A) 20 N
B) 10 N
C) 12 N
D) 15 N
Correct Answer: A
Solution :
Free body diagram (FBD) of the block (shown by a dot) is shown in figure. For vertical equilibrium of the block \[N=mg+F\sin {{60}^{o}}=\sqrt{3}g+F\times \frac{\sqrt{3}}{2}\] ...(1) For no motion, force of friction \[f\ge F\cos {{60}^{o}}\] Or \[\mu N\ge F\cos {{60}^{o}}\] Or \[\frac{1}{2\sqrt{3}}\left( \sqrt{3}g+F\frac{\sqrt{3}}{2} \right)\ge \frac{F}{2}\] Or \[g\ge \frac{F}{2}\]or\[F\le 2g\]or\[20\,N\]You need to login to perform this action.
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