Deformable solids : Traction / Compression

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The discussion revolves around calculating the normal force N(x) for a hanging cable under its own weight. The user derives the equation N = -∫_0^x qdx but struggles to determine the value of q, suggesting it might be q = (ρgSx)/L based on dimensional analysis. Other participants clarify that q should represent the weight per unit length of the cable, expressed in Newtons per meter. They recommend breaking down the problem into steps to find expressions for the cable's length, volume, mass, and weight at a point x. The conversation emphasizes the need for correct units and understanding of linear weight in this context.
Lantean
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Homework Statement
I need to find the normal force N(x) of a cable. L, suspended in x=0 and free at the end, in x=L. I have the following values : ##\rho## the density (homogen) , S the section of the cable, ##\vec{g} = g \vec{e_x}## the acceleration due to gravity and L the length of a cable. We assume that there is just the weight to considerate.
Relevant Equations
##N' + q = 0## (where q is the lineic effort)
Hi,

I use the equation above to isolate N(x), I get ##N = -\int_0^x qdx## I don't know how to get the value of q. The dimensionnal analysis give me :
## q = \frac{\rho g S x}{L}##, but I'm not sure.

Thanks for your help !
 
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Lantean said:
Homework Statement:: I need to find the normal force N(x) of a cable. L, suspended in x=0 and free at the end, in x=L. I have the following values : ##\rho## the density (homogen) , S the section of the cable, ##\vec{g} = g \vec{e_x}## the acceleration due to gravity and L the length of a cable. We assume that there is just the weight to considerate.
Relevant Equations:: ##N' + q = 0## (where q is the lineic effort)

Hi,

I use the equation above to isolate N(x), I get ##N = -\int_0^x qdx## I don't know how to get the value of q. The dimensionnal analysis give me :
## q = \frac{\rho g S x}{L}##, but I'm not sure.

Thanks for your help !
The value of q should be the weight of 1 meter of cable, and its units should be Newton/meter, if I am not wrong.
 
I think you're right, but I have to consider a weight in ##N.m^{-1}## So it would be a lineic weight, and I don't know how to express that.
 
Lantean said:
I think you're right, but I have to consider a weight in ##N.m^{-1}## So it would be a lineic weight, and I don't know how to express that.
Perhaps these members could help:
@kuruman
@Steve4Physics
@haruspex
 
Lnewqban said:
Perhaps these members could help:
@kuruman
@Steve4Physics
@haruspex
I’ll have a go but there is a language problem. I'll assume the question is this:

A cable has length L, cross-sectional area S and is made of material with density ρ. Acceleration due to gravity is g.
Taking the x-axis as vertical, the cable hangs suspended at x=0 and with its free end at x=L.
Find an expression for T(x), the tension in the cable as a function of x.
If that’s correct, I don’t see any need for calculus.

For a point (P) on the cable, a distance x below the top, the OP should answer these questions (and post them for us to check):

a) What is an expression for the length of cable below P?
b) What is the volume of this length of cable?
c) What is the mass of this length of cable??
d) What is the weight of this length of cable?
e) What is the vale value of T(x) at P?
(Some/all of these steps can be combined.)

Edit: spelling corrected.
 
Last edited:
the value of q is not correct,
 
Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

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