FAcos30∘=μ(mg+FAsin30∘)
FBcos60∘=μ(mg−FBsin60∘)
FA(cos30∘−μsin30∘)=μmg
FB(cos60∘+μsin60∘)=μmg
Divide the equation,
we get,
⇒FBFA=μmg/(cos60∘+μsin60∘)μmg/(cos30∘−μsin30∘)=32

JEE Main 2014 — Physics Mechanics
A heavy box is to be dragged along a rough horizontal floor. To do so, the person A pushes it at an angle 30∘ from the horizontal and requires a minimum force FA, while the person B pulls the box at an angle 60∘ from the horizontal and needs minimum force FB. If the coefficient of friction between the box and the floor is 53, the ratio FBFA is
Held on 19 Apr 2014 · Verified 6 Jul 2026.
23
32
3
35
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