Find the angle of contact on the small pulley of an open belt drive system. The pulley diameters are 6 in. and 9 in. and the center distance is 36 in.
Identify the parameters
Let's identify the given parameters:
- Small pulley diameter: $D_1 = 6$ in
- Large pulley diameter: $D_2 = 9$ in
- Center distance: $C = 36$ in
Calculate the angle of contact using the formula
For an open belt drive, the angle of contact on the smaller pulley is given by:
$\theta = 180° - 2\sin^{-1}\left(\frac{D_2-D_1}{2C}\right)$
Substituting the values:
$\theta = 180° - 2\sin^{-1}\left(\frac{9-6}{2 \times 36}\right)$
$\theta = 180° - 2\sin^{-1}\left(\frac{3}{72}\right)$
$\theta = 180° - 2\sin^{-1}(0.04167)$
$\theta = 180° - 2 \times 2.39°$
$\theta = 180° - 4.78°$
$\theta = 175.22°$
Convert to radians
Converting to radians:
$\theta = 175.22° \times \frac{\pi}{180°} = 3.058$ radians
The angle of contact on the small pulley for the open belt drive system is $\boxed{\theta = 175.22° \text{ or } 3.058 \text{ radians}}$
Find the belt length for the system of Problem 1.
Use the belt length formula for open belt drive
For an open belt drive, the belt length is given by:
$L = 2C + \frac{\pi}{2}(D_1 + D_2) + \frac{(D_2 - D_1)^2}{4C}$
Substituting the values:
$L = 2 \times 36 + \frac{\pi}{2}(6 + 9) + \frac{(9 - 6)^2}{4 \times 36}$
$L = 72 + \frac{\pi}{2} \times 15 + \frac{9}{144}$
$L = 72 + 7.5\pi + 0.0625$
$L = 72 + 23.562 + 0.0625$
$L = 95.625$ inches
The belt length for the open belt drive system is $\boxed{L = 95.625 \text{ inches}}$
Solve problems 1 and 2 for a cross-belt drive.
Calculate the angle of contact for cross-belt drive
For a cross-belt drive, the angle of contact on the smaller pulley is given by:
$\theta = 180° + 2\sin^{-1}\left(\frac{D_2-D_1}{2C}\right)$
Substituting the values:
$\theta = 180° + 2\sin^{-1}\left(\frac{9-6}{2 \times 36}\right)$
$\theta = 180° + 2\sin^{-1}\left(\frac{3}{72}\right)$
$\theta = 180° + 2\sin^{-1}(0.04167)$
$\theta = 180° + 2 \times 2.39°$
$\theta = 180° + 4.78°$
$\theta = 184.78°$
Converting to radians:
$\theta = 184.78° \times \frac{\pi}{180°} = 3.225$ radians
Calculate the belt length for cross-belt drive
For a cross-belt drive, the belt length is given by:
$L = 2C + \frac{\pi}{2}(D_1 + D_2) + \frac{(D_2 + D_1)^2}{4C}$
Note the difference in the formula compared to open belt drive (+ instead of - in the last term).
Substituting the values:
$L = 2 \times 36 + \frac{\pi}{2}(6 + 9) + \frac{(9 + 6)^2}{4 \times 36}$
$L = 72 + \frac{\pi}{2} \times 15 + \frac{225}{144}$
$L = 72 + 7.5\pi + 1.5625$
$L = 72 + 23.562 + 1.5625$
$L = 97.125$ inches
For a cross-belt drive:
- The angle of contact on the small pulley is $\boxed{\theta = 184.78° \text{ or } 3.225 \text{ radians}}$
- The belt length is $\boxed{L = 97.125 \text{ inches}}$
In Problem 1, the 6 in. diameter drive pulley rotates at 1000 rpm. If combined slip and creep amount to 3 percent, find the speed of the 9 in. diameter pulley.
Calculate the theoretical speed ratio
For belt drives, the theoretical speed ratio is inversely proportional to the diameters:
$\frac{n_1}{n_2} = \frac{D_2}{D_1}$
Where:
- $n_1$ = speed of driving pulley = 1000 rpm
- $D_1$ = diameter of driving pulley = 6 in
- $D_2$ = diameter of driven pulley = 9 in
- $n_2$ = speed of driven pulley (to be calculated)
Theoretically:
$n_2 = n_1 \times \frac{D_1}{D_2} = 1000 \times \frac{6}{9} = 1000 \times \frac{2}{3} = 666.67$ rpm
Account for slip and creep
With slip and creep of 3%, the actual speed will be 3% less than the theoretical speed:
Actual speed = Theoretical speed × (1 - slip percentage)
$n_2 = 666.67 \times (1 - 0.03) = 666.67 \times 0.97 = 646.67$ rpm
The speed of the 9 in. diameter pulley is $\boxed{n_2 = 646.67 \text{ rpm}}$
- Angle of contact on small pulley (open belt): $\boxed{\theta = 175.22° \text{ or } 3.058 \text{ radians}}$
- Belt length (open belt): $\boxed{L = 95.625 \text{ inches}}$
- For cross-belt drive:
- Angle of contact: $\boxed{\theta = 184.78° \text{ or } 3.225 \text{ radians}}$
- Belt length: $\boxed{L = 97.125 \text{ inches}}$
- Speed of 9 in. pulley: $\boxed{n_2 = 646.67 \text{ rpm}}$
Answered
