Question 12.6: A block brake with a short shoe is shown in Fig. 12.8. It is...

\text { Given }(p v)=2 \quad \mu=0.2 \quad p_{\max }=1 N / mm ^{2} .

Step I Brake shoe force
The following notations are used in this example:

n_{1}=\text { speed of rotation of cable drum }( rpm ) .

n_{2}=\text { speed of rotation of brake drum (rpm) } .

\left(M_{t}\right)_{1}=\text { torque on cable drum }( N – mm ) .

\left(M_{t}\right)_{2}=\text { torque on brake drum (N-mm) } .

It is also the velocity with which the mass is lowered

v_{2}=\text { peripheral velocity of brake drum }( m / s ) .

The brake drum rotates four times as fast as the cable drum.

n_{2}=4 n_{1}               (a).

Referring to Fig. 12.8(a),

\left(M_{t}\right)_{1}=(m g) \times 150=(500 \times 9.81) \times 150 N – mm                ( b ).

Referring to Fig.12.8(b),

\left(M_{t}\right)_{2}=\mu P \times 200=(0.2 P) \times 200 N – mm         (c)

\frac{2 \pi n_{1}\left(M_{t}\right)_{1}}{60 \times 10^{6}}=\frac{2 \pi n_{2}\left(M_{t}\right)_{2}}{60 \times 10^{6}} .

\therefore \quad n_{1}\left(M_{t}\right)_{1}=n_{2}\left(M_{t}\right)_{2} .

\therefore \quad n_{1}(500 \times 9.81) \times 150=4 n_{1}(0.2 P) \times 200 .

∴        P = 4598.44 N          (i)
Step II Area of friction lining

P=p_{\max .} \times \text { Area of lining } .

= 4598.44 mm²            (ii)
Step III Uniform velocity at which mass can be lowered

v_{2} p_{2}=2 \quad \therefore v_{2}=\frac{2}{p_{2}}=\frac{2}{1}=2 m / s             (d).

\omega_{2}=\frac{v_{2}}{R_{2}}=\frac{2}{0.2}=10 rad / s .

\text { and } \quad \omega_{1}=\frac{\omega_{2}}{4}=\frac{10}{4}=2.5 rad / s .

v_{1}=\omega_{1} R_{1}=2.5(0.15)=0.375 m / s .

\text { or } \quad v_{1}=0.375 \times 60=22.5 m / min             (iii).

When the coefficient of friction is constant, the rate of heat generated is proportional to the product pv. Therefore, at higher speeds the brake drum will be overheated.