Question

A V-belt is used to transmit 5 KW at 17 m/s, the included angle of the belt is 38°, the density of the belt is 1.25 g/cm^3, contact angle on smaller pulley is 150°, tension on the belt is adjusted by providing idlers. Find tension at tight and slack side of belt.

(a) 300 N, 5 N

(b) 400 N, 6 N

(c) 500 N, 7 N

(d) 600 N, 8 N

I have been asked this question in an online quiz.

My doubt is from Mechanical Power Transmission topic in chapter Mechanical Power Transmission of Farm Machinery

Answer 1

Correct choice is (a) 300 N, 5 N

Explanation: P = (T1 – T2) * v

5000 = (T1 – T2) * 17

(T1 – T2) = 294.117

\(\frac{T1}{T2}=e^{\frac{μθ}{sinβ}} \)

\(\frac{T1}{T2}\) = e^[(0.5*150*(\(\frac{π}{180}\))/(sin⁡19)]

\(\frac{T1}{T2}\)=55.73

T1 = 55.73 T2

55.73 T2 – T2 = 294.117

T2 = 5.38 N (slack side)

T1 = 299.82 N (tight side).