A heavy box is to be dragged along a rough horizontal floor. To do so, person A pushes it at an angle \(30^\circ\) from the horizontal and requires a minimum force \(F_A\), while 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 \(\frac{\sqrt3}{5}\), the ratio \(\frac{F_A}{F_B}\) is :
If the coefficient of friction between the box and the floor is \(\frac{\sqrt3}{5}\), the ratio \(\frac{F_A}{F_B}\) is :
Show Hint
- \(\sqrt3\)
- \(\frac{5}{\sqrt3}\)
- \(\sqrt{\frac32}\)
- \(\frac{2}{\sqrt3}\)
The Correct Option is D
Solution and Explanation
Person A pushes (force directed downward at \(30^\circ\)): \(N = mg + F_A\sin30^\circ\) and \(F_A\cos30^\circ = \mu N\).
So \(F_A = \frac{\mu mg}{\cos30^\circ - \mu\sin30^\circ} = \frac{\mu mg}{\frac{\sqrt3}{2}-\frac{\sqrt3}{10}} = \frac{\mu mg}{\frac{2\sqrt3}{5}}\).
Person B pulls (force directed upward at \(60^\circ\)): \(N = mg - F_B\sin60^\circ\) and \(F_B\cos60^\circ = \mu N\).
So \(F_B = \frac{\mu mg}{\cos60^\circ + \mu\sin60^\circ} = \frac{\mu mg}{\frac12+\frac{3}{10}} = \frac{\mu mg}{\frac45}\).
\(\frac{F_A}{F_B} = \frac{4/5}{2\sqrt3/5} = \frac{2}{\sqrt3}\).
Answer: (4) \(\frac{2}{\sqrt3}\)
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