The question:
In a different arrangement, the string is cut so that the lengths of the two parts are $0.5m$ and $2.3m$. Describe how the block hangs in equilibrium in this case and state the tensions in the two strings.
My attempt:
I used $a^2 = b^2 + c^2 - 2bc\cos{A}$ with $A =\alpha$: $$2.3^2 = 2^2 + 0.5^2 - 2(2)(0.5)\cos{\alpha}$$ $$\cos{\alpha} = \frac{2.3^2-(2^2+0.5^2)}{-2(2)(0.5)}= \frac{1.04}{-2} = -0.52$$ $$\alpha = \arccos{(-0.52)} = 121.33^\circ$$
However the answer to this question states:
Tension in AC is 50 N (it takes all the weight)
Tension in BC is zero (it is slack)
Which implies $\alpha = 90^\circ$ as Tension in $AC$ only has a vertical component
What am i doing wrong?
Since the cosine of $\alpha$ is negative, $\alpha$ is an obtuse angle, so the diagram's a bit of a con: to have both strings taught, $AC$ will be beyond the vertical, further to the left. But that doesn't make physical sense: $BC$ can't provide a force to hold the block beyond the vertical line through $A$ because it's a string. Therefore BC is actually slack, and $AC$ is vertical.