F.B.D. for A block

F.B.D. for B block

for block A

mg sin θ – f 1 = ma .........

for motion w.r.t. block B

mg sin θ – µmg cos θ = ma.........

for limiting case

a = 0

and a = b = 0

⇒ mg sin θ = µmg cos θ

µ = tan θ

θ = tan –1 µ

for block B

mg sin θ + f 1 – f 2 = mb

for motion w.r.t wedge

f 2 = 2µmg cos θ

mg sin θ + f 1 2µmg cos θ = mb..........

for no relative motion between A and B block from equation & : a = b

2mg sin θ – 2 µmg cos θ = 2ma

for limiting case a = 0

⇒ θ = tan –1 (µ)

for motion θ ≥ tan –1 (µ)

when block B is moving w.r.t wedge

mg sin θ + f 1 – 2µ mg cos θ = mb

But f 1 = µmg cos θ ⇒ mg sin θ – µmg cos θ = mb

for block A

mg sin θ – µmg cos θ = ma ⇒ a = b.