Hello jegues Originally Posted by

**jegues** **...**Are the tensions A and B suppose to be the same?

Yes. You have given the correct reason why:
tension should be constant over the smooth pulley.

...and the additional clue is in the question, which you quote:There's only one tension asked for, so it must be the same in both parts of the cable.

Now, as to your working:

- You have the angle at $\displaystyle A$ wrong. $\displaystyle \tan A = \tfrac25$, not $\displaystyle \tfrac 23$.

- Your $\displaystyle \Sigma f_x$ and $\displaystyle \Sigma f_y$ equations are OK apart from this (and, of course, you need to put the same tension in both parts of the cable). Then:

- You need to take moments to solve this problem. If you take moments about $\displaystyle C$, you'll get the tension immediately. Then use the $\displaystyle \Sigma f_x$ and $\displaystyle \Sigma f_y$ equations to find the reaction at $\displaystyle C$.

Here's the $\displaystyle \Sigma M$ equation:$\displaystyle T\times AC\sin\angle A+T\times BC\sin\angle B - 40\times 5.5=0$

Can you complete it now?

Grandad