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. $(\frac{3}{2\mu })mg$
• B. $(\frac{5}{2\mu })mg$
• C. $(\frac{5}{2})\mu mg$
• D. $(\frac{3}{2})\mu mg$

Answer 1

The correct option is B $(\frac{5}{2\mu })mg$

$\text{Step 1: Drawing free body diagram}$ $\text{[Ref. Fig.]}$

$f\to$ Frictional force between $B$ and $C$.

$N_1\to$ Normal contact force between $A$ and $B$

$N_2\to$ Normal contact force between $B$ and $C$

$\text{Step 2: Applying Newton's second law}$

$\text{On all three blocks Combined:}$

From figure, we see that all three blocks will move with same acceleration(say $a$).

Therefore, Applying Newton's second law on the complete system(Blocks A, B & C combined)

(Positive Rightwards)

$\sum F_x=ma$

$F=(2m+m+2m)a$

$=5ma$

$\therefore$ Common Acceleration of the blocks, $a=\frac{F}{5m}$

$\text{On block}C\text{:}$

$N_2=2ma$

$=2m\frac{F}{5m}=\frac{2F}{5}$ $....\left(1\right)$

$\text{Step 3: Condition of Minimum force}$

To prevent block $B$ from slipping downward

$f\ge m_{B}g$ $....\left(2\right)$

For minimum value of F, Maximum frictional force should act

So, $f_{max}=\mu N_2$

$=\frac{2}{5}\mu F$ $....\left(3\right)$ [Using Equation $\left(1\right)$]

$\text{Step 4: Equation solving}$

Using equation $\left(2\right)$ and $\left(3\right)$

$\frac{2}{5}\mu F\ge mg$

$\Rightarrow F\ge \frac{5}{2\mu }mg$

Hence minimum value of $F$ to prevent block $B$ from downward slipping is $\frac{5}{2\mu }mg$

Henc Option B is correct