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Why does an object tip over instead of sliding

  1. Dec 12, 2013 #1
    What causes an object to tip over instead of slide when a sufficient force is applied to an object? If you apply a force of 7N to an object that requires a force of 6N to tip it over and a force of 3N for it to slide it will always tip over. As long as the force applied is more than the force required to tip the object over it will tip no matter how low the force of static friction is. Why is this? The Fnet for the object to slide is much greater than the Fnet for the object to tip so what makes the object tip?

    What theory explains this and could you give a brief explanation of why it happens?
     
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  3. Dec 12, 2013 #2

    Andrew Mason

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    What is your source for this? Can you give us a specific example of this?

    AM
     
  4. Dec 12, 2013 #3

    Simon Bridge

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    The simple answer is "because sufficient force was applied." That's what "sufficient" means.
    But I know what you mean :)

    A better answer is because staying upright means that there must be a balance of torques.
    But the model you have about "minimum force to tip over" is actually too simplistic.

    i.e. if a box has a rated 6N tippover force - and I apply 6N for 10^{-23} seconds ... the box still won't tip.

    The general rule is: something will not fall over so long as a vertical line through it's center of mass passes through it's base.

    So to understand the force to tip you need to take a closer look: how would that be calculated exactly?

    For a box mass m, square base area A and height h, initially stationary on a surface with friction coefficients ##\mu_s## and ##\mu_k## ... when you apply a horizontal force F a height y<h up one side ...

    how big does F have to be for the box to slide at all?
    if it doesn't slide - how big does F have to be to tip the box over?

    ... but if it does slide?
     
  5. Dec 12, 2013 #4
    Hmm, you need to know to some rigid body statics to understand this. The tipping over isn't just a matter of how much force you apply, but a conjugation of that and the point of application of the force. Let's say you have a large rectangular box, upright, and you apply some horizontal force F to it. If you gradually apply the force higher and higher, assuming it's a big enough force, at some height, the box will tip over.

    If you draw a free body diagram, you have the applied horizontal force F at a heigh h, a vertical reaction force N, its weight W, and a static friction force Fa. To have a static situation, you need to have equilibrium on both axes, that is F = Fa, and N = W, and additionally, zero net moment of the forces. To guarantee that, the point of application of N needs to be such that it balances out the torque created by the other forces. The situation of tipping over is when the point of application that you would need to balance out that torque is outside the box, so that it can no longer be applied, and so a net torque is created which tips over the box.

    EDIT: Only noticed Simon Bridge's post after posting this. Anyway, we address slightly different things, so I'm leaving it be :)
     
  6. Dec 12, 2013 #5

    AlephZero

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    Statics will only tell you how much force you can apply before the box moves. To find out whether it tips or slides for a bigger force, you need dynamics.

    And if you apply a enough force low down, the box will tip over backwards, not forwards. Try it with something like a cereal packet.
     
  7. Dec 12, 2013 #6
    Yes, I generally agree with you that statics can only get you so far. You'll eventually have to include horizontal and angular acceleration to figure out how it actually moves. However, the problem is first easier to visualize in terms of equilibrium of forces and torques, and I find the picture of the point of application of the normal reaction moving out of the body to be quite intuitive and satisfying.

    Of course, but for other objects (things which you visualize of when you think of a box), they'll start to slide before tipping over backwards.
     
  8. Jul 7, 2015 #7
    Along with the above answers, it also depends on where the force is applied. For example, get a stick of deodorant, and align it so that the thin part is facing you. Next, get your finger (preferably the nail for less friction), and place it near the bottom while pointing to the right (along the positive x- direction). Push slightly until it slides at a constant acceleration. While sliding, move your finger up along the center, and notice what happens when you reach past the middle of the stick. It tips! Why does this happen? When you push on the stick anywhere, it changes the position of the normal force and "x amount" away from the center of the z-axis. When you reach the middle of the stick, the normal force is applied exactly at the furthest edge of contact along the x-axis. Past that point, the normal force goes past the furthest point and has no effect of the stick. Because of this the applied force from your finger and the friction force acting in the opposite direction create an instant moment, and because there is no normal force acting to counter it, the moment causes a rotation, and thus, the tipping.
     
  9. Jul 8, 2015 #8

    Andrew Mason

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    Welcome to PF mormonastroscou!

    Things tip because of moments or torques. So one has to analyse the torques. There are two torques about the centre of mass. Friction between the bottom of the stick and the table surface is generating a torque about the centre of mass acting in a clock-wise direction. Initially, your finger is generating a torque in a counter-clockwise direction so it balances the torque due to friction. When the finger moves above the centre what happens to the direction of that torque?

    AM
     
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