Clutches

18.4 There are 6 friction interfaces, so N = 6.  Each interface has 
to provide 12,800/6 Nm = 2133 Nm.  For uniform wear, F = 79,000 N.  
For uniform pressure, F = 76.2 kN.

18.15 Tan(alpha) = (D - d)/2b = 0.18, so alpha = 10.2°.  For uniform 
wear, F = 896 N.  For uniform pressure, F = 895 N.

Band Brake

18.35  P1 = 975N, P2 = 300N, T = 54Nm.
Using Torque = J * Deceleration = J * (Change in omega)/(Delta time),
get braking time = 5.82 seconds.