# Another pulley problem/resolving

• May 10th 2011, 09:42 AM
chartsy
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?
• May 11th 2011, 08:31 AM
Unknown008
Where did you get 0.6T?

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

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