# Tension when two connected blocks are pulled in opposite directions [duplicate]

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Suppose there exist two blocks A and B connected by a massless, inextensible cord. Both A and B have equal masses and are acted upon by equal forces in opposite directions. So, is the tension in the string the sum of the two forces, the resultant of the two forces or something else?

      <-F--  A _________ CORD__________B   --F->

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## marked as duplicate by John Rennie, Qmechanic♦Jun 15 '14 at 13:37

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Of course. If we neglect any kind of friction force on the floor then we can say that the tension in the string is equal to the sum of the force applied on both side. – Gamma Jun 15 '14 at 7:20
More on forces and factors of two: physics.stackexchange.com/q/41291/2451 and links therein. – Qmechanic Jun 15 '14 at 7:32

## 1 Answer

The tension in the string is $F$. Why? Consider the entire system as a whole. Two equal forces acting on either side, so the net force is zero and the center of mass of the system does not move. Now consider just the object A. There's a leftward force $F$ acting on the object. We now have two possibilities: A moves to the right, or A stays where it is (moving to the left is not an option because the string is inextensible).

Case 1: A stays where it is. This implies there's a force on A cancelling the leftward force $F$. Which means the force applied by the string must be $F$.

Case 2: A moves to the right. This implies the tension in the string will be more than the force you've applied. Since the string doesn't have any means to exert a force of its own on the objects, this can not be the case.

Bottom line: Tension in the string is $F$.

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