The maximun value of the applied force $F$ such that the block as shown in the arrangement does not move is (Acceleration due to gravity, $g=10 \mathrm{~ms}^{-2}$ )

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Aicharya · Accepted Answer

Applied force on the block on a horizontal surface, as shown in the figure, we have

As we know that, force of friction, $f=\mu R$ From the diagram, $\begin{aligned} & f=\mu\left(W+F \sin 60^{\circ}\right) \\ & F \cos 60^{\circ}=\frac{1}{2 \sqrt{3}}\left(10 \sqrt{3}+F \sin 60^{\circ}\right) \\ & \Rightarrow F \times \frac{1}{2}=\frac{1}{2 \sqrt{3}}\left(10 \sqrt{3}+F \times \frac{\sqrt{3}}{2}\right) \quad\left\{\begin{array}{l} \cos 60^{\circ}=\frac{1}{2} \\ \sin 60^{\circ}=\frac{\sqrt{3}}{2} \end{array}\right\} \\ & \Rightarrow \frac{F}{2}=\frac{1}{2 \sqrt{3}} \times \sqrt{3}\left(10+\frac{F}{2}\right) \\ & \Rightarrow \frac{F}{2}=\frac{1}{2} \left(10+\frac{F}{2}\right) \Rightarrow F=\frac{20+F}{2} \\ & \Rightarrow 2 F=20+F \Rightarrow F=20 \\ & \therefore \quad F=20 \mathrm{~N} \\ \end{aligned}$

The maximun value of the applied force \(F\) such that the block as shown in the arrangement does not move…

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