F.B.D. of the block is shown in the diagram

Since block is at rest therefore
fr−mg=0...(1)
F−N=0...(2)
fr≤μN
In limiting case
fr=μN=μF…(3)
Using eq. (1) and (3)
∴μF=mg
⇒F=0.20.5×10=25N
JEE Main 2021 — Physics Mechanics
The coefficient of static friction between a wooden block of mass 0.5kg and a vertical rough wall is 0.2. The magnitude of the horizontal force that should be applied on the block to keep it adhere to the wall will be______N. ⌈g=10ms−2]
Held on 24 Feb 2021 · Verified 6 Jul 2026.
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