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--- 183_notes:examples:statics [2016/03/18 15:57] – klinkos1
+++ 183_notes:examples:statics [2016/03/18 16:13] – klinkos1
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-{{ 183_projects:screen_shot_2016-03-18_at_11.15.44_am.png }} This picture clearly hates me
+{{ 183_projects:screen_shot_2016-03-18_at_11.15.44_am.png }} (This picture clearly hates me)
 If a sign were hung like the one above, what would the tension forces acting on both of the ropes?
 ==== Setup ====
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   *I have a drawing but I'm having a hard time adding pictures
 ==== Solution ====
+We assume the net force is zero, so
+$$F_{net} = \sum F = \sum F_{x} + \sum F_{y} = 0.$$
+To find the forces acting the the x direction, we have to decompose the tension of both ropes.
+For rope 1, $$T_{1,x} = T_{1}\sin{\alpha}$$
+For rope 2, $$T_{2,x} = T_{2}\sin{\beta}.$$
+One important note is since the force of tension of the two ropes point in opposite directions, their signs must be different. As convention, the force pointing to the left will be negative in these calculations.
+Now we know the net force in the x direction
+$$\sum F_{x} = T_{2}\sin{\beta} - T_{1}\sin{\alpha}=0.$$
+Next, we will use the same technique to decompose the forces in the y direction.
+For rope one, $$T_{1,y} = T_{1}\cos{\alpha}$$
+For rope 2, $$T_{2,y} = T_{2}\cos{\beta}.$$
+When finding the net force in the y direction, we cannot forget our assumption that gravity also works in the dimension in the opposite direction as our tension forces. The net force in the y direction is,
+$$\sum F_{y} = T_{1}\cos{\alpha}+T_{2}\cos{\beta}-Mg = 0.$$