Why Does Tension Vary Along the Rope in a Hanging Superhero Scenario?

AI Thread Summary
In the scenario of a Superhero and Trusty Sidekick hanging from a rope, the tension varies due to their differing weights, with the Superhero weighing 90.0 kg (882 N) and the Sidekick 55.0 kg (539 N). The total tension in the rope above the Superhero is calculated as 1421 N, which is the sum of their weights. However, the tension between the Superhero and Sidekick is only 539 N, as it only needs to support the Sidekick's weight. This discrepancy arises because the rope can be conceptualized as two segments, with the tension being equal along each segment only under ideal conditions, such as in massless, frictionless scenarios. Understanding these principles clarifies how tension can differ at various points along a non-ideal rope system.
ThePiGeek314
Messages
10
Reaction score
0

Homework Statement


[/B]
Superhero and Trusty Sidekick hang motionless from a rope. Superhero’s mass is 90.0 kg, while Trusty Sidekick’s is 55.0 kg, and the mass of the rope is negligible. (a) Draw a free-body diagram of the situation. (b) Find the tension in the rope above Superhero. (c) Find the tension in the rope between Superhero and Trusty Sidekick.

Homework Equations



F = ma
W = ga

The Attempt at a Solution



In the free-body diagram, I have Superhero being pulled down by gravity = 882 N, and Trusty Sidekick being pulled down by gravity = 539 N. The rope is pulling 882 N up on Superhero and 539 N up on Trusty Sidekick (definition of equilibrium, since they're motionless). I'm pretty sure the diagram is correct.

The total tension on the rope, since the mass of the rope itself is negligible = 882 + 539 = 1421 N. Because a rope pulls with equal tension on each object, the tension in the rope above Superhero is just 1421 N - the total weight it's being pulled down by.

Tension in the rope between Superhero and Trusty Sidekick should also be 1421, because tension is equal along the whole rope...but the answer is 539 N. How are the different forces at different points along the rope consistent with the tension remaining equal throughout the whole rope?
 
Physics news on Phys.org
Hi Pi,
What if Sidekick let's go ? Would the tension in the lower part of the rope also be the same as in the top part ?

Can you post the free body diagram ?
 
515960-4-34PEEI1.png
 
ThePiGeek314 said:
That is not a free body diagram.
ThePiGeek314 said:
tension is equal along the whole rope
The fact that it is one continuous rope is confusing you. Think of it as two ropes with a join where Superhero is holding on.
Try to answer BvU's question. If Sidekick were not there, would the tension still be the same all along the rope?
 
No, because the only weight on the rope would be Superhero's weight of 882 N. Using Newton's Third Law, the tension in the rope above Superhero would also be 882 N, right?

Using common sense, the tension in the rope below where Superhero is hanging on should be 0...but isn't tension supposed to be the same all along a rope?
 
ThePiGeek314 said:
Nbut isn't tension supposed to be the same all along a rope?
Only if you refrain from pulling on the rope somewhere in the middle. If you've pulled on a real rope, you might notice that it is taut on one side of your hands and slack on the other.

In physics textbooks, one often uses massless, frictionless pulleys which exert zero longitudinal force along a rope. The ropes are also massless. The tension on a massless rope is indeed the same on either side of an ideal pulley. But we're not dealing with ideal pulleys here.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top