Question

What is the formula for tension??

Answer 1

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:

![http://i.stack.imgur.com/]()

`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(5^o) = (T_y)/(T_R) = (T_y)/(T_L)`

`=> T_y = T_Rsintheta = T_Lsintheta`

where `T_y` is each individual upward contribution of the tension.

`sum F_y = T_(y,"left") + T_(y,"right") - F_g`

`= T_Lsintheta + T_Rsintheta - F_g = 2Tsintheta - F_g`

`F_g = 2Tsintheta`
`T = T_L = T_R = F_g/(2sintheta)`

So, if the person's mass was `60 kg`, then:

`F_g = mg = (60kg)(9.807m/(s^2)) ~~ 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:

`color(blue)(T) = (588.42 N)/(2sin(5^o)) color(blue)(~~ 3375.7 N)`

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