Question

The system is pushed by a force F as shown in fig. All surfaces are smooth except between B and C. Friction coefficient between B and C is \[\mu \] . Minimum value of F to prevent block B from down ward slipping is:

\[
  A.{\text{ }}\left( {\dfrac{3}{{2\mu }}} \right)mg \\
  B.{\text{ }}\left( {\dfrac{5}{{2\mu }}} \right)mg \\
  C.{\text{ }}\left( {\dfrac{5}{2}} \right)\mu mg \\
  D.{\text{ }}\left( {\dfrac{3}{2}} \right)\mu mg \\
 \]

Answer 1

Hint- In order to deal with this question we have to keep in mind that B will not slide down if frictional force is more than the weight of block B, so first we will find the normal forces by using the simple formula as product of mass and acceleration.

Formula used- $a = \dfrac{F}{M},N = M \times a,W = mg,{\text{frictional force}} = {\text{frictional coefficient}} \times {\text{normal force}}$.

Complete step-by-step solution -
Given that there are three blocks having masses as 2m, m and 2m respectively.
Then horizontal acceleration of the system is
$
  \because a = \dfrac{F}{M} \\
   \Rightarrow a = \dfrac{F}{{2m + m + 2m}} \\
   \Rightarrow a = \dfrac{F}{{5m}} \\
 $
Now, we will calculate the normal force between B and C which will be calculated as
$
  \because N = M \times a \\
   \Rightarrow N = 2m \times \dfrac{F}{{5m}} \\
   \Rightarrow N = \dfrac{{2F}}{5} \\
 $
As we know that the frictional force is given as the product of frictional coefficient and the normal force.
As for this case as we know the frictional coefficient and the normal force so the frictional force is given as:
\[
  {\text{frictional force}} = {\text{frictional coefficient}} \times {\text{normal force}} \\
   \Rightarrow {F_f} = \mu \times N \\
   \Rightarrow {F_f} = \mu \times \dfrac{{2F}}{5} \\
 \]
From the figure it has been cleared that B will not slide down if frictional force is more than the weight mg of block B.
So, minimum value of F to prevent block B from down ward slipping is given by the inequality:
$
  {F_f} \geqslant W \\
   \Rightarrow \mu \dfrac{{2F}}{5} \geqslant mg \\
   \Rightarrow F \geqslant \dfrac{5}{{2\mu }}mg \\
 $
Hence, the minimum value of F to prevent block B from down ward slipping is $\dfrac{5}{{2\mu }}mg$
So, the correct answer is option B.

Note- Frictional Force refers to the force generated by two surfaces that contact and slide against each other. These forces are mainly influenced by the structure of the surface and the amount of force which needs them together. The object's angle and location affect the amount of frictional force.