_k=0.2*435
    f_k=87N

    Assuming an (F) applied force of 101N, Fcos30 (horizontal component) would be 87.46N (only 0.46N off f_k due to rounding).

    Pushing found F=128N by the same trial and error as above.

    Now to put an equation (or two) together to show how I got this is the next priority for me. :wink:
     
  10. Mar 28, 2007 #9
    well i have to confess I've resorted to calculator twiddling, but it needn't be so here. And on a test in a limited time under pressure, not so good.

    lets assume one is lazy and will use the minimum force required.

    we know that we have to overcome the horizontal frictional force. so

    F*cos(30)=Fn* 0.2

    take the case of pulling, you now understand that since the force is applied partially upward, it reduces the normal force by an amt Fsin(30). (Since there is no actual vertical displacement, no work is done in the Y direction). AGAIN Normal force is reduced by F*sin(30)

    F*cos(30)=(mg-Fsin(30))0.2 collecting force terms we have,

    F(cos30+.2*sin30)=mg*0.2. or F= 50*0.2*9.8/(cos 30+0.2*sin30)=98/(0.866+.1)=101
     
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