# Push a box in a plane with friction. How to deal with the rotation?

Suppose I have a box (say, length-1m, width-1m, height-0.5m) on the plane with friction. I can apply a horizontal force in on the surface of the box.

If the force doesn't pass through the center of the box, then the box will rotate. My question is, how to solve the angular velocity, angular acceleration and the center of rotation?

The difficult thing in this problem is that we need to consider the torque of the friction. In my opinion, the friction should be distributed in the bottom surface of the box. So in order to get its torque, we need to do the integral within the bottom surface.

My idea is: move all the forces to the center of mass, and add the corresponding torques to keep the equivalence before and after moving the force. So the box's movement can be divided into two parts: (1) the translation of the center of mass (2) the rotation around the center of mass.

I'm not sure about my idea. The problem is that the box doesn't necessarily rotate around its C.O.M, right?

Another question is about the direction of the friction. Suppose that the friction is distributed on the bottom surface. If the box rotates, then the movement of the point on bottom surface is a superposition of translation and rotation, which means the absolute velocities in different positions on the bottom surface are different. Because we know friction always has the opposite direction to the relative movement, does that mean the frictions in different positions on the bottom surface have different orientations?

Who can give me the right approach to think about this problem?

Thank you very much!

• If the box is ideal and symmetric there would be no friction torque. Friction would be applied through the center of pressure which is in the center of the box. – John Alexiou Feb 11 '14 at 20:14
• And yes to solve the problem you add up all the equipollent forces and moments at the center of mass. – John Alexiou Feb 11 '14 at 20:16
• Thanks for your comments. (1) Why is the friction applied in the center of the box? Isn't it distributed in the bottom surface of the box? And in my opinion, we need to do the integral within the bottom surface to calculate the torque. Is that correct? (2) Does that means that the center of rotation is the center of the box? – Zijian Feb 11 '14 at 20:36
• To calculate a distribution of forces you need to consider the elasticity of the part, which ups the complexity immensely. I guess you can consider a linear distribution of pressure to counter act any applied force/moment keeping the box flat. – John Alexiou Feb 11 '14 at 21:23
• Yes I want to use the linear distribution of pressure, which means the friction magnitude is equal at each point on the bottom surface. However, I'm concerning the orientation of the friction at different points since different points may have different absolute velocity(translation + rotation). Is this true? – Zijian Feb 11 '14 at 23:04

## 1 Answer

Simplifying the problem slightly, I think it can be solved. Assumptions:

• Pushing force is applied at the bottom of the box - so there is no net torque about the horizontal axis
• Weight distribution in the box is even
• Force distribution (normal force) is even - imagine 1000's of tiny springs touching the ground
• Coefficient of friction is constant

Now we can draw a picture of the box at some point: It is easy to see that $$d = \frac{w}{2} \sin \theta\\ \Gamma = F_{friction} d$$

Now your question was - "what is the force of friction, and how do I deal with the fact that the apparent origin of rotation may not be the center of the box?" That deserves some more thought.

You are right that the force of friction is in the opposite direction of the motion at every point. Thus, when you describe the motion of the box as the sum of a translation and a rotation (at a given instant in time), the net force of friction can be computed as the sum of a torque (due to rotation of the box) and a linear force through the center of the box (since for translation all points in the box move at the same speed in the same direction).

This should make the problem more tractable in terms of the equations of motion - at every moment in time, you can write down the total torque on the box as the sum of the applied torque (from the pair of forces shown in my picture) minus the frictional torque (due to the rotation of the box) plus the net force on the box (after taking account of the torque) accelerating the center of mass. I will see if I have time later to write down the equations more fully.