Will a Rope Break if Each End is Pulled with 500 N?

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Homework Help Overview

The discussion revolves around the mechanics of a rope subjected to forces from two individuals pulling on opposite ends, each exerting a force of 500 N. The question posed is whether the rope will break given its strength of 750 N.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the relationship between tension, applied forces, and net force, questioning how these concepts apply to the scenario. Some participants discuss the implications of equilibrium and the nature of forces acting on the rope.

Discussion Status

The discussion is active, with various interpretations being explored regarding the tension in the rope and the forces involved. Some participants offer insights into the mechanics of materials and static equilibrium, while others express confusion about the relationship between applied force and tension.

Contextual Notes

Participants are considering the assumptions related to static versus dynamic conditions and the definitions of tension and net force in the context of the problem. There is an acknowledgment of the rope's strength limit and the conditions under which it may break.

Coldie
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"A rope is strong enough to withstand a 750 N force without breaking. If two people pull on opposite ends of the rope, each with a force of 500 N, will it break? Explain."

Am I correct in assuming that the tension force will be 500N?
 
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No, the total force would be 1000N since you have one force pulling on the left, and one force pulling on the right, EACH with 500N. The rope would break.
 
UrbanXrisis said:
No, the total force would be 1000N since you have one force pulling on the left, and one force pulling on the right, EACH with 500N. The rope would break.
How does one explain that however, since the net force is still 0 N ?
 
Net force is zero, since
force from the left = -500N
force from the right = 500N

Fnet=Fl+Fr
=-500N+500N =0 N

All this is telling you is that there is constant velocity and zero acceleration.

However, total force of the two forces is 1000N, since it is the combined force.
 
I completely forgot about this thread. Sorry for the delay and seeming lack of gratitude on my part:p

Thanks Urban, your response helped, but I'm confused as to why, if you isolate one mass and show the forces like so, Fa - Ft = ma, the tension force will turn out to be 500N? Fa = 500N, the applied force by one team, and ma will be 0, since acceleration is zero. Therefore, the tension force would seem to be the same as the applied force?...
 
imo tension force would be same as the applied force...

From what I've learned Fapp-Ff= ma ( where ff= frictional force )

Hence Fapp = Ff

Imagine such a diagram where the arrows show the direction in which the people are pulling... they do not coincied with other
<<<<<<.>>>>>>>

i would use Fapp-Ff=fnet if it were like this... eg. if someone is pulling a box and theere is frictional force against it... the (.) is the point of origin

.>>>>>>>>><<<<<<<<

dunno if my explanation sounded stupid.. hopefully it helped.. LOL... Btw you using the Glencoe Physics textbook coz I've seen the problem in mine..
 
If you think about the problem from an equilibrium point of view it is a typical mechanics of material exercise ... the "net force" == 0 applies to the system, but at a section of the rope stresses equal to 500 N / Area arise. So I'd take this as a static strength of materials problem rather than a dynamics problem.
 

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