question_answer

The system is pushed by a force \[F\] as shown in figure. 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 downward slipping is

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

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

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

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

Correct Answer: B

Solution :

[b] Horizontal acceleration of the system is \[a=\frac{F}{2m+m+2m}=\frac{F}{5m}\] Let N be the normal reaction between B and C. Free body diagram of C gives \[N=2ma=\frac{2}{5}F\] Now B will not slide downward if \[\mu N\ge {{m}_{B}}g\] Or \[\mu \left( \frac{2}{5}F \right)\ge mgorF\ge \frac{5}{2\mu }mg\] \[{{F}_{\min }}=\frac{5}{2\mu }mg\]