A braking system has its braking lever inclined at an angle of 30^{\circ} to the horizontal plane, as shown in Fig. 19.7. The mass and diameter of the brake drum are 218 kg and 0.54 m respectively.
At the instant the lever is pressed on the brake drum with a vertical force of 600 N, the drum is found to rotate at 2400 r.p.m. clockwise. The coefficient of friction between the brake shoe and the brake drum is 0.4. Assume that the lever and brake shoe are perfectly rigid and possess negligible weight. Find :
1. Braking torque, 2. Number of revolutions the drum will make before coming to rest from the instant of pressing the lever, and 3. Time taken for the drum to come to rest from the instant of pressing the lever.
Question : A braking system has its braking lever inclined at an angle ...
Given : m = 218 kg ; d = 0.54 m or r = 0.27 m ; P = 600 N ; N = 2400 r.p.m.;\mu = 0.4
1. Braking torque
Let R_{ N } = Normal force pressing the block to the brake drum, and
F_{t}= Tangential braking force.
The various forces acting on the braking system are shown in Fig. 19.8.
Taking moments about the fulcrum O,
600 \cos 30^{\circ}\times 1.2=R_{N}\times 0.4 or 623.5=0.4 R_{N}
\therefore R_{ N }= 623.5/0.4 = 1560 N
and F_{t}=\mu R_{N}=0.4\times 1560 = 624 N
We know that braking torque,
T_{B}=F_{t}\times r = 624 \times 0.27 = 168.5 N-m
2. Number of revolutions the drum will make before coming to rest
Let n = Required number of revolutions.
We know that kinetic energy of the brake drum
=\frac{m \cdot v^{2}}{2}=\frac{218}{2}\left(\frac{\pi d . N}{60}\right)^{2}=109\left(\frac{\pi \times 0.54 \times 2400}{60}\right)^{2} N – m
=502 \times 10^{3} N – m …(i)
and work done by the brake drum due to braking torque
=T_{ B } \times 2 \pi n=168.5 \times 2 \pi n=1060 n N – m …(ii)
Since the kinetic energy of the brake drum is used to overcome the work done due to braking torque, therefore equating equations (i) and (ii),
n=502 \times 10^{3} / 1060=474
3. Time taken for the drum to come to rest
We know that time taken for the drum to come to rest i.e. time required for 474 revolutions,
t=\frac{n}{N}=\frac{474}{2400}=0.2 \min =12 s
