$R_1 = 72/2 = 36 ~ \text{cm}$

$R_2 = 36/2 = 18 ~ \text{cm}$

$c = 137 ~ \text{cm}$

**Part 1: Pulleys will rotate in the same direction**

$\cos \left( \dfrac{\beta}{2} \right) = \dfrac{R_1 - R_2}{c} = \dfrac{36 - 18}{137}$

$\dfrac{\beta}{2} = 82.45^\circ$

$\beta = 164.9^\circ$

$\alpha = 360^\circ - \beta = 360^\circ - 164.9^\circ$

$\alpha = 195.1^\circ$

Solve for S_{1}

$S_1 = \dfrac{2\pi R_1 \alpha}{360^\circ} = \dfrac{2\pi (36)(195.1^\circ)}{360^\circ}$

$S_1 = 122.58 ~ \text{cm}$

Solve for S_{2}

$S_2 = \dfrac{2\pi R_2 \beta}{360^\circ} = \dfrac{2\pi (18)(164.9^\circ)}{360^\circ}$

$S_2 = 51.80 ~ \text{cm}$

Solve for z by Pythagorean Theorem

$z^2 + (R_1 - R_2)^2 = c^2$

$z^2 + (36 - 18)^2 = 137^2$

$z = 135.81 ~ \text{cm}$

Length of belt, L

$L = S_1 + S_2 + 2z = 122.58 + 51.80 + 2(135.81)$

$L = 446 ~ \text{cm}$ *answer*

**Parts 2 and 3: Pulleys will rotate in opposite directions**

$\cos \left( \dfrac{\beta}{2} \right) = \dfrac{R_1 + R_2}{c} = \dfrac{36 + 18}{137}$

$\beta = 133.57^\circ$

$\alpha = 360^\circ - \beta = 360^\circ - 133.57^\circ$

$\alpha = 226.43^\circ$

Solve for S_{1}

$S_1 = \dfrac{2\pi R_1 \alpha}{360^\circ} = \dfrac{2\pi (36)(226.43^\circ)}{360^\circ}$

$S_1 = 142.27 ~ \text{cm}$

Solve for S_{2}

$S_2 = \dfrac{2\pi R_2 \alpha}{360^\circ} = \dfrac{2\pi (18)(226.43^\circ)}{360^\circ}$

$S_2 = 71.14 ~ \text{cm}$

Solve for z by Pythagorean Theorem

$z^2 + (R_1 + R_2)^2 = c^2$

$z^2 + (36 + 18)^2 = 137^2$

$z = 125.91 ~ \text{cm}$

Length of belt, L

$L = S_1 + S_2 + 2z = 142.27 + 71.14 + 2(125.91)$

$L = 465.23 ~ \text{cm}$ *answer for Part 2*

By Ratio and Proportion

$\dfrac{x}{R_1} = \dfrac{c}{R_1 + R_2}$

$\dfrac{x}{36} = \dfrac{137}{36 + 18}$

$x = 91.33 ~ \text{cm}$ *answer for Part 3*