Force on 70kg Person Holding Rope onto Pully

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Homework Help Overview

The discussion revolves around a scenario involving two individuals connected by a rope over a pulley, with one person holding the rope and the other suspended. The problem examines the forces acting on each person, particularly in a static situation where both individuals are at rest.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss free body diagrams and the application of Newton's laws to analyze the forces acting on each person. There are attempts to clarify the relationship between tension and gravitational forces, as well as the implications of the system being static.

Discussion Status

There is ongoing exploration of the forces involved, with some participants questioning the assumptions about the system's behavior under static conditions. Guidance has been offered regarding the interpretation of tension and gravitational forces, but multiple interpretations of the problem are still being discussed.

Contextual Notes

Participants are navigating the complexities of the problem, including the distinction between static and dynamic scenarios, and the implications of each person's position relative to the pulley. There is a focus on understanding the forces exerted on the pulley and the individuals involved.

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If a person of mass 70 kg is holding onto a rope that is connected to a pully and is suspending a person of mass 60 kg on the other side, what are the forces acting on each person. Oh, and both persons are still (0 velocity).
 
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Assuming that the system is not accelerating, just draw some free body diagrams and use Newton's third law for tension.
 
Ja4Coltrane said:
Assuming that the system is not accelerating, just draw some free body diagrams and use Newton's third law for tension.
Thanks for posting. So the forces that act on person one are 600 N upward and 700 N downard; and for person two the forces are 700 N upward and 600 N downward. Does that sound right?
 
sorry but you made an error. The forces on each person have nothing to do with the other person. (this is because they are not moving, so it is just like hanging on a hook)

Think of tension as the reaction to gravity.
 
So then the forces for person one are 700 N upward and 700 N downward; and for person two 600 N upward and 600 N downward. I don't understand why they don't affect each other since they are both pulling down on the rope.
 
You are right. Your confusion is the same one that I used to have. You see, tension is not how much the rope is stretched, but rather, how hard the rope is pulling up. In fact, I suggest you think of it as some random other force instead of the confusing term tension. Does that clear it up a little?
 
Yes, thank you for simplifying things. The only thing I don't understand is, if person one (mass 70 kg) is pulling down on the rope with 700 N, shouldn't person two (mass 60 kg) be accelerated upward because of the net force of 100 N?
 
Of course, but you said that both persons are still. If this is not true tell me so that I can explain what you need to do.
 
Oh yes, both persons are still. But if one person is pulling with a force of 700 N, shouldn't that accelerate the person on the other side of the pully of mass 60 kg?
 
  • #10
Oh haha. I thought that you meant that both people are held still. Okay this is interesting, it's called the Atwood Machine (originaly used to measure the value of g).

So first, you need to forget that gravity exists and instead just think of a 600N force and a 700 N force acting on two objects. Now you know that since they are connected, they both have to accelerate at the exact same magnitude. Now because they are connected, for the purpose of finding acceleration, it is okay to think of the system as one mass (M1+M2 or 60kg +70kg). Now you said earlier that the net force on this system is 100 N, and you know from Newtons second law that a=f/m. So plug it in:
100N/130kg= 10/13m/s(sq). general formula is of course A=(Mg-mg)/(M+m)

Now you have acceleration, but you need tension. Think of the heavy person. you know that the net force on this person is Tension minus 700N. Does that make sense? Also you know that the persons net force must also equal the mass times the acceleration of that person (F=MA) You know acceleration and you know mass. Solve for T! T-700N=(70kg)(10/13m/s(sq)) I hope that made some sense!
 
  • #11
NO IM SORRY!

Not T-700
700-T

the person is falling! I'm so sorry for that confusion!
 
  • #12
Ja4Coltrane said:
NO IM SORRY!

Not T-700
700-T

the person is falling! I'm so sorry for that confusion!

you could also use T-600N
 
  • #13
Sorry, I don't think I was clear on what the problem is. Here it is: There are two people connected to a pully. One is holding the rope and touching the ground, the other is suspended in the air and hanging on to the pully. Both are perfectly still -- there is zero velocity.

Now, what I need to know is, what are the forces that each people are exerting on the pully (preferably in a FBD).
 
  • #14
okay, use the atwood machine information for finding tension on each person. That is the upward force on BOTH people. Now the downward force on both people is simply the gravity on each person. So for the 60 kg person, 600 N down and T up(Im too lazy to do the calculations for T) and for the 70 kg person, 700N down and T up.
 
  • #15
HEY!
I read your question again! the seventy kg person is on bottom so that means that there is no acceleration! Forget the atwood stuff, that's if it was accelerating.
600 up and 600 down for the suspended person and for the grounded, 600 N down and 600N up (because he is holding it, so not 700N)
 
  • #16
you could also say in your free body diagram that gravity acts down on the person with 700 Newtons, the tension acts up with 600, and the normal force from the ground is 100, but either way, the pulley pulls up with 600, and he pulls down on it with 600 (that's 700-100)
 
  • #17
Thanks again, Ja4Coltrane!
 

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