∑H = 0; F = P1 cosθ
∑V = 0; R = W – P1 sinθ
Also F = µR
µ(W – P sinθ) = P1 cosθ
or P1 = µ W / (cosθ + µsinθ) ...(i)
With reference to the free body diagram (Fig)when push is applied)
∑H = 0; F = P2cosθ
∑V = 0; R = W + P2sinθ
Also F = µR
µ(W + P2sin θ) = P2cosθ
P2 = µW/(cosθ – µsinθ)......(ii)
From expression (i) and (ii),
P1/ P2 = (cos θ – µsin θ )/ (cos θ + µsin θ)
24/30 = (cos 30°– µ sin 30°)/ (cos 30°+ µsin 30°)
= (0.866 – 0.5µ)/(0.866 + 0.5µ)
0.6928 + 0.4µ = 0.866-0.5u
On solving
µ = 0.192
Putting the value of µ in equation (i) we get the value of W
= 120.25N