1. ## vector problem

Hi guys,
This problem has been bothering me for a while now, and seems to contradict what I had learned in physics, that is, that the tension in a strong is not related to the length of the string. It goes:

A clothesline is tied between two poles 6 m apart. The line is taut and has negligible sag. When a wet shirt with mass 0.6 kg is hung at the middle of the line, the midpoint is pulled down 0.1 m. Find the tension in each half of the clothesline.

Although technically not a calculus problem, this was on a test in my calculus III class over the unit on vectors and the geometry of space. Normally with these problems we are given the angle between the horizontal and the string, with that angle you can easily find the vertical and horizontal components of the tension vector and solve the problem. Here we are not given an angle. Here we can find are the lengths of the string and the displacement downward. I thought tension has nothing to do with the length of the string? So, is it even possible the do this problem without finding the angle. I suppose you could take $\displaystyle tan^{-1}\left(\frac{.1}{30}\right)$ but my teacher said this problem involves no trigonometry at all.

James

2. Originally Posted by james121515
Hi guys,
This problem has been bothering me for a while now, and seems to contradict what I had learned in physics, that is, that the tension in a strong is not related to the length of the string. It goes:

A clothesline is tied between two poles 6 m apart. The line is taut and has negligible sag. When a wet shirt with mass 0.6 kg is hung at the middle of the line, the midpoint is pulled down 0.1 m. Find the tension in each half of the clothesline.

Although technically not a calculus problem, this was on a test in my calculus III class over the unit on vectors and the geometry of space. Normally with these problems we are given the angle between the horizontal and the string, with that angle you can easily find the vertical and horizontal components of the tension vector and solve the problem. Here we are not given an angle. Here we can find are the lengths of the string and the displacement downward. I thought tension has nothing to do with the length of the string? So, is it even possible the do this problem without finding the angle. I suppose you could take $\displaystyle tan^{-1}\left(\frac{.1}{30}\right)$ but my teacher said this problem involves no trigonometry at all.

James
it may not involve trig, but it still uses side ratios with similar right triangles.

one of the two congruent triangles formed has vertical leg = 0.1 m and horizontal leg = 3 m

from Pythagoras ... hypotenuse = $\displaystyle \sqrt{9.01}$

let T = tension force in the rope (either side)

use $\displaystyle g = 10 \, m/s^2$

shirt weighs $\displaystyle 6 \, N$

force vectors have the same ratio ...

$\displaystyle \frac{T}{3} = \frac{\sqrt{9.01}}{0.1}$

$\displaystyle T \approx 90 \, N$

3. Dear james121515,

Of course this problem needs no trignometry. If you consider a cartesian axes system with the origin as the midpoint of the string, you could represent the two forces acting on both sides of the string as in vector notation,

$\displaystyle (\frac{3i+0.1j}{\sqrt(3^2+0.1^2)})\times{F}\mbox{ and }(\frac{-3i+0.1j}{\sqrt(3^2+0.1^2)})\times{F}$

Now the force due to the cloth could be represented by, $\displaystyle (-0.6\times{9.81})j$

Since the cloth is in equlibrium,

$\displaystyle (\frac{3i+0.1j}{\sqrt(3^2+0.1^2)})\times{F}+(\frac {-3i+0.1j}{\sqrt(3^2+0.1^2)})\times{F}+(-0.6\times{9.0}1)j=0$

By simplification you would get, F=88.25N (of course if youuse $\displaystyle g=10ms^{-2}$ you would get F~90N

Hope this helps.