Question

The braking system shown in the figure uses a belt to slow down a pulley rotating in the clockwise direction by the application of a force P. The belt wraps around the pulley over an angle \(\alpha = 270\) degrees. The coefficient of friction between the belt and the pulley is 0.3. The influence of centrifugal forces on the belt is negligible. During braking, the ratio of the tensions \(T_1\) to \(T_2\) in the belt is equal to ................ (Rounded off to two decimal places)
Take \(\pi = 3.14\).

Hint:

The most common mistake in belt friction problems is forgetting to convert the wrap angle from degrees to radians. Always ensure your angle \(\theta\) is in radians before using it in the exponent \(e^{\mu \theta}\).

Answer 1

Step 1: Understanding the Concept:
This problem involves the analysis of belt friction. When a belt is wrapped around a rotating pulley and is about to slip (or is slipping), the tensions in the tight side (\(T_1\)) and slack side (\(T_2\)) are related by the capstan equation. We need to identify the tight and slack sides and apply the formula.
Step 2: Key Formula or Approach:
The belt friction formula (capstan equation) is: \[ \frac{T_1}{T_2} = e^{\mu \theta} \] where:
- \(T_1\) is the tension on the tight side.
- \(T_2\) is the tension on the slack side.
- \(\mu\) is the coefficient of friction.
- \(\theta\) is the angle of wrap in radians.
Step 3: Detailed Calculation:
1. Identify Tight and Slack Sides: The pulley is rotating clockwise. The belt acts as a brake, resisting this motion. Therefore, the friction force on the pulley from the belt is counter-clockwise. This means the belt is being pulled more on the left side and less on the right side. Thus, \(T_1\) is the tight side tension, and \(T_2\) is the slack side tension. The formula \(\frac{T_1}{T_2}\) can be used directly.
2. Convert the Angle of Wrap to Radians:
The angle is given in degrees: \(\alpha = 270^\circ\).
To convert to radians, use the conversion factor \(\pi \text{ radians} = 180^\circ\). \[ \theta = 270^\circ \times \frac{\pi}{180^\circ} = \frac{3}{2}\pi \text{ radians} \] Using the given value \(\pi = 3.14\): \[ \theta = 1.5 \times 3.14 = 4.71 \text{ radians} \] 3. Apply the Belt Friction Formula:
Given:
- Coefficient of friction, \(\mu = 0.3\)
- Angle of wrap, \(\theta = 4.71\) radians
The ratio of tensions is: \[ \frac{T_1}{T_2} = e^{\mu \theta} = e^{(0.3 \times 4.71)} = e^{1.413} \] 4. Calculate the Final Value: \[ e^{1.413} \approx 4.1078 \] Rounding off to two decimal places, we get 4.11. Step 4: Final Answer:
The ratio of the tensions \(T_1\) to \(T_2\) is 4.11.
Step 5: Why This is Correct:
The solution correctly identifies the tight and slack sides based on the direction of rotation and the braking action. It correctly converts the wrap angle to radians before applying the standard belt friction formula. The calculation is straightforward and leads to the correct numerical result.