Finding the Mass of a Hanging Rope (Wave Problem)

Click For Summary
SUMMARY

The discussion focuses on calculating the mass of a hanging rope in a wave problem involving a 20.0 kg box suspended from an 80.0 m vertical rope. The wave pulse initiated by the geologist takes 1.26 seconds to travel to the top of the rope. Key equations include the wave speed formula v = √(F/μ) and the equilibrium condition ∑F = T - mBoxg - mRope = 0. The tension in the rope at any point y is expressed as Ty = mBoxg + μyg, where μ represents the mass per unit length of the rope.

PREREQUISITES
  • Understanding of wave mechanics and wave speed equations
  • Knowledge of forces in equilibrium and tension in ropes
  • Familiarity with mass linear density (μ) concepts
  • Ability to solve differential equations related to wave propagation
NEXT STEPS
  • Study the derivation of wave speed in different mediums
  • Learn about differential equations in physics, particularly in wave mechanics
  • Explore the concept of mass linear density and its applications in tension calculations
  • Investigate the relationship between tension and wave speed in strings and ropes
USEFUL FOR

Physics students, educators, and anyone interested in wave mechanics and tension analysis in ropes will benefit from this discussion.

Chansudesu
Messages
3
Reaction score
0

Homework Statement


"...you were presented with a geologist at the bottom of a mineshaft next to a box suspended from a vertical rope. The geologist sent signals to his colleague at the top by initiating a wave pulse at the bottom of the rope that would travel to the top of the rope. The mass of the box is 20.0 kg and the length of the rope is 80.0 m. If a wave pulse initiated by the geologist takes 1.26 s to travel up the rope to his colleague at the top, find the mass of the rope."

mBox = 20.0kg

L = 80.0m

T = 1.26

Homework Equations


v = √(F/μ)

∑F = ma

v = λƒ

The Attempt at a Solution


I'm still trying to figure out how to approach the problem (set up my equations)...

I know that because the rope hangs vertical, the restoring force (weight UNDER a given point) is changing as the wave travels upwards, thus, the velocity is changing as well. So

dv = √(dFRestoring/μ)

and then we need to find the restoring force:

∑F = T - mBoxg - mRopeg = 0 (because it's in equilibrium in the y-direction)

T = mBoxg + mRopeg

At the beginning, the restoring force is only the weight of the box; and at the end, the restoring force is the weight of the box AND the weight of the rope. So

dFRestoring = dT = wBox + (dmRope)g

And then I tried turning μ = m/L into:

μ = dm/dL -> dm = μdL

but this doesn't get me anywhere.. I do not know what μ is equal to since we don't know the weight and when I put the equations together I get

dv = √[(wBox + μdL)/μ] -> dv = √[L(wBox + μdL)/ m]

I'm kinda stuck here and I don't think it's in the right direction because I haven't figured out why I was given the period...

Any help would be appreciated ^^
 
Physics news on Phys.org
You have one equation that relates ##\dot y## to ##F##. You can easily write one relating ##y## to ##F##, and you have almost done so, but this equation is incorrect:
Chansudesu said:
dFRestoring = dT = wBox + (dmRope)g
Pick a point ##y## on the rope. What is the tension on the rope at that point?
When you get that figured out, you can eliminate the restoring force from the two equations and solve the resulting differential equation for the time the wave takes to travel up the rope.
 
tnich said:
You have one equation that relates ##\dot y## to ##F##. You can easily write one relating ##y## to ##F##, and you have almost done so, but this equation is incorrect:

Pick a point ##y## on the rope. What is the tension on the rope at that point?
When you get that figured out, you can eliminate the restoring force from the two equations and solve the resulting differential equation for the time the wave takes to travel up the rope.

Hmm... I can't see the tension in a rope at point y being other than Ty = mBoxg + μyg because the the mass at a point y in the rope would be the mass linear density times y...
 
Chansudesu said:
Hmm... I can't see the tension in a rope at point y being other than Ty = mBoxg + μyg because the the mass at a point y in the rope would be the mass linear density times y...
Sounds right to me.
 
Ahhhh! I see now. Thank you!
 

Similar threads

  • · Replies 10 ·
Replies
10
Views
3K
Replies
46
Views
4K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 4 ·
Replies
4
Views
4K
Replies
15
Views
3K
Replies
6
Views
2K
Replies
8
Views
2K
Replies
13
Views
5K
Replies
2
Views
2K
  • · Replies 27 ·
Replies
27
Views
1K