# I Box Tipping on Inclined Plane

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1. Sep 14, 2016

### siyujiang81

I found the following problem on the 2009 F = ma exam.

24. A uniform rectangular wood block of mass M, with length b and height a, rests on an incline. The incline and the wood block have a coefficient of static friction, μs. The incline is moved upwards from an angle of zero through an angle θ. At some critical angle the block will either tip over or slip down the plane. Determine the relationship between a, b, and μs such that the block will tip over (and not slip) at the critical angle. The box is rectangular, and a ̸= b (a is not equal to b).

After trying the problem myself, I looked at some other solutions online, including one presented by an AoPS user. The link to that document can be found here: https://services.artofproblemsolvin...kYmZlNjIz&rn=MjAwOSBGPW1hIHNvbHV0aW9ucy5wZGY=. However, I am not sure about the concepts used in that solution.

First, that solution specified the top corner of the box as the point of rotation. However, the box not actually rotate about that point; it will rotate about the lowest point on the box. Second, the solution completely neglects the torque caused by the gravitational force. Even if we do specify the top corner of the box as the rotational axis, gravity will still provide a torque. The answer still appears to be correct.

I think (and let me know if I am wrong) that a better method would be to specify the rotational axis at the bottom corner of the box. In this way, at the critical angle, the normal force will not produce any torque, and neither does the frictional force. Only the gravity can provide a torque.

It can be easily seen that at the critical angle, the maximum possible frictional force is μs * mg * cos(θc), which in order to counter gravity (and NOT slip), must be greater than mg * cos(θc). From this, we see that μs > tan(θc).

We can break up the weight into its components. The horizontal component has magnitude mg * sin(θc) and acts at a distance a / 2 from the bottom edge. The vertical component has magnitude mg * cos(θc) and acts at a distance b / 2 from the right (lower vertical) edge. In order for the box to tip, the clockwise torque must be larger than the counterclockwise torque, so the inequality mg * sin(θc) * (a / 2) > mg * cos(θc) * (b / 2) must hold. This simplifies nicely to tan(θc) > b / a.

Combining the two simple inequalities gives us our final answer, μs > b / a. Does this solution seem clear?

2. Sep 14, 2016

### haruspex

If a rigid object rotates, it will rotate equally about any axis you choose. As long as contributions from all forces are considered correctly, the choice of axis does not matter in statics. (In dynamics it can.)
The solver is considering the block as being on the point of tipping. In that arrangement, where is the centre of gravity in relation to the highest and lowest corners? What moment will the gravitational force have about those points?

But I agree it is easier your way.