ProblemÂ : Â A wedge of mass M with a smooth quarter circular plane, is kept on a rough horizontal surface.Â A particle of mass m is released from rest from the top of the wedge as shown in the figure.Â When the particle slides along the quarter circular plane, it exerts a force on the wedge.Â 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 (F_{x})_{max} exerted on the wedge by the particle given by the formula

Î¼Â = Â f_{maxÂ }/NÂ Â . . . (i)

Where f_{max} = maximum static friction between the wedge & ground.

(limiting friction ) that must be equal to the maximum horizontal force

(F_{x})_{Âmax }for prevalence of sliding of the wedge & N = normal force offered by the horizontal surface on the wedge.

â‡’ (F)_{max} & N can be calculated as follows.Â Let the particle attain a speed v at the angular position Î¸ as shown in the free body diagram. Since the particle accelerates towards the centre O with an acceleration a = v^{2}/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} $ Â Â . . .Â (ii)

Conserving energy of the particle between position 1 & 2 we obtain

(Î”KE)_{1â†’}_{2} = (Î”PE)_{ 1â†’}_{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 Î¸ Â Â Â Â Â Â Â Â Â Â . .. (iv)

The horizontal force acting on the wedgeÂ = F_{x} = R cos Î¸

F_{x} =(3 mg sin Î¸) cos Î¸

F_{x} = 3 m g sin Î¸ cos Î¸

$ \displaystyle F_x = \frac{3}{2}mg sin2\theta $ Â .. . (v)

For F_{x} to be maximum sin 2Î¸ = 1

â‡’ Î¸Â = 45^{0}Â Putting Î¸ 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,

f_{max} â€“ (F_{x})_{max} = Ma = 0

f_{max} = (F_{x})_{max}

f_{max} = (F_{x})_{max} =Â mg

N â€“ Mg â€“R sin Î¸ = Ma_{y} = 0

N = Mg + R sin Î¸

N = Mg + (3 mg sin Î¸ ) sin Î¸

$ \displaystyle N = (M + \frac{3}{2}m) g $ (since Î¸ = Ï€/4)

Putting the values of f_{max} & N in equation (i) we obtain

$ \displaystyle \mu = \frac{3 m}{2M + 3m} $