# Newton's 3rd law in lami's theorem

I'm a high school physics student dealing with the following problem: Despite solving the problem correctly, I don't understand the diagram (b) supplied with the problem: These are my questions:

• Why are vectors $${ F }_{ RB }$$ and $${ F }_{ RC }$$ in opposite directions in diagram (b)? The textbook explains that it used Newton's 3rd law of motion, but I have no idea how.
• Newton's 3rd Law states that after the student applies a force on the rope, the rope would apply an equal but opposite force on the student. Where is that force in the diagram? Do we not consider it because it's acting on the student and not the rope?

The first diagram (a) shows 3 forces acting on 3 different objects: $$\vec{F}_P$$ is the force of the person pushing on the rope, $$\vec{F}_{BR}$$ is the force on the boulder from the rope, and $$\vec{F}_{CR}$$ is the force on the car from the rope. To see how these forces relate we look at the free-body-diagram for the rope (b).

Newton's 3rd law says $$\vec{F}_{RB} = -\vec{F}_{BR}$$. The force on the rope from the boulder is equal in magnitude to and in the opposite direction of the force on the boulder from the rope. Similarly, $$\vec{F}_{RC} = -\vec{F}_{CR}$$. For consistency, I would have labeled the force on the rope from the person as $$\vec{F}_{RP}$$. All three of these forces ($$\vec{F}_{RP}$$, $$\vec{F}_{RB}$$, $$\vec{F}_{RC}$$) are acting on the rope. That's why they appear on the free-body-diagram for the rope.

To solve the problem, you probably applied Newton's 2nd law to the rope: $$\vec{F}_\mathrm{net} = \vec{F}_{RP} + \vec{F}_{RB} + \vec{F}_{RC} = m a = 0.$$ You can add up the forces in diagram (b) because they all act on the same object, the rope.

The 3rd law companion to $$\vec{F}_{RP}$$ is $$\vec{F}_{PR}$$ the force on the person from the rope. You correctly identify that this force does not act on the rope. It would appear on the free-body-diagram for the person.

There's a tension in the rope that exerts a force in both directions.

Think about it this way: suppose you're participating in a tug-of-war with your friend. You pull him, and he pulls you. Although you're exerting a force on the rope, the rope is certainly also exerting a force on you (that's why you feel you are pulled forward). Similarly, your friend also feels a force pulling him forward. What's happening is that the tension in the rope is exerting a force in both directions.

You see something similar here. For the midpoint of the rope, it's being tugged 1) by the rock and 2) by the car, and it's being pushed by the student. Therefore there are three forces, all of them leading out of the midpoint. This explains the second diagram.

Again, the tension in the rope acts in both directions, which is why $$F_{RB}$$ and $$F_{BR}$$ can seem to act in opposite directions (but they are certainly of equal magnitude).