JEE Mains Model Paper 2

Length of rod AB = 1 m
N1 = normal reaction of the wall
N2 = normal reaction of the floor
The frictional force (f) between the rod and the floor acts along AD. The weight mg of the
rod acts at its midpoint (centre of mass) C so that AC = AB/2.
For translational equilibrium N2
= mg and N1
= f
Taking moments of forces about A, we have for rotational equilibrium
N2 × 0 - mg × AD + N1 × BE = 0
⇒ 0 - N2 × AD + f × BE = 0 (∴ mg = N2 , N1 = f)
⇒

f

N2

=

AD

BE

=

ACcos30°

ABsin30°

⇒ µ =

1

2

×

√3 3

1 2

=

√3

2

= 0.87