Question

For a band brake is shown in figure below. The drum diameter is 50 cm. The coefficient of friction is 0.25, The drum rotates at 72 rpm, and F = 3 kN.


The braking torque for counter clockwise rotation of the drum is

Answer 1

$\text{Taking moment about fulcrum}$

$F\times 100+T_2\times 20=T_1\times 60......\left(i\right)$

$\frac{T_1}{T_2}=2.85.....\left(ii\right)$

$\text{Solving (i) and (ii)}$

$\text{we get}{T}_1=5.66kN$

$T_2=1.987kN$

$\text{Braking Torque}=(T_1-T_2)\times r$

$=0.92kN-m$