Question

The tensions on the tight side and slack side of the flat belt are 750N and 250N

respectively. If the speed of the belt is 2m/s and the coefficient of friction between the

belt and the pulley is 0.3, neglecting centrifugal tension,

Determine; (i) Power transmitted at this speed in kW.

(ii) Angle of lap in degrees

Answer 1

Power transmitted, P $\frac{T_1-T_2 \times v}{1000}kW$

$Power =\frac{750-250 \times v}{1000}kW$

$Power =1kW$

Angle of lap

$\frac{T_1}{T_2}=e^{\mu \theta}$

$\ln({\frac{T_1}{T_2}})= \mu \theta$

$1.0986 =0.3 \times \theta$

$\theta= \frac{1.0986}{0.3}=3.662$

$\theta =3.662 \times \frac{180}{\pi}=209.78^0$