Hanging Cable (Projectile Motion?)

Click For Summary
SUMMARY

The discussion focuses on the dynamics of a hanging cable modeled as a projectile motion problem. The initial setup involves a flexible cable of length L, with a portion x_0 hanging over a table edge. The solution to the motion is derived using the equation x = x_0 cosh(√(g/L) t), contrasting with the initial assumption of x = x_0 + (1/2)gt². Key insights include the treatment of the entire mass of the cable and the acceleration being uniform due to gravity, leading to a deeper understanding of the forces acting on the cable.

PREREQUISITES
  • Understanding of basic physics concepts, particularly projectile motion.
  • Familiarity with hyperbolic functions, specifically cosh.
  • Knowledge of Newton's second law of motion.
  • Basic calculus for solving differential equations.
NEXT STEPS
  • Study the derivation of hyperbolic functions in physics applications.
  • Learn about differential equations in the context of mechanical systems.
  • Explore the concept of linear density and its implications in cable dynamics.
  • Investigate the effects of friction in similar mechanical problems.
USEFUL FOR

Students and educators in physics, particularly those focusing on mechanics and dynamics, as well as engineers dealing with cable systems and projectile motion analysis.

TopCat
Messages
55
Reaction score
0
(Solved) Hanging Cable (Projectile Motion?)

Homework Statement


A perfectly flexible cable has length L. Initially, the cable is at rest, with a length x_0 of it hanging vertically over the edge of a table. Neglecting friction, compute the length hanging over the edge after a time t. Assume that the sections of the cable remain straight during the motion.


The Attempt at a Solution


I figured that I could take a reference point, that being the very tip of the cable hanging over the edge. Since friction is neglected the only force on the cable is gravity. If I take the table edge to be x=0 then the position of the tip of the cable over the edge will also give the length of cable behind it hanging. So then it should simply be x = x_0 + \frac{1}{2}gt^2 but the book lists the answer as x = x_0 cosh (\sqrt\frac{g}{L} t). I have no idea where to even start. Please help.
 
Last edited:
Physics news on Phys.org
Assume the linear density, that is, mass per unit length is d. The weight of the hanging portion is dxg, which is moving the whole mass of the chain. The accn of the whole chain is the same at time t. So,

dxg = Md^x/dt^2. Now, solve.
 
Outstanding! Thanks for the help. I've gotten the solution though I have one last question.

When describing the force on the portion of cable on the table why is the mass M used d*L and not d*(L - x)?
 
I've never taken into account the force on the mass on the table separately, because the whole mass, the mass on the table and the hanging portion both, is moved with the same accn by the weight of the hanging portion only.

So, where you have to write mass*accn, it's the whole mass M=d*L.
 
Okay, thank you very much. It all makes sense now.
 

Similar threads

  • · Replies 19 ·
Replies
19
Views
2K
Replies
13
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 15 ·
Replies
15
Views
8K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 5 ·
Replies
5
Views
8K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 9 ·
Replies
9
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K