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 α = 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 T1 to T2 in the belt is equal to __________. (Rounded off to two decimal places)

Take π = 3.14.

The correct answer is 4.11.

Step-by-Step Explanation:

1. Identify the parameters from the problem description and the provided diagram:
As shown in the diagram:
- The coefficient of friction between the belt and the pulley is: μ = 0.3
- The belt wrap angle is: α = 270° = 3π / 2 radians
- The tension ratio formula is: T₁ / T₂ = e^μα

2. Convert the wrap angle �� to radians:
Using the approximation of π = 3.14 as specified:

$$\alpha =\frac{3\times 3.14}{2}=4.71\text{radians}$$

3. Calculate the ratio of the tensions using the belt friction equation:
Substitute the values of the friction coefficient (μ = 0.3) and the wrap angle (α = 4.71 radians) into the tension ratio equation:

$$\frac{T_1}{T_2}=e^{0.3\times 4.71}$$

$$\frac{T_1}{T_2}=e^{1.413}$$

Evaluating the exponential function:

$$e^{1.413}\approx 4.10825$$

4. Round off to two decimal places:
Rounding 4.10825 to two decimal places gives:

$$\frac{T_1}{T_2}\approx 4.11$$

Thus, the ratio of the tensions T₁ to T₂ in the belt is 4.11.