# Is there a better method to solve the Box Tipping on Inclined Plane problem?

• siyujiang81
In summary, the problem involves a rectangular block on an incline with a coefficient of static friction. The question is to determine the relationship between the block's dimensions and the coefficient of friction in order for the block to tip over, rather than slip, at a critical angle. An online solution uses the top corner of the block as the point of rotation, but this may not be accurate as the block actually rotates about the lowest point. Another solution considers the torque caused by gravity, and suggests using the bottom corner as the rotational axis. This leads to the conclusion that in order for the block to tip, the coefficient of friction must be greater than the ratio of the block's width to height. This solution may be clearer and more accurate.
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?

siyujiang81 said:
that solution specified the top corner of the box as the point of rotation. However, the box not actually rotate about that point;
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.)
siyujiang81 said:
the solution completely neglects the torque caused by the gravitational force.
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.

## 1. What is box tipping on an inclined plane?

Box tipping on an inclined plane is a scientific experiment that involves placing a box on a sloped surface and observing its movement as it slides down the slope. This experiment helps to understand the effects of gravity and friction on an object's motion.

## 2. What are the factors that affect box tipping on an inclined plane?

The main factors that affect box tipping on an inclined plane are the angle of the slope, the mass of the box, and the coefficient of friction between the box and the inclined surface. These factors can be manipulated to observe the changes in the box's motion.

## 3. How does the angle of the slope affect box tipping?

The steeper the slope, the greater the force of gravity acting on the box, causing it to accelerate faster down the slope. As the angle of the slope decreases, the force of gravity decreases, resulting in slower acceleration and a longer distance travelled by the box.

## 4. What is the significance of box tipping on an inclined plane?

Box tipping on an inclined plane is a fundamental experiment in physics that helps to understand the concepts of force, motion, and energy. It also demonstrates the relationship between gravity, friction, and the angle of the slope on an object's movement.

## 5. How is box tipping on an inclined plane used in real life?

The principles learned from box tipping on an inclined plane have practical applications in everyday life, such as designing roller coasters, calculating the speed and trajectory of objects rolling down a hill, and understanding the motion of objects on an inclined surface.

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