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 wedge of mass m, lying on a rough horizontal plane, is acted upon by a horizontal force F1 and another force F2, inclined at an angle θ to the vertical. The block is in equilibrium, then the minimum coefficient of friction between it and the surface is
PhysicsLaws of MotionJEE Main
Options:
  • A F2sinθ(mg+F2cosθ)
  • B (F1cosθ+F2)mg-F2sinθ
  • C (F1+F2sinθ)(mg+F2cosθ)
  • D (F1sinθ-F2)(mg-F2cosθ)
Solution:
2743 Upvotes Verified Answer
The correct answer is: (F1+F2sinθ)(mg+F2cosθ)

We know f=μN and N=mg+F2cosθ

Now, since wedge is in equilibrium,

(F1+F2sinθ) = μ(mg+F2cosθ)

μ=F1+F2sinθmg+F2cosθ

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.