Question

Two blocks A and B of masses ${{m}_{A}}=1kg$ and ${{m}_{B}}=3kg$ are kept on the table as shown in figure. The coefficient of friction between A and B is 0.2 and between B and the surface of the table is also 0.2. The maximum force F that can be applied on B horizontally, so that the block A does not slide over the block is: (Take $g=10m{{s}^{-2}}$)

  A. 16N
  B. 40N
  C. 12N
  D. 8N

Answer 1

Hint: The acceleration of the system is calculated by μg. Then we calculate frictional force by μmg. Then we use the equation of motion to calculate force applied on the system. Equation of motion used in the solution is
$\Rightarrow F-\mu mg=ma$

Complete step by step answer:
You have been given the following data
$\mu =0.2$
${{m}_{A}}=1kg$
${{m}_{B}}=3kg$
Acceleration a = $\mu g=2m{{s}^{-2}}$
So you can calculate frictional force using the following formula:
Frictional force = $\mu mg=0.2\times 4\times 10=8N$
Let us assume F is the force applied on the system,
$\Rightarrow F-\mu mg=ma$
$F-8=4\times 2$
F=16N.
Hence the correct option is A.

Additional information:
The coefficient of friction demonstrates the relation between two objects and the normal reaction force between the objects. It is the ratio between frictional force and normal reaction force that gives us a coefficient of friction. There are two types of coefficient of friction. They are the coefficient of static friction and the coefficient of Dynamic friction. Frictional force between two objects when both the objects are at rest is indicated by coefficient of static friction. Frictional force between two objects when either one body or both the bodies are in motion is indicated by coefficient of dynamic friction.

Note:
The coefficient of friction is dimensionless as it is the ratio between two forces. Normal reaction force is always opposite to the direction in which the mass of the system acts. Object on a rough surface is set in motion only if the force applied on it overcomes the frictional force between object and the surface.