Applying Newton's 3rd Law to Tension in a Tug of War

[kara123]

Homework Statement

Two students are playing tug of war. Student A has a mass of 62 kg and is pulling with a force of 410 N to the left. Student B has a mass of 65 kg and is pulling with a force of 430 N to the right. Determine the tension in the rope between them.

a. 840 N
b. 610 N
c. 420 N
d. 360 N

Relevant Equations

T=ma

the only formula related to tension that is provided is T=ma I have found an equation on the internet that I believe would work but we have never been introduced to this equation so I'm unsure if I'm able to use it. i entered the given values and got an answer of 419.47 which is one of the answers provided but i can't even explain what i did. any suggestions is appreciated!
T=f1+(m1/m1+m2)(f2-f1)
=419.76 N

 

[jbriggs444]

Homework Statement:: Two students are playing tug of war. Student A has a mass of 62 kg and is pulling with a force of 410 N to the left. Student B has a mass of 65 kg and is pulling with a force of 430 N to the right. Determine the tension in the rope between them.

a. 840 N
b. 610 N
c. 420 N
d. 360 N
Relevant Equations:: T=ma

the only formula related to tension that is provided is T=ma I have found an equation on the internet that I believe would work but we have never been introduced to this equation so I'm unsure if I'm able to use it. i entered the given values and got an answer of 419.47 which is one of the answers provided but i can't even explain what i did. any suggestions is appreciated!
T=f1+(m1/m1+m2)(f2-f1)
=419.76 N

Unfortunately, this sounds a lot like the "pick an equation that fits the knowns you have and throw some values at it" approach to physics. I am not sure how you got from $T=ma$ to $T=f_1 + (\frac{m_1}{m_2}+m_2)(f_2-f_1)$

Edit: Possibly you decided that you needed to calculate acceleration $a$, so you divided net force by total mass. But since the only force numbers you had were rope tensions, and the only mass numbers you had were student masses, you used those and divided the net force on the rope by the total mass of the students. But that just gives you a nonsense figure. You divided the net force on one thing by the total mass of another.

Let's start with some easy questions:

What is the tension at the student A end of the rope?
What is the tension at the student B end of the rope?
[Groping for the Eureka moment where you realize that this is either a trick problem or a stupid problem]

What is the net force on the rope?

 

[kara123]

Unfortunately, this sounds a lot like the "pick an equation that fits the knowns you have and throw some values at it" approach to physics. I am not sure how you got from $T=ma$ to $T=f_1 + (\frac{m_1}{m_2}+m_2)(f_2-f_1)$

Edit: Possibly you decided that you needed to calculate acceleration $a$, so you divided net force by total mass. But since the only force numbers you had were rope tensions, you used those and divided the net force on the rope by the total mass of the students. But that just gives you a nonsense figure. You divided the net force on one thing by the total mass of another.

Let's start with some easy questions:

What is the tension at the student A end of the rope?
What is the tension at the student B end of the rope?
[Groping for the Eureka moment where you realize that this is either a trick problem or a stupid problem]

What is the net force on the rope?

i don't know how to find the tension because i don't have the acceleration, that was just an equation i found to find tension because i didn't see how i could use T=ma.

 

[jbriggs444]

i don't know how to find the tension because i don't have the acceleration, that was just an equation i found to find tension because i didn't see how i could use T=ma.

I have reconsidered the problem and have an interpretation in mind. It does not match what the author wrote, but may match what he meant:

Student A on the left digs in his feet and exerts a rightward force of 410 N on the ground. He masses 62 kg.
Student B on the right digs in his feet and exerts a leftward force of 430 N on the ground. He masses 65 kg.
The rope is negligibly massive.
No other external forces are relevant.

Now let us try to determine the tension in the rope. Your formula may well be correct.

Mumble. Net force. Total mass. Acceleration.
Net force on (pick a student): Single mass. Acceleration. Giving net force. Subtract known force. Retrieve tension.

Yep, you got it right.

The interpretation I originally had in mind was that Student A pulled on the rope with a force of 410 N and Student B pulled on the rope with a force of 430 N. The "tension in the rope" is then ill defined, but demands a massive rope. On the assumption that rope is uniform, the tension then varies linearly from one end to the other and is best approximated by a tension of 420 N in the middle.

The phraseology: "pulling with a force" led me to this original interpretation.
The fact that ropes are usually assumed to be massless leads to your interpretation.

 

[Steve4Physics]

Since I’d already drafted this before @jbriggs444 beat me to it, may I add...

The question is badly posed. I’ll repose it. (I guess I’m a poser.)

Two students, A and B, are playing tug-of-war. $m_A=62kg, m_B= 65kg$.

A---------B​

Student A pushes against the ground with a force of 410 N to the right. (This is equivalent to saying A pulls with a force of 410N to the left.)

Student B pushes against the ground with a force of 430 N to the left. (This is equivalent to saying B pulls with a force of 430N to the right.)

The tension in the massless rope is T.

Maybe these questions can now guide you:
a) what is the resultant force on the system (A+B+rope)?
b) what is the acceleration of the system?
c) what is the resultant force on A alone?
d) what is T?

 

[haruspex]

found an equation on the internet

Presumably the equation is actually $T=f_1+\frac{m_1}{m_1+m_2}(f_2-f_1)$. (Your casual omission of parentheses leads to the dimensionally incorrect form in post #2.)
I would have written it more symmetrically as $T=\frac{m_1f_2+m_2f_1}{m_1+m_2}$.
Even though this gave the right answer to the corrected question, it is a terrible idea to use equations found on the net without understanding how they are obtained. You learn nothing that way.
Please follow the guide @Steve4Physics lays out in post #5.

 

[kuruman]

Student A pushes against the ground with a force of 410 N to the right. (This is equivalent to saying A pulls with a force of 410N to the left.)

I think it is better to re-repose by using Newton's 3rd: The ground pushes student A to the left with a force of 410 N and student B to the right with a force of 430 N.

As you mention farther down, the system is the two students plus the rope. It might be misleading to imply that part of the system acts on itself.

 

[Steve4Physics]

I think it is better to re-repose by using Newton's 3rd: The ground pushes student A to the left with a force of 410 N and student B to the right with a force of 430 N.

As you mention farther down, the system is the two students plus the rope. It might be misleading to imply that part of the system acts on itself.

In fact I considered doing that. But in the end I decided that applying N3L is an important part of the solution-process. I wanted the OP to use N3L for themself (with prompting if needed) and work out that

"Student A pushes against the ground with a force of 410 N to the right"
means that
"The ground pushes the student with a force of 410 N to the left"