Question

A flat belt of negligible mass and thickness drives an output pulley of 0.6 m diameter, rotating at 500 rpm without slip. The arc of contact is \(190^{\circ}\) for this pulley, while it is \(170^{\circ}\) for the smaller input pulley. The coefficient of friction between the belt and the pulleys is 0.30. The measured slack side tension is 250 N. Neglecting other losses, the power transmitted by the belt drive, in kW, is nearest to (take \(\pi = 3.14\))

• A. 4.8
• B. 5.6
• C. 6.5
• D. 7.3

Hint:

Identify which pulley has the smaller angle of contact, since that one governs belt slip.

Answer 1

The Correct Option is B

Step 1: Identify which pulley governs slipping.
The smaller pulley has the smaller angle of wrap, \(170^{\circ}\), so belt slip starts there first; this angle sets the tension ratio for the whole drive.
Convert to radians: \(\theta = 170 \times \dfrac{3.14}{180} = 2.971\ \text{rad}\).

Step 2: Find the ratio of tight side to slack side tension.
The belt friction relation gives \(\dfrac{T_1}{T_2} = e^{\mu\theta}\).
\(\mu\theta = 0.30 \times 2.971 = 0.8913\), so \(\dfrac{T_1}{T_2} = e^{0.8913} = 2.438\).
With \(T_2 = 250\ \text{N}\), \(T_1 = 2.438 \times 250 = 609.6\ \text{N}\).

Step 3: Find the belt speed at the output pulley.
\(v = \dfrac{\pi D N}{60} = \dfrac{3.14 \times 0.6 \times 500}{60} = 15.70\ \text{m/s}\).

Step 4: Find the power transmitted.
\(P = (T_1 - T_2)v = (609.6 - 250) \times 15.70 = 359.6 \times 15.70 = 5646\ \text{W} \approx 5.65\ \text{kW}\).

Step 5: Check the wrong options.
Using the larger \(190^{\circ}\) angle by mistake would push the answer up near 6.5 kW (option C), which ignores that the smaller pulley slips first.

Final Answer:
The power transmitted is nearest to 5.6 kW. \[ \boxed{P \approx 5.6\ \text{kW}} \]