Question

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

• A. $\sqrt{3}$
• B. $\frac{5}{\sqrt{3}}$
• C. $\sqrt{\frac{3}{2}}$
• D. $\frac{2}{\sqrt{3}}$

Answer 1

Answer: (D)

$\frac{2}{\sqrt{3}}$

For First case, where the person pushes the box,

$N=F_{A}sin30+mg$ ............... (1)

$F_{A}cos30=f_r$

$F_{A}cos30=\mu N$

$F_{A}cos30=\mu (mg+F_{A}sin30)$

$F_A(cos30-\mu sin30)=\mu mg$

$F_A=\frac{\mu mg}{(cos30-\mu sin30)}$ ..............(2)

For the second case where person pulls the box,

$N=mg-F_{B}sin60$ .................(3)

$F_{B}cos60=f_r$

$F_{B}cos60=\mu N$

$F_{B}cos60=\mu (mg-F_{B}sin60)$

$F_B(cos60+\mu sin60)=\mu mg$

$F_B=\frac{\mu mg}{(cos60+\mu sin60)}$ ...............(4)

From (2) and (4), we can say that

$\frac{F_A}{F_B}=\frac{(cos60+\mu sin60)}{(cos30-\mu sin30)}$

On solving we will get,

$\frac{F_A}{F_B}=\frac{2}{\sqrt{3}}$