Recent content by virtuoso735

Sorry, the question is: relate the traction force F2 to the weight mg and the angle theta. I tried it again, hopefully this is closer. Horizontal component: -F1sinθ + F2 = 0 Vertical component: F1cosθ - mg = 0 Substitute F1 and F2 with mg and sum them, which leaves: -mgsinθ + F2 +...

Sorry, the question is: relate the traction force F2 to the weight mg and the angle theta. I tried it again, hopefully this is closer. Horizontal component: -F1sinθ + F2 = 0 Vertical component: F1cosθ - mg = 0 Substitute F1 and F2 with mg and sum them, which leaves: -mgsinθ + F2 + mgcosθ -...

Does that work out to F3 = 2mgcosθ ? Here is another similar problem, which I used the same process on: [PLAIN]http://img708.imageshack.us/img708/5553/pulley2q.jpg There are horizontal and vertical components to this problem as well. For the horizontal component I get: -F1cosθ + F2 = 0...

Sorry, I'm not understanding what you mean. Which was the right equation I had originally? Should I not have added the two cosine terms together? So is -mgcosθ - mgcosθ + F3 - mg = 0 any good?

Is this right? I replace the F1 and F2 with T. So I get -Tcosθ - Tcosθ + F3 - mg = 0. -2Tcosθ + F3 = mg Then I replace the T with mg, so -2mgcosθ + F3 = mg Is this the relationship?

Homework Statement http://img849.imageshack.us/img849/3716/staticeqpulley.jpg Hopefully the image showed up. If not, here's the link: [PLAIN]http://img707.imageshack.us/img707/3716/staticeqpulley.jpg" Sorry for the low quality pictures. Hopefully it will do though. The question is: Relate...