A heavy box is to be dragged along a rough horizontal floor. To do so, person A pushes it at an angle `30^(@)` from the horizontal and requires a minimum force `F_(A)`, while person B pulls the box at angle `60^(@)` from the horizontal and needs minimum force `F_(B)`. If the coefficient of friction between the box and the floor is `sqrt(3)/(5)` , the ratio is `(F_(A))/(F_(B))`

Options

A. `sqrt3`
B. `(5)/(sqrt3)`
C. `sqrt((3)/(2))`
D. `(2)/(sqrt3)`

Answer

D

Explanation

`F_(A)=(mug )/(sin theta - mucos theta)`
Similarly,
`F_(B)=(mu mg)/(sin theta+ mu cos theta)`
`:. (F_(A))/(F_(B))=((mu mb)/(sin theta- mu cos theta))/((mu mg)/(sin theta+ mu cos theta))=((mumg)/(sin60^(@)-(sqrt(3))/(5)cos60^(@)))/((mu mg)/(sin30^(@)+(sqrt(3))/(5) cos 30^(@)))=(2)/(sqrt(3))[mu=(sqrt(3))/(5)` given]