# If two people pull on either side of the rope with a force of 65N, why is the tension in the rope 65N?

Jan 19, 2017

Apply Newton's third law. See below.

#### Explanation:

The rope experiences the same pulling force on both sides, and therefore it is in a state of static equilibrium (i.e. it is at rest consequently has a net force of zero).

A force diagram:

Where ${\vec{F}}_{1}$ and ${\vec{F}}_{2}$ are the pulling forces on the rope and $\vec{T}$ is the tension force. $R$ will represent the rope later.

Here we consider Newton's third law.

Newton's third law states that every force occurs as a member of an action/reaction pair of forces. You may know it as the familiar phrase, "every action has an equal and opposite reaction." Most importantly, two members of an action/reaction pair are equal in magnitude but opposite in direction.

${\vec{F}}_{A o n B} = - {\vec{F}}_{B o n A}$

In this situation, each person pulls on the rope with $65 N$ of force, so:

${\vec{F}}_{1 o n R} = 65 N$ and ${\vec{F}}_{2 o n R} = 65 N$.

This tells us that the pulling force exerted by each person on the rope is $65 N$, and consequently by NIII, the rope exerts an equal but opposite tension force of $65 N$ on each person.

This is an action reaction pair. Therefore:

${\vec{F}}_{R o n 2} = {\vec{F}}_{2 o n R} = {\vec{F}}_{1 o n R} = {\vec{F}}_{R o n 1} = 65 N$

The tension force in the rope, given by ${\vec{F}}_{R o n 1}$ and ${\vec{F}}_{R o n 2}$, is therefore $65 N$.