As shown in the figure, a block of mass
Question:

As shown in the figure, a block of mass $\sqrt{3} \mathrm{~kg}$ is kept on a horizontal rough surface of coefficient of friction $\frac{1}{3 \sqrt{3}}$. The critical force to be applied on the vertical surface as shown at an angle $60^{\circ}$ with horizontal such that it does not move, will be $3 \mathrm{x}$. The value of $\mathrm{x}$ will be

$\left[\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2} ; \sin 60^{\circ}=\frac{\sqrt{3}}{2} ; \cos 60^{\circ}=\frac{1}{2}\right]$ (round off to nearest integer)

Solution:

$\mathrm{N}=\mathrm{Mg}+\mathrm{F} \sin 60^{\circ}$

$N=\sqrt{3} g+\frac{F \sqrt{3}}{2}$

For No slipping

$\mathrm{F} \cos 60^{\circ}=$ Friction

$\frac{F}{2}=\mu N=\frac{1}{3 \sqrt{3}}\left(\sqrt{3} g+\frac{F \sqrt{3}}{2}\right)$

$\frac{F}{2}=\frac{g}{3}+\frac{F}{6}$

$\frac{F}{2}-\frac{F}{6}=\frac{g}{3}$

$\frac{6 F-2 F}{12}=\frac{g}{3}$

$4 \mathrm{~F}=4 \mathrm{~g}$

$\mathrm{~F}=10$

$\mathrm{~F}=3 \mathrm{x}$

$\mathrm{X}=\frac{\mathrm{F}}{3}=\frac{10}{3}=3.33$

$\mathrm{X}=3.33$

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