# Problem on friction

## Homework Statement

##A## and ##B## are two objects with mass ##m_1## and ##m_2## respectively. The floor is frictionless. The kinetic friction coefficient between ##A## and ##B## is ##\mu## ; A constant force ##F## is applied on ##A##. Assume, ##F## is greater than the limiting static frictional force. So, ##B## will apply a frictional force, ##f = \mu m_2 g## on ##A##. According to Newton's third law, ##A## will also apply the same force ##f## on ##B## in the opposite direction. So, ##A## and ##B## both will be accelarated (Suppose, ##f < F##). Let, the accelaration of ##A## and ##B## be ##a_1## and ##a_2## respectively.
If, ##a_1 > a_2##, at some moment the situation will be like this:

And, if ##a_2 > a_1## , it will be like this:

What will happen afterwards?

## Homework Equations

##\mu = \frac {f}{R} ##

## The Attempt at a Solution

In the first case, ##B## will rotate and fall down. But what will be the axis of rotation and angular velocity?
In the second case, after a moment, the center of gravity of ##B## will be ahead of the edge of ##A##. So, the normal reaction is ##0## on the surface between ##A## and ##B##, so the frictional force is also ##0##. Hence, B will not be accelerated forward anymore, but ##A## will. And so, ##A## will catch the center of gravity of ##B## again in a moment. So, I think ##B## will not fall in this situation and will remain as it is in the third picture.

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## Answers and Replies

ehild
Homework Helper
In the first case, the upper block would turn over the edge of A, as gravity exerts a positive torque about it as soon as the CM moves over the edge.
You are right about the second case. Friction opposes relative motion of the surfaces in contact. If it happened that a2>a1, block B would move forward with higher speed then block A. Then friction would retard B and would accelerate A, till they moved with the same velocity, and then the friction between them became static.