Answer to Question #182107 in Mechanical Engineering for Brian

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

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

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

$1.0986 =0.3 \times \theta$

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

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