question_answer

The system of three blocks as shown in figure is pushed by a force F. All surfaces are smooth except between B and C. Coefficient of friction between B and C is \[\mu \]. Minimum value of F to prevent block B from slipping is

A) \[\left( \frac{3}{2\mu } \right)mg\]

B) \[\left( \frac{5}{2\mu } \right)mg\]

C) \[\frac{3}{2}\mu mg\]    

D) \[\frac{5}{2}\mu mg\]

Correct Answer: B

Solution :

[b] : Total mass of the system \[=2m+m+2m=5m\] \[\therefore \]Horizontal acceleration of the system,\[a=\frac{F}{5m}\]..(i) Let N be the normal reaction between blocks B and C. From the free body diagram of C, as shown in figure. \[N=2ma=2m\times \frac{F}{5m}=\frac{2}{5}F\](Using (i))      ...(ii) Block B will not slide (downwards), if force of friction; \[f\ge {{m}_{B}}g,\]i.e.,\[\mu N\ge mg\] \[\mu \left( \frac{2}{5}F \right)\ge mg\] \[F\ge \frac{5}{2\mu }mg\] Hence,\[{{F}_{\min }}=\frac{5}{2\mu }mg\]