To calculate the ratio of tensions in the belt (\( T_1 \) and \( T_2 \)), we can use the formula derived from the belt friction theory, which relates the tension on the driving and driven sides of a belt drive to the angle of contact, coefficient of friction, and the centrifugal tension. The formula is:
\[ \frac{T_1}{T_2} = e^{\mu \theta} \]
Where:
- \( T_1 \) is the tension on the tight side of the belt.
- \( T_2 \) is the tension on the slack side of the belt.
- \( \mu \) is the coefficient of friction.
- \( \theta \) is the angle of contact in radians.
- \( e \) is the base of the natural logarithm (approximately 2.71828).
Given Data
- Coefficient of friction, \( \mu = 0.25 \)
- Angle of contact, \( \theta = 192° = \frac{192 \times \pi}{180} = \frac{32\pi}{30} \text{ radians} \approx 3.356 \text{ radians} \)
Calculation
Substituting the values into the formula:
\[ \frac{T_1}{T_2} = e^{0.25 \times 3.356} \]
Calculating \( 0.25 \times 3.356 \):
\[ 0.25 \times 3.356 \approx 0.839 \]
Now, calculating \( e^{0.839} \):
Using a calculator or table, we find:
\[ e^{0.839} \approx 2.315 \]
Conclusion
Thus, the ratio of tensions in the belt is:
\[ \frac{T_1}{T_2} \approx 2.315 \]
This means that the tension on the tight side of the belt (\( T_1 \)) is approximately 2.315 times the tension on the slack side of the belt (\( T_2 \)).
1 answer