# Thread: Tension Statics Problem

1. ## Tension Statics Problem

1. The problem statement, all variables and given/known data
See figure, "CraneProblem".

2. Relevant equations:

Sum Fx = 0,
Sum Fy = 0.

3. The attempt at a solution

See figure, "CraneAttempt". I said for the maximum weight to be held the ropes must be at maximum tension then used Fy: to compute m1 and m2. My solution causes problems in the equilibrium of the X-Components of the tensions. Any ideas?

2. Hello jegues
Originally Posted by jegues
1. The problem statement, all variables and given/known data
See figure, "CraneProblem".

2. Relevant equations:

Sum Fx = 0,
Sum Fy = 0.

3. The attempt at a solution

See figure, "CraneAttempt". I said for the maximum weight to be held the ropes must be at maximum tension then used Fy: to compute m1 and m2. My solution causes problems in the equilibrium of the X-Components of the tensions. Any ideas?
Your equations are fine. I think you then need to work out which cable has the greatest tension. So, using:
$\cos45^o=\sin45^o =\frac{1}{\sqrt{2}}$

$\cos30^o = \frac{\sqrt3}{2}$

$\sin30^o = \frac12$
we get, by eliminating $T_5$:
$T_3\cos45^o = T_4\cos30^o$
$\Rightarrow T_3 =T_4\sqrt{\frac32}$

$\Rightarrow T_3>T_4$
and
$T_5 = T_3\frac{1}{\sqrt2}$

$\Rightarrow T_3>T_5$
Similarly
$T_3 > T_1$ and $T_4 > T_2$
So $T_3$ is the greatest tension. The maximum value of the combined masses occurs, then, when $T_3 = 100,000$ N.

Substitute this value into the equations to work out $T_1$ and hence $M_1$. Then work out $T_4$; and hence $T_2$ and $M_2$.