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

AI Thread Summary
The discussion centers on whether a rope will break when each end is pulled with a force of 500 N. It is clarified that while the net force on the system is zero, the total tension in the rope is 1000 N due to the combined forces from both ends. This exceeds the rope's breaking strength of 750 N, indicating that the rope will break. The conversation also touches on the distinction between net force and tension, emphasizing that tension can be equal to the applied force in static scenarios. Overall, the analysis concludes that the rope cannot withstand the combined forces applied.
Coldie
Messages
84
Reaction score
0
"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?
 
Last edited:
Physics news on Phys.org
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.
 
Back
Top