Question 183 of 210

A heavy box is to be dragged along a rough horizontal floor. To do so, the person $A$ pushes it at an angle $30^{\circ}$ from the horizontal and requires a minimum force ${F}_{A}$, while the person $B$ pulls the box at an angle $60^{\circ}$ from the horizontal and needs minimum force ${F}_{B}$. If the coefficient of friction between the box and the floor is $\dfrac{\sqrt{3}}{5}$, the ratio $\dfrac{{F}_{A}}{{F}_{B}}$ is

A$\sqrt{\dfrac{3}{2}}$B$\dfrac{2}{\sqrt{3}}$C$\sqrt{3}$D$\dfrac{5}{\sqrt{3}}$

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Answer: (B) $\dfrac{2}{\sqrt{3}}$

${F}_{A}\cos 30^{\circ}=\mu \left(mg+{F}_{A}\sin 30^{\circ}\right)$
${F}_{B}\cos 60^{\circ}=\mu \left(mg-{F}_{B}\sin 60^{\circ}\right)$
${F}_{A}\left(\cos 30^{\circ}-\mu \sin 30^{\circ}\right)=\mu mg$
${F}_{B}\left(\cos 60^{\circ}+\mu \sin 60^{\circ}\right)=\mu mg$
Divide the equation,
we get,
$\Rightarrow \dfrac{{F}_{A}}{{F}_{B}}=\dfrac{\mu mg/\left(\cos 30^{\circ}-\mu \sin 30^{\circ}\right)}{\mu mg/\left(\cos 60^{\circ}+\mu \sin 60^{\circ}\right)}=\dfrac{2}{\sqrt{3}}$