Here is what I think. If everyone stands exactly vertically and the force on the rope is applied exactly horizontally then it seems to be possible. But this is totally ideal. In reality, I don't think its possible.

If there is anything wrong in my understanding please explain it in the comment section.

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## Comments

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TopNewestHmm... I think everyone would slip (because there is no friction) and the side with the most weighted people will win because the force of the more weighted team will be more than the less weighted team

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Yes I agree with you.

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Lets say there are two teams. As friction is absent you can treat all the players on one side as one block and the players on the other side as one block. Let the total masses be ${\color{#3D99F6}m_1} < {\color{#20A900}m_2}$. Then the scenario will be : $\large {\color{#3D99F6}F_1} \leftarrow \quad \quad {\color{#20A900}F_2} \rightarrow$

$\large {\color{#3D99F6}m_1a} \leftarrow \quad \quad {\color{#20A900}m_2a} \rightarrow$

$\large \text{acceleration and displacement} \rightarrow$ As, initially and finally the string is horizontal and taut if one team moves by $x$ distance the other team will also move by $x$ distance.

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I also have a question : What if the teams had equal weight with no other factors disrupting their 'evenness', then will the rope be taut and no one falls down or another thing will happen? I believe the rope will be taut and no one will fall down

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Yes you are correct the rope will always be taut whatever the weight may be on both sides.

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I agree with Mohmmad Farhan

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Let's analyze what happens if friction is present.

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