# Thread: Finding the angles of a pulley using vector

1. ## Finding the angles of a pulley using vector

Find $\displaystyle \phi$ and $\displaystyle \theta$

2. ## Re: Finding the angles of a pulley using vector

Break the force at w2 into horizontal and vertical components. The horizontal component is $\displaystyle T_1cos(\theta)- T_2cos(\phi)= 0$ while the vertical force is $\displaystyle T_1 sin(\theta)+ T_2 sin(\phi)= w2$. In terms of the weights, $\displaystyle T_1= w_1$ and $\displaystyle T_2= w_3$. Solve the two equations $\displaystyle w_1cos(\theta)- w_3cos(\phi)= 0$ and $\displaystyle w_1 sin(\theta)+ w_3sin(\phi)= w_2$ for $\displaystyle \theta$ and $\displaystyle \phi$.

3. ## Re: Finding the angles of a pulley using vector

Originally Posted by HallsofIvy
Break the force at w2 into horizontal and vertical components. The horizontal component is $\displaystyle T_1cos(\theta)- T_2cos(\phi)= 0$ while the vertical force is $\displaystyle T_1 sin(\theta)+ T_2 sin(\phi)= w2$. In terms of the weights, $\displaystyle T_1= w_1$ and $\displaystyle T_2= w_3$. Solve the two equations $\displaystyle w_1cos(\theta)- w_3cos(\phi)= 0$ and $\displaystyle w_1 sin(\theta)+ w_3sin(\phi)= w_2$ for $\displaystyle \theta$ and $\displaystyle \phi$.
$\displaystyle \theta = sin^{-1}( \frac{w_1^2+w_2^2-w_3^2}{2w_1w_2})$
$\displaystyle \phi = sin^{-1}( \frac{w_2^2+w_3^2-w_1^2}{2w_2w_3})$