To calculate the ratios of the belt tensions in an open flat belt drive system, we can use the following formula derived from the belt tension equation:
\[ \frac{T_1}{T_2} = e^{\mu \theta} \]
where:
- \( T_1 \) = tension in the tight side of the belt
- \( T_2 \) = tension in the slack side of the belt
- \( \mu \) = coefficient of friction
- \( \theta \) = angle of contact in radians
Given data:
- \(\mu = 0.37\)
- Angle of contact for the driven pulley = 176° (we need to convert this to radians).
First, convert the angle of contact from degrees to radians:
\[ \theta = 176° \times \frac{\pi}{180} = \frac{176 \pi}{180} \approx 3.073 , \text{radians} \]
Now, substitute the values of \( \mu \) and \( \theta \) into the tension ratio formula:
\[ \frac{T_1}{T_2} = e^{0.37 \times 3.073} \]
Calculating \(0.37 \times 3.073\):
\[ 0.37 \times 3.073 \approx 1.137 \]
Now, applying the exponential function:
\[ \frac{T_1}{T_2} = e^{1.137} \approx 3.115 \]
Thus, the ratio of the belt tensions is approximately:
\[ \frac{T_1}{T_2} \approx 3.115 \]
Conclusion
The ratio of the belt tensions \( T_1 \) to \( T_2 \) is approximately 3.115.
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