In a belt drive the mass of the belt is 1 kg/m and its speed is 6m/s. The drive transmits 9.6 KW of power. Determine the strength of the belt when the coefficient of friction is 0.25 and wrap angle is 220°.
a) 2629.78 N
b) 2404.73 N
c) 2003.09 N
d) 1829.78 N

Answer: a
Explanation: T_c = mv² = 1*6² = 36 N
\(\frac{T1}{T2}\)=e^μθ
\(\frac{T1}{T2}=e^{0.25*220*\frac{\pi}{180}}\) = 2.6102
P = (T1-T2) * V
9.6*10³ = (T1-T2) * 6
T1-T2 = 1600
T1 = 2593.78 N
T2 = 993.78 N
Strength = T1 + T_c = 2629.78 N