How large is the frictional torque about the axle when a force of 42 N is exerted in the brake pedal?

Question

In a hydraulic brake system, the area of the piston in the master cylinder is 7 cm^2, and that of the piston in the brake cylinder is 1.5 cm^2. The coefficient of friction between shoe and wheel drum is 0.45. If the wheel has a radius of 30 cm, how large is the frictional torque about the axle when a force of 42 N is exerted on the brake pedal? Answer in units of N-m.

Answer 1

Torque T= R x f (vector cross product)
or T=R f since f is perpendicular
R- Radius of the wheel drum
f - force of friction

f = uN

N - Normal force
u - coefficient of friction
T=R u N

N= F(A1/A2)
finally
T=u R F (A1/A2)=
T=0.45 x 0.30 x 42(7/1.5)=
T=26.5 m N (we definitely need power breaks! :-) )

I hope it helps