Question

A 2 kg brick is placed on an inclined plane of inclination 45°. The brick is at rest. The minimum co-efficient of static friction is

(1) 0.5 

(2) \(\sqrt3\) 

(3)  1

(4) \(\frac{1}{\sqrt3}\)

Answer 1

Correct Option is (3) 1 

N = mgcos45°

f_s = mgsin45°

\(\Rightarrow\) mgsin45°≤  \(\mu\)mgcos45°

\(\Rightarrow\) \(\mu\) ≥ 1. 

Answer 2

mgsinø = F

mgcosø = N

f= uN where u is coeff of friction 

as block is at rest F = f

hence, umgcosø= mgsinø

tanø= u

as given ø is 45°

therefore, u = 1