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Equal work lifting versus pushing up a plane?

  1. Feb 28, 2007 #1
    Equal work lifting versus pushing up a plane??

    I don't understand anything in physics having to do with work and/or energy. I drew a picture of the problem I'm having now:

    http://img80.imageshack.us/img80/9834/physicsoc8.png

    According to every physics book/website ever, the work in getting the object up there is the same in both cases.

    I can't see why. Can someone please explain this to me? And why is it that just because they get to the same height does it automatically mean the work done is EXACTLY the same?
     
  2. jcsd
  3. Feb 28, 2007 #2
    plane => less force but proportionally greater distance => equal work
     
  4. Feb 28, 2007 #3
    Because the only force being considered is your force pushing on it and gravity, which is a conservative force. Gravity acts straight downward on the object, so moving it upward against the gravitational field raises it to a higher gravitational "potential".

    If you want to prove it mathematically...

    In the first case (straight up):
    W = Fd
    W = mgd
    W = mgh

    In the second case (along the plane):
    W = Fd
    W = mgsin(theta)d
    sin(theta) = h/d
    d = h/sin(theta)
    W = mgsin(theta) * h/sin(theta)
    W = mgh
     
    Last edited: Feb 28, 2007
  5. Feb 28, 2007 #4
    But why is the force equal to [tex]mg\sin(\theta)[/tex]? Are you only considering the horizontal component of the force? (If so, why?) And is that 'd' you wrote just a general d or does it refer specifically to the d I drew in the picture (the length of the ramp)?
     
  6. Mar 1, 2007 #5
    I used your drawings to choose the names for the variables. The d is your d up the ramp (the distance the object moves) and the h is the vertical displacement. mgsin(theta) is the component of gravity up the incline (mgcos(theta) cancels with the normal force exerted by the plane on the object). Therefore, your applied force must equal mgsin(theta) in order to move the object with constant velocity up the plane.
     
  7. Mar 1, 2007 #6

    russ_watters

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    Staff: Mentor

    If by horizontal, you mean parallel to the plane, then yes. The reason is - that's the direction the object is moving, so that's the direction you must push it. If you push it at a lower angle, you are simply pushing it into the plane and if you push it at a higher angle, you are lifting it unnecessarily.

    It is very helpful/useful to break-up forces into directional components for specifically this purpose.
     
  8. Mar 2, 2007 #7
    Thanks everyone I've read what you've all written and I understand this triangle thing now but I still don't understand work at all.

    I'm really confused on this: Say there's a box of mass m on the ground. I pick it up to my chest (h meters) and so have done mgh joules of work on the box. Now, holding the box tightly to my chest, I step to the side.

    According to everyone I've talked to, I did NOT do work on the box when moving to the side... But this in my opinion is BS because I originally held the box still, and then I moved it with me to the left (or right). So that means it accelerated from rest and so by definition there was a force, RIGHT?!? And obviously there was a displacement too, right!? And work = force * distance and yet no work is done?!? THAT MAKES NO SENSE!
     
  9. Mar 3, 2007 #8

    russ_watters

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    Staff: Mentor

    You do no work against gravity in moving a box laterally. You most certainly do work against the box's inertia.
     
  10. Mar 3, 2007 #9
    SH86 your friends are looking at the total picture, work is the total cahnge in energy over some interval.

    in the case you described above, you can easily see that the total change in the energy of the box is zero, as it was at rest originally, and at rest at teh end of the interval.

    you can also think about it as if you broke the interval into two halfs, in the first interval your exerting a force F on the box, and as such you do Fx work on the box over the first half of the distance. Now consider the second half of the interval where you are exerting a force in the opposite direction -F, this does -Fx work over the second half of the distance.

    add them together and you get Fx-FX=0 or no work performed
     
  11. Mar 4, 2007 #10
    So there actually really was work, but just no net work? Is that right?
     
  12. Mar 4, 2007 #11

    Doc Al

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    Staff: Mentor

    You are absolutely correct. As Russ pointed out, if you accelerate the box as you move it to the side you are definitely doing work on it. Of course, if you bring the box to rest again as you move it to the side, you did positive work accelerating it, then an equal amount of negative work slowing it down. So the net work done on the box in moving it to the side is zero.

    Note that in your original problem it was assumed that you did just the minimum work to "overcome" gravity and get the box to a height "h"--without giving the box any kinetic energy.
     
  13. Mar 6, 2007 #12
    Yes...not sure if you've figured this out already but the actual definition of work is [itex]W = \int_a^b F \dot dl[/itex] (notice it's a dot product); along a straight line this reduces to [itex]W = F d \cos(\theta)[/itex] where theta is the angle between the force and the direction of displacement. Therefore, if you are exerting an upward force on the box equal and opposite to its weight mg, and you move the box to the right, the angle between your force and the box's displacement is 90 degrees. W = Fdcos(theta) = Fd(0) = 0. Another way to think about it is what the others have said: you are not raising it to a higher gravitational potential by moving it sideways. If the object's total mechanical energy is made up of only kinetic energy and gravitational potential energy, then obviously you didn't change its total energy by just moving it sideways.
     
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