# What is the formula for tension??

Aug 5, 2015

There is no explicit formula for tension; it is basically a reaction force that occurs on strings, ropes, etc. in the opposite direction when you apply a force in some direction. You kind of have to consider the context first. Let's take this as an example.

If the Free-Body Diagram is drawn as follows:

$W$ is the same as, let's say ${F}_{g}$, for the force due to gravity.

When ${F}_{g}$ acts on the person, it weighs down the string, and creates tension along it in both directions. This person is weighing down the string by ${5}^{o}$ from the horizontal.

Assuming static equilibrium, examining only the part of the string with the man on it (the exact center), and summing the forces in the y-direction (up = positive y, right = positive x):

$\sin \left({5}^{o}\right) = \frac{{T}_{y}}{{T}_{R}} = \frac{{T}_{y}}{{T}_{L}}$

$\implies {T}_{y} = {T}_{R} \sin \theta = {T}_{L} \sin \theta$

where ${T}_{y}$ is each individual upward contribution of the tension.

$\sum {F}_{y} = {T}_{y , \text{left") + T_(y,"right}} - {F}_{g}$

$= {T}_{L} \sin \theta + {T}_{R} \sin \theta - {F}_{g} = 2 T \sin \theta - {F}_{g}$

${F}_{g} = 2 T \sin \theta$
$T = {T}_{L} = {T}_{R} = {F}_{g} / \left(2 \sin \theta\right)$

So, if the person's mass was $60 k g$, then:

${F}_{g} = m g = \left(60 k g\right) \left(9.807 \frac{m}{{s}^{2}}\right) \approx 588.42 N$

Thus, to counter a downwards force of $588.42 N$ with only a ${5}^{o}$ sag, the tension along the string in each direction is:

$\textcolor{b l u e}{T} = \frac{588.42 N}{2 \sin \left({5}^{o}\right)} \textcolor{b l u e}{\approx 3375.7 N}$

Other more detailed examples can be found here. The ones with pulleys are the most difficult.