litelboi said:
It often helps to consider a special case variant of the problem. (This is one of the many advantages of working purely algebraically as far as possible. Never plug in numbers until you need to!)
Here's how you might have gone:
##T_1\cos(\alpha)+T_2\sin(\beta)=W##
##T_1\sin(\alpha)=T_2\cos(\beta)##
##T_1=T_2\cos(\beta)/\sin(\alpha)##
##T_2(\cos(\beta)/\sin(\alpha))\cos(\alpha)+T_2\sin(\beta)=W##
##T_2=\frac{W\sin(\alpha)}{\cos(\beta)\cos(\alpha)+\sin(\beta)\sin(\alpha)}##
##=\frac{W\sin(\alpha)}{\cos(\beta-\alpha)}##
To check, what would happen if you change 40° to 30°, so that the two tensions are in a straight line? There would be no way for the string to remain straight when a weight is hung on it.
Substituting the angle values 60° and 30°, nothing special happens, showing the above answer is wrong.
After correcting the error in the first line, you arrive at
##T_2=\frac{W\sin(\alpha)}{\cos(\beta+\alpha)}##.
Now plugging in those angles gives ##T_2\rightarrow\infty##.
PeroK said:
note that you should always include units in your answer
The instructions in post #1 say not to …
Steve4Physics said:
your answers should be rounded to 2 significant figures
… and to round the answers to whole numbers.