A heavy box is to be dragged along a rough horizontal floor. To do so, the person A pushes it at an angle 30…

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 35, the ratio FAFB is
  1. 32
  2. 23
  3. 3
  4. 53

Solution

FAcos30°=μmg+FAsin30°

FBcos60°=μmg-FBsin60°

FAcos30°-μsin30°=μmg

FBcos60°+μsin60°=μmg

Divide the equation,

we get,

⇒FAFB=μmg/cos30°-μsin30°μmg/cos60°+μsin60°=23

Asked in: JEE Main 2014 (19 Apr Online)

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