A pulley system uses a flat belt of c. s.a. 4x100 mm2 and density 11x100 kg/m3. The angle of lap is 165o on the smaller wheel. The coefficient of friction is 0.39. The maximum force allowed in the belt is 637 N. Calculate the power (kW) transmitted when the belt runs at 13 m/s and centrifugal force is included.

Note that the angle of lap is 165 degrees at the question above

So the power (kW) transmitted when the belt runs at 13 m/s and centrifugal force is included = 3.79782kW

Explanation:

Centrifugal force, Fc = pAv^2

Where p = density, A = c.s.a.  v = veelocity

= 11x100 kg/m3 X 4x100 mm2 X (13 m/s)^2

= 1100 kg/m3 X 400 mm2 x 10^-6 X 169m/s

= 74.36N

P v (f - Fc) (1-e^-μθ)

The θ = (165 degrees ÷ 180 degrees) X π = 2.878 rads  (π = 3.14)

So power p, = 10 (637 N - 74.36N) (1 - e^-0.39 x 2.878)

= 10 X 562.64 X 0.675

P = 3797.82 watts = 3.79782kW

answer; how far out of place which the object finished relative to the starting location;