Finding the angles of a pulley using vector

**BabyMilo** · October 10th 2011, 11:25 AM

Find $\phi$ and $\theta$

**HallsofIvy** · October 11th 2011, 07:17 AM

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

**BabyMilo** · October 11th 2011, 09:41 AM

Originally Posted by HallsofIvy

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

The answer book gives me

$\theta = sin^{-1}( \frac{w_1^2+w_2^2-w_3^2}{2w_1w_2})$

and

$\phi = sin^{-1}( \frac{w_2^2+w_3^2-w_1^2}{2w_2w_3})$

any idea?

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