Rope, tension and Newtons third law

AI Thread Summary
The discussion centers on understanding tension forces in the context of Newton's third law. A problem is presented where segment R exerts a force T on segment L, and participants explore the force exerted on segment R by segment L, denoted as FRL. The concept of massless ropes is highlighted, leading to confusion about incorporating gravitational force (g) into the equations. Participants clarify that according to Newton's third law, the forces are equal and opposite, and the mention of g may not be necessary for the solution. Ultimately, the conclusion is reached that the force can be expressed simply as |-T|, with the reference to g being a potential distraction.
DakE_FeatH
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This is more of a conceptual question, meant to help us understand how tension forces work.

Homework Statement



Assume that segment R exerts a force of magnitude T on segment L. What is the magnitude FRL of the force exerted on segment R by segment L?
Give your answer in terms of T and other constants such as g.

Homework Equations



Fnet = ma

The Attempt at a Solution



Fnet = FR on L + FL on R = 0 (I assume it's 0)

I also know that Fnet = ma, but I don't know if I should use any mass, since it's a rope, and in first year undergraduate class, I understood that ropes are massless.

The question says in terms of g and I know that FG=m*g (g = -9.81)

I don't know how to indroduce the g in the equations.
 

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Consider Newton's 3rd law. (It's way easier than you think.)
 
Ok, Newtons third law states that "every force occurs as one member of an action/reaction pair of forces". So the two objects have forces acting on them of the same magnitude but opposite directions.

I think I understand how the tension forces work horizontally, but I have no idea how to add the g in there. Is there also a normal force that is equal and opposite to g? If so, how? It's in the air so I don't understand where that would come from :(
 
Welcome to PF!

DakE_FeatH said:
Ok, Newtons third law states that "every force occurs as one member of an action/reaction pair of forces". So the two objects have forces acting on them of the same magnitude but opposite directions.

I think I understand how the tension forces work horizontally, but I have no idea how to add the g in there. Is there also a normal force that is equal and opposite to g? If so, how? It's in the air so I don't understand where that would come from :(

Hi DakE_FeatH! Welcome to PF! :smile:

What makes you think g is involved? :wink:
 


tiny-tim said:
What makes you think g is involved? :wink:

They say in the problem to give my answer in terms of T and "constants such as g"

So I can't just use Fnet = FR on L + FL on R = 0
 


DakE_FeatH said:
They say in the problem to give my answer in terms of T and "constants such as g"
I think you are overly constraining yourself. I'm sure they meant: "... in terms of T and constants such as g if you need them".

If A exerts a force on B, then B exerts an equal and opposite force on A. That's Newton's 3rd law. Case closed. :wink:
 
Ok I got it, it was |-T|, they just put in the g to mess me up. Thank you guys!
 
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