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## On what is described in your

On what is described in your problem, it would look like this:

Upon closer understanding of the problem, it would look like this:

To get the tension of the strings AB and BC, we can consider the string AB as vector $U$, the string BC as a vector $V$ and the resultant force

would be the weight of the box (symbolized as vector $W$)

So...

$$U+V=W$$

Then we set the point B as the origin of the rectangular coordinate system, so the new equation would be:

$$[X_1, Y_1] + [X_2, Y_2] = [X_3, Y_3]$$ $$[r_1\cos 60^o, r_1 \sin 60^o]+[r_2\cos 60^o, r_2\sin 60^o]=[100\cos 270^o, 100\sin 270^o]$$ $$[r_1\cos 60^o, r_1 \sin 60^o]+[r_2\cos 60^o, r_2\sin 60^o]=[0, -100]$$

Notice that the box was hanged on the middle of the string AC, so the tension in the strings AB and BC would be the same, so $U = V $

$$U+V=W$$ $$V+V=W$$ $$2V=W$$ $$2[r_2\cos 60^o, r_2\sin 60^o]=[0, -100]$$ $$[2r_2\cos 60^o, 2r_2\sin 60^o]=[0, -100]$$

Then doing this now:

$$2r_2\cos 60^o = 0$$ $$2r_2\sin 60^o = -100$$

Now getting the $r_2$ in $2r_2\cos 60^o = 0$:

$$r_2 = 0$$

That's not right....

Now getting the $r_2$ in $2r_2\sin 60^o = -100$:

$$r_2 = 57.7$$

We now conclude that the tension in one of the strings would be $57.7$ pounds.

To elaborate...tension in string AB would be $57.7$ pounds and the tension in the string BC would be $57.7$ pounds too...

Alternate solutions are encouraged:-)