Since the weight W impend vertical motion in the down ward direction, the tension in the two sides of the pulley will be as shown in fig. Given date:

T_{1} = W, µ = 0.3, θ = 90° = π/2 rad

Using the relation of Ratio of belt tension, T_{1}/T_{2 }= e^{µ.θ}

W/T_{2} = e^{(0.3).(p/2) }= 1.6

W = 1.6 × T_{2 }.......(i)

Considering the equilibrium of block:

∑V = 0

R = 300N

∑H = 0 ......(ii)

T_{2 }= µR = 0.3 × 300 = 180N ......(iii)

Equating equation (i) and (iii), we get

W = 1.6 × 180

**W = 288N**