**Question:** A leather belt transmits 30 KW from a pulley 750 mm diameter which runs at 500 rpm. The belt is in contact with the pullry over an arc of 160° and coefficient of friction, μ = 0.3. If the permissible stress in the belt is not to exceed 2 MN / m^2, find the maximum cross sectional area of the belt?

**Ans:**

Given:

Power transmitted, P = 30 KW

Diameter of the pulley, d = 750 x 10^-3m

Speed of the pulley, N = 500 rpm

Angle of contact, Θ = 160° = 160 . π/180 = 2.7925 rad

Coefficient of friction, μ = 0.3

Permissible Stress, f_{t} = T_{max}/Area_{max} = 2Mn/m^{2}

Here, T_{max} = T1, tension on the tight side.

We have to find out the maximum cross-sectional area of the belt – Δ_{max}

We have, T1/T2 = e^{μΘ}, where T2 is tension on slack side.

= e^{0.3 x 2.7925} = 2.31118

or, T1 = T2 . 2.31118

Again, we have, Power transmitted, P = (T1-T2) . π . d . N /60.

Lets substitute the values of T1 and T2, thus solving for T1 and T2, we get –

(2.31118 . T2 – T2) . π . 750 . 10^{-3} . 500 / 600 = 30 . 10^{3}

i.e,

T2 = 30 . 10^{3} / 25.745 = 1165.277 N

T1 = T2 . 2.31118

= 2693.16 N = T_{max}

We have,

Permissible Stress, f_{t} = T_{max}/Area_{max} = 2 x 10^{6}

∴ Maximum cross sectional area,

Δ_{max} = T_{max} / f_{t} = 2693.16 / 2 x 10^{6} = 0.0013465 m^{2}