Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
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
PhysicsLaws of MotionJEE MainJEE Main 2014 (19 Apr Online)
Options:
  • A 32
  • B 23
  • C 3
  • D 53
Solution:
2014 Upvotes Verified Answer
The correct answer is: 23

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

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.