# Thread:

1. ## Another pulley problem/resolving

Hey, another question whereby i don't get how my method gets the wrong answer.... solution is herehttp://clip2net.com/s/VKcK, question is here:part 1 and question i need help with
tension in the string is 6g, where g=9.8, sin theta is 3/5.

I resolve tension from the left so that i get R= T(0.6) + T , because one tension is sloped (therefore needs verticall resolved) and one is already completely downward, right?

I don't understand why the mark scheme suggests that the vertically resolved downward compenent X2 is the force acting down, can someone please explain this to me?

2. Where did you get 0.6T?

If you want to resolve the tension on the left, you have a vertical component of $T\cos\alpha$ and a horizontal component of $T\sin\alpha$

Then, teh net force downwards is $T\cos\alpha + T$ while the net left force is $T\sin\alpha$. The net force on the pulley is then the vector sum of those two forces.
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