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Rolling a Box

  1. Nov 18, 2008 #1
    I have been trying to figure out this problem for a while. I am in AP Physics now, preparing for Higher Level IB Physics next year. A few physics friends and I gave this problem to my teacher but he was unable to give a solution, so I decided to come here. I love physics and my perpetual wondering about it has given birth to this (probably simple) problem.

    1. The problem statement, all variables and given/known data
    A square, with sides h is upright on a flat surface, and the coefficient of kinetic friction between it and the surface is mu. It's mass is m and it is on Earth at sea level. If a person pushes on the box at the same height as the center of gravity and parallel to the floor, how hard must the person push to get it to tip over?
    I'm not sure if the final statement is even specific enough, but I want to understand how a problem like this would work. My attempt describes several solutions my friends and I have come up with that are variations of this question.

    2. Relevant equations
    I know this is definitely a torque problem, T=Fd
    obviously these...

    3. The attempt at a solution
    We concluded that as soon as any force even acts, as long as it is greater than friction (I think), ignoring static friction, the bottom of the box will begin to tip to a certain theta with the floor. To make things simpler, we decided to allow the force to remain parallel to the floor and at the same height as the center of gravity. This allowed us to have normal and friction act at one point and the force at another, but I think there is still a force missing, or else this object will spin no matter what. I think it's the weight, but the center of gravity isn't on the torque that goes through this box.
    If our premise is correct, one of us tried, arbitrarilly, drawing a torque from the center of mass to the center of the bottom of the box, saying friction acts there (before it tips), and another torque from the middle of the bottom of the box to the end of the box, where he said normal would act. Then he said the pushing force is basically acting on the center of mass (is this true?) and set the torque of friction and pushing force (on the first torque diagram) equal to the torque of the normal (on the second diagram). Does double torque like this exist?
    I thought that you should ignore the process of actually tipping the box, and draw a torque from the pushing force to the corner where friction and normal act, but this diagram would be missing something that stops it from rotating indefinitely. I also thought that there is some way to draw a triangular torque going from the point of pushing, to the center of mass, to the corner where the box is rested, but I doubt that is correct.

    Can someone just push me in the right direction for this simple variation of a rolling box problem? I would like to expand on this while I have free time in school to look at circles and squares on inclined planes, electromagnetic forces that change when the box is pushed (rather than constant pushing) and other stuff like that.
    I was also wondering if it would be easier to make the pushing force always normal to the side of the box, and if this would involve beloved Calculus.

    Thank you
  2. jcsd
  3. Nov 19, 2008 #2


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    [EDIT: responding to deleted post ... ]

    I didn't follow this completely, but agree with the idea of assuming the box will tilt--or at least begin to tilt at a small angle. This puts the normal and friction forces acting at a bottom corner.

    This is a problem, since in most situations ms<1.
    Last edited: Nov 19, 2008
  4. Nov 19, 2008 #3
    Why was a post deleted? I still don't understand how to do this problem.
  5. Nov 19, 2008 #4


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    I would suggest looking at the problem in a different way. The problem presents the box as a square.

    What about the cases where the box is tall and skinny? What if it was really tall? How much force is required to tip it?

    Take the other extreme where it is really squat and long. Is there any speed that the box can be tipped over?

    At what point is it that the box goes from easily tipped to impossible to tip?
  6. Nov 19, 2008 #5
    I've already considered that. It becomes impossible to tip when it has no height. Also, it is impossible when the surface is frictionless. Can someone tell me what forces would be on the torque diagram? Or direct me to a site that shows how to solve a problem even remotely similar to this one?
  7. Nov 19, 2008 #6


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    Write an equation then for the one you know tips.

    What is the horizontal moment arm about the corner that creates the forward directed torque? How is that force limited?

    Balanced against that is what? The weight acting over what opposing moment arm.
  8. Nov 19, 2008 #7
    What tips do I know? I just don't understand how to include weight if the center of mass does not lie on the torque from the point of pushing to the pivot point on the floor, so I will assume you push exactly on the center of mass. I don't understand the terminology, but I think you have the pushing force and gravity at one end, and the friction and normal forces at the other.

    | ' ' ' ' ' ' ' ' ' ' '|
    | ' ' ' 'o--> ' ' '|
    | ' ' ' ' '''-,,' ' ' |

    So the torque would be (if that line of `'-,) is 45 degrees


    Am I right? What do I do from here? F(push)sinx=mgsinx ? That narrows it down to one variable. Should I include the theta between the bottom of the box and the floor already? So it would be F(push)sin(x+y)=mgsin(x-y), where y is that angle? This can't be right, because that would mean friction is irrelevant.
    Last edited: Nov 19, 2008
  9. Nov 19, 2008 #8


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    In the horizontal direction you have friction resisting your moving force.

    Projecting that force to the perpendicular above the pivot you have a moment about the front corner of ... Height/2 right?

    And what happens when that force is greater than the frictional resistance? Doesn't it simply serve to accelerate the block?

    Now in the other direction that one is pretty easy isn't it? Weight acting through what distance to the pivot? ... Width/2 maybe?

    So doesn't it mean that when the frictional force times height/2 is greater than weight times width/2 it will tend to tip?

    As to tips: My earlier statement should more clearly be read as:
    Last edited: Nov 19, 2008
  10. Nov 19, 2008 #9
    (moment about the front corner=where you push?)
    I thought pushing force times distance would have to be greater than friction times distance. I'm just not able to visualize this yet, sorry. Is it this?

    (center of box)
    .l..........................l h/2
    ...........................l (pivot point)
    . . . . . . . . . . . <-- Friction


    Then how would I find the angle between the floor and the bottom of the box?
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