Problem :  A wedge of mass M with a smooth quarter circular plane, is kept on a rough horizontal surface. A particle of mass m is released from rest from the top of the wedge as shown in the figure. When the particle slides along the quarter circular plane, it exerts a force on the wedge. If the wedge begins to slide when the particle exerts a maximum horizontal force on it, find the coefficient of friction between the wedge & the horizontal surface.

Solution :Â

In order to find the coefficient of friction m at the time of maximum horizontal thrust (Fx)max exerted on the wedge by the particle given by the formula
μ =  fmax /N  . . . (i)
Where fmax = maximum static friction between the wedge & ground.
(limiting friction ) that must be equal to the maximum horizontal force
(Fx)Âmax for prevalence of sliding of the wedge & N = normal force offered by the horizontal surface on the wedge.
⇒ (F)max & N can be calculated as follows. Let the particle attain a speed v at the angular position θ as shown in the free body diagram. Since the particle accelerates towards the centre O with an acceleration a = v2/r , the force exerted on it must be radially inwards.
$ \displaystyle R – mg sin\theta = \frac{m v^2}{r} $
$ \displaystyle R = mg sin\theta + \frac{m v^2}{r} $   . . . (ii)
Conserving energy of the particle between position 1 & 2 we obtain
(ΔKE)1→2 = (ΔPE) 1→2
$ \displaystyle \frac{1}{2}m v^2 = m g h $
$ \displaystyle \frac{1}{2}m v^2 = m g (r sin\theta) $
$ \displaystyle v = \sqrt{2 g r sin\theta} $ Â . . . (iii)
Elimination of v between (ii) & (iii) yields
R = 3 mg sin θ           . .. (iv)
The horizontal force acting on the wedge = Fx = R cos θ
Fx =(3 mg sin θ) cos θ
Fx = 3 m g sin θ cos θ
$ \displaystyle F_x = \frac{3}{2}mg sin2\theta $ Â .. . (v)
For Fx to be maximum sin 2θ = 1
⇒ θ = 450 Putting θ in equation (v) horizontally
we obtain $ \displaystyle F_{x(max)} = \frac{3}{2}mg $ Â Â Â Â . . . (vi)
Resolving forces acting on the wedge for its equilibrium along horizontal & vertical we obtain,
fmax – (Fx)max = Ma = 0
fmax = (Fx)max
fmax = (Fx)max =Â mg
N – Mg –R sin θ = May = 0
N = Mg + R sin θ
N = Mg + (3 mg sin θ ) sin θ
$ \displaystyle N = (M + \frac{3}{2}m) g $ (since θ = π/4)
Putting the values of fmax & N in equation (i) we obtain
$ \displaystyle \mu = \frac{3 m}{2M + 3m} $