How does tension work in a rope or cable?

[kkps]

Let's examine and interpret the equations in your drawing. You have

View attachment 316230
Equation 1: Ft - 10 = 1*a
Here Ft is the tension, 10 (N) is the weight of the mass on the left and 1*a is mass*acceleration

Equation 2: Ft - 20 = 1*(-a)

Here Ft is the tension, 20 (N) is the weight of the mass on the right and 1*(-a) $\dots~$ is what ? If the mass is 2 kg, the term should be 2*(-a) to be correct. Then one has
Ft - 10 (N) = a
Ft - 20 (N) = -2a
If the first equation is doubled and added it to the second, one gets
3 Ft - 40 (N) = 0 ⇒ Ft = 40/3 (N) = 13.3 (N) and that is the correct tension in the string connecting the masses. More generally, $$F_T=\frac{2m_1m_2}{m_1+m_2}g.

Thanks for pointing out the mistake
What I wanted to show by this example is that the following statement made by PeroK is wrong "B cannot be pulling on the string with 20N and the string pulling back with 15N. That violates Newton's third law."

In my FBD of B, there is 20N down and 13.3N up. This is similar to the question of the original poster - just that where the force comes from is different in the Atwood machine example vs the original poster's question

 

[berkeman]

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