# Tension Statics Problem

• Jan 31st 2010, 07:35 PM
jegues
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?
• Feb 1st 2010, 08:04 AM
Hello jegues
Quote:

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:
$\displaystyle \cos45^o=\sin45^o =\frac{1}{\sqrt{2}}$

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

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

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

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

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