A block A of mass M is placed on horizontal table. Another block B of mass m is places on block A as shown in figure. Now F force is applied on B. The conditions are explained below.
1. If there is no friction between A and B and between A and table, then B will move with acceleration aB = 
. There is no frictional force to move A on the table, so A remain rest.
If L is the length of A, then B takes t time to fall from A. L = 
 = 
Â
t = 
.
2. Friction presents in between A and B only and the coefficient of static friction between A and B is 
. The limiting friction between A and B is f = mg. B will not slide on A till F < f i.e. F < mg.
This f also acts on A by B. So, both will move with acceleration a = aA = aB = 
.
If F > f i.e. F > mg, then A and B will move in same direction but with different acceleration. The net force on B is
F – f = (F – mg). The acceleration of B is aB = 
.
The net force on A is f = mg. The acceleration of A is aA = 
.
3. Find the time when B falls from A: The acceleration of B with respect to A is
aBA = aB – aA = () – () = 
.
If L is the length of A, then B takes t time to fall from A, then, L = 
 t = 
 = 
.
4. The coefficient of static friction between A and B is . Friction present in between block A and table. The coefficient of static friction between A and table is 
. The limiting friction between A and B is f = mg. The limiting friction between A and table is f/ = (M+m)g. Block A will move if f > f/.
The acceleration of A is 
 = 
 = 
.
The acceleration of B with respect to A is aBA = aB – . If L is the length of A, then B takes t time to fall from A.
L = 
Â
 t = 
.
If A remain rest on the table, then the friction between A and table is f not f/. [Because friction is a self-adjusted force and if the applied force is less than the limiting friction, then the value of static friction is equal to the applied force]
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