Answer:
f = 628.32 lb
t = 2513.28 lb-inc
Explanation:
given data:
θ = 45°
outside radius = 6 inch
inside radius = 4 inch
coefficient of friction = 0.4
max pressure = 100 psi
a) determine force required for applying one pad
f = [tex] \frac{\theta }{360}* 2\pi *p_{max}*r_{i}(r_{o}-r_{i})[/tex]
f = [tex] \frac{45 }{360}* 2\pi *100*4(6-4)[/tex]
f = 628.32 lb
b) torque capacity (t)
t = [tex]\mu *f*r_{average}^{}[/tex]
t = 0.4 *628.32*5
torque = 1256.64 lb-inc
for both pad = 2 * 1256.64 =2513.28 lb-inc
