Question

The coefficient of static friction between the two blocks is $0.5$ and the table is smooth. The maximum horizontal force that can be applied to move the blocks together is _____$\text{N}$.
$(\text{Take}g=10{\text{ms}}^{-2})$

Answer 1

The free body diagram of block $A$ is as shown below:


The maximum frictional force that can act on block $A$ is given by,

$f_L={\mu }_{s}N={\mu }_{s}mg=\left(0.5\right)(1\times 10)=5\text{N}$

The maximum acceleration possible for the block $A$ is given by,

$a_{max}=\frac{f_L}{m}=\frac{5}{1}=5{\text{m/s}}^2$

Now, if the two blocks are to move together, the maximum possible accelaration of both the blocks should be $5{\text{m/s}}^2$

Under this scenario, there is no slipping between the two blocks, and they should move as a single system with acceleration $a=5{\text{m/s}}^2$

The free body diagram of the system is shown below:


Using Newton's second law, we have:

$F_{max}-f_L=m{a}_{max}$

$\therefore F=F_{max}=5+(3\times 5)=20\text{N}$