The tension of a rope in a tug of war with unequal forces is equal to the least force on the rope. Am I right? (If group A pulls the wire with 20 newton leftwards and the group B pulls with 30 newton. I expect that the tension is 20 newton.)
But the tension on the rope pulling a mass vertically upward is said to be equal to $mg+ma$.( where $m$ is the mass of the object being pulled, $a$ is the acceleration of the object)
How come the same situations bearing different answers?
Best Answer
Short answer.
In your example, the rope can not be massless (otherwise its acceleration would be infinite), but if it has mass, then tension is different on each point of the rope. Therefore you have to assume there is an object connected to it. Depending on which side of the rope you assume the object is, you get tension 20N or 30N.
This does not contradict the vertically lifted object by a rope. In that case the acceleration would be $a=\frac{T-w}{m}=\frac{T-mg}{m}$, that is $ma=T-mg$, that is $T=ma+mg$. What you are missing is that in this case $mg$ is the weight of the object, which does not affect the tug of war case (since its horizontal).
Long answer
Tug of war
If you pull the leftmost piece with 20N and rightmost with 30N, you would get 10N on a 0 mass object, meaning infinite acceleration. Therefore you have to assume there are two bodies (actually only one would be enough). So, one person pulling on each side of the rope.
Assuming the rope is massless, and is consisted of lots of tiny pieces, we can see that each tiny piece has two forces on it. $F_l$ from the left and $F_r$ from the right. Since acceleration of each tiny piece is $a=\frac{F_r-F_l}{m}$ and m->0 then $F_r$ = $F_l$, otherwise $a$ would be infinite.
Also, this means that any piece of the rope, including the ones that connect to the bodies, pull adjacent pieces with the same force. That is, body A and body B are both pulled by the rope with the same force $T$, which is the tension.
There is $F_A=20N$ on the left body and $F_B=30N$ on the right, plus $T$ and $-T$ respectively. Since bodies A, B and the rope have the same acceleration (if they didn't, they'd move apart from each other), we get:
$a=\frac{T-F_A}{m_A}=\frac{F_B-T}{m_B}$ , meaning $T=\frac{m_AF_B+m_BF_A}{m_A+m_B}$.
If $m_B$->0 then $F_B-T=0$, and $T=F_B=30N$. Likewise if $m_A$->0 then $T=F_A=20N$.
You can not assume that $m_B=m_A=0$ without $F_A=F_B$, therefore you need at least one non 0 mass object attached to the rope.
Upwards pulling
If $m_B=0$ then you get the same result as the vertically pulled body, excluding the weight $w$ of the object ($w=mg$), because when pulling horizontally its weight has no effect.