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,\[{{F}_{\min }}=\frac{5}{2\mu }mg\]You need to login to perform this action.
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