# Would this box on the floor rotate based on friction?

There is a floor that friction is proportional to its velocity (like $F=-kv$) and there is a box with its width as $l$ and its height as $h$. (you may assume that $l$ is longer than $h$). It is on the floor with the initial velocity $v$. Then, if $v$ is big, I think it would rotate. But I don't know how to analyze this situation as differential equation.

This figure shows the situation it rotate as the $\angle \theta$. But, as the initial time, $\theta$ is $0$ so thus I think its friction would not be exerted at the left-bottom corner. So it will be different with this figure.

If there is no floor, the rotate pivot is just the center of mass I think. But there is a floor. It makes me crazy. • You have the normal force incorrect. (and therefore, the friction force is incorrect too). The normal force should ensure that the acceleration of the corner is zero in the y direction (it is at position $(-\cos(\theta+\theta_2)r+x,\sin(\theta+\theta_2)r+y)$, where $\theta_2$ is some constant angle made between the bottom of the rectangle and the line between the corner and the center, and r is the distance between the corner and the center). Solving this gives you three forces, and you will need to convert them to torque and force on the center of mass to get the final eqs. – user12029 Oct 5 '13 at 0:17
• – ja72 Apr 3 '14 at 15:10

Velocity cannot produce torque. Only forces acting from a distance can produce torque.The friction acts on the entire base of the box to prevent its relative slipping. Here, the frictional force is proprotional to the velocity applied. Hence considering the horizontal forces $$F=-kv$$ $$ma=-kv$$ $$mdv/dt=-kv$$ $$mdv/v=-kdt$$ Integrate with proper limits to get the velocity at any time t.