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JEE Main 2014 · 19 Apr · Shift 1, Physics question 3

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 FAF_A, while person B pulls the box at an angle 60° from the horizontal and needs minimum force FBF_B. If the coefficient of friction between the box and the floor is 35\dfrac{\sqrt3}5, the ratio FAFB\dfrac{F_A}{F_B} is :

Choose one option
Answer and worked solution

Official answer: D

Pushing at 30° below the horizontal: FAcos⁡30∘=μ(mg+FAsin⁡30∘)F_A\cos30^\circ=\mu(mg+F_A\sin30^\circ), so FA=μmgcos⁡30∘−μsin⁡30∘F_A=\dfrac{\mu mg}{\cos30^\circ-\mu\sin30^\circ}.

Pulling at 60° above: FB=μmgcos⁡60∘+μsin⁡60∘F_B=\dfrac{\mu mg}{\cos60^\circ+\mu\sin60^\circ}.

With μ=35\mu=\frac{\sqrt3}5:
FAFB=12+35⋅3232−35⋅12=0.80.43=23.\frac{F_A}{F_B}=\frac{\frac12+\frac{\sqrt3}5\cdot\frac{\sqrt3}2}{\frac{\sqrt3}2-\frac{\sqrt3}5\cdot\frac12}=\frac{0.8}{0.4\sqrt3}=\frac2{\sqrt3}.

Topics:FrictionFree body diagrams
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