Answer to Question #222233 in Classical Mechanics for tonny419

Question

Two crossed belts on pulleys of diameters 3.6 m and 2.4 m connect two parallel shafts with centres 4.2 meters apart. The maximum tension in the belts is limited to 1200 N and friction between the belts and the pulley, μ = 0.26. The smaller pulley has a speed of 300 rev/min.

2.1. Find the power that can be transmitted. (8)

2.2. What would be transmitted if open belts were used. (8)

Answer 1

$d_1 = 2.4m;r_1=\frac{d_1}{2}=1.2m;n_1= 300rpm$

$d_2= 3.6m;r_1=\frac{d_2}{2}=1.8m;$

$l= 4.2;\mu =0.26$

$T_1= 1200$

$P= (T_1-T_2)v$

$v= \frac{\pi d_1n_1}{60}\approx37.7\frac{m}{s}$

$\text{by Euler's formula:}$

$\frac{T_1}{T_2}=e^{\mu\theta}$

$\theta - \text{pulley wrap angle in radians}$

$2.1 \text{ for crossed belts}$

$\theta = 2\pi-2\arccos(\frac{r_1+r_2}{l})=4.73$

$T_2 = \frac{T_1}{e^{\mu \theta}}=\frac{1200}{e^{0.26 *4.73}}=350.82$

$P= (T_1-T_2)v=(1200-350.82)*37.7=32014W=35kW$

$2.2 \text{ for open belts}$

$\theta = 2\pi-2\arccos(\frac{r_1-r_2}{l})=3.43$

$T_2' = \frac{T_1}{e^{\mu \theta}}=\frac{1200}{e^{0.26 *3.43}}\approx492$

$P'= (T_1-T'_2)v=(1200-492)*37.7=26691W=26.7kW$

$\text{Answer: power that can be transmitted:}$

2.1 P=35kW

2.2 P'=26.7kW