A) \[\mu \,mg\]
B) \[\mu \,\left[ mg-\left( \frac{\sqrt{3}}{2} \right)F \right]\]
C) \[\mu \,\left[ mg-\left( \frac{F}{2} \right) \right]\]
D) \[\mu \,\left[ mg+\left( \frac{F}{2} \right) \right]\]
Correct Answer: C
Solution :
\[R+F\sin 30=mg\] So, \[R=mg-\frac{F}{2}\] Force of limiting friction\[=P=\mu R\] \[=\mu \left( mg-\frac{F}{2} \right)\]
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