Can we solve the functions describing the bend of a cable at rest fixed at two positions?
Assume we have a cable which endpoints is attached to two points at $(x,h)$ and $(x+Delta_x,h)$.
Further assume it has some mass density distribution, $rho(m),m in [0,l]$ and is of some length $l > Delta_x$.
Let us call its position at $m$ : $(f_x(m),f_y(m))$.
In other words $f_x$, $f_y$ are such functions so that if only external (thanks to Andrei) force is acting downwards $(-g{bf hat y})$ (gravity force) everywhere and of course whatever forces needed at the endpoints. Then we seek the equilibrium.
How can we find $f_x,f_y$ given the above arrangement?
Own work: I think we can maybe try to express it as an energy minimization problem. However I don't know quite how to set it up.
(Other solutions not using energy minimization are also welcome!)
optimization physics classical-mechanics applications
asked 19 hours ago
mathreadler
14.7k72160
add a comment |
Assume we have a cable which endpoints is attached to two points at $(x,h)$ and $(x+Delta_x,h)$.
Further assume it has some mass density distribution, $rho(m),m in [0,l]$ and is of some length $l > Delta_x$.
Let us call its position at $m$ : $(f_x(m),f_y(m))$.
In other words $f_x$, $f_y$ are such functions so that if only external (thanks to Andrei) force is acting downwards $(-g{bf hat y})$ (gravity force) everywhere and of course whatever forces needed at the endpoints. Then we seek the equilibrium.
How can we find $f_x,f_y$ given the above arrangement?
Own work: I think we can maybe try to express it as an energy minimization problem. However I don't know quite how to set it up.
(Other solutions not using energy minimization are also welcome!)
optimization physics classical-mechanics applications
asked 19 hours ago
mathreadler
14.7k72160
add a comment |
Assume we have a cable which endpoints is attached to two points at $(x,h)$ and $(x+Delta_x,h)$.
Further assume it has some mass density distribution, $rho(m),m in [0,l]$ and is of some length $l > Delta_x$.
Let us call its position at $m$ : $(f_x(m),f_y(m))$.
In other words $f_x$, $f_y$ are such functions so that if only external (thanks to Andrei) force is acting downwards $(-g{bf hat y})$ (gravity force) everywhere and of course whatever forces needed at the endpoints. Then we seek the equilibrium.
How can we find $f_x,f_y$ given the above arrangement?
Own work: I think we can maybe try to express it as an energy minimization problem. However I don't know quite how to set it up.
(Other solutions not using energy minimization are also welcome!)
optimization physics classical-mechanics applications
asked 19 hours ago
mathreadler
14.7k72160
Assume we have a cable which endpoints is attached to two points at $(x,h)$ and $(x+Delta_x,h)$.
Further assume it has some mass density distribution, $rho(m),m in [0,l]$ and is of some length $l > Delta_x$.
Let us call its position at $m$ : $(f_x(m),f_y(m))$.
In other words $f_x$, $f_y$ are such functions so that if only external (thanks to Andrei) force is acting downwards $(-g{bf hat y})$ (gravity force) everywhere and of course whatever forces needed at the endpoints. Then we seek the equilibrium.
How can we find $f_x,f_y$ given the above arrangement?
Own work: I think we can maybe try to express it as an energy minimization problem. However I don't know quite how to set it up.
(Other solutions not using energy minimization are also welcome!)
optimization physics classical-mechanics applications
optimization physics classical-mechanics applications
asked 19 hours ago
mathreadler
14.7k72160
asked 19 hours ago
mathreadler
14.7k72160
edited 16 hours ago
asked 19 hours ago
mathreadler
14.7k72160
asked 19 hours ago
mathreadler
14.7k72160
asked 19 hours ago
mathreadler
14.7k72160
14.7k72160
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
First of all, you cannot have equilibrium with only one force. You also need the tension in the cable.On any piece of the cable you have three forces: the tension to the left, the tension to the right, and the gravity.
What you are describing is the caternary problem.
answered 17 hours ago
Andrei
11.3k21026
Yes sorry, I mean the only force from outside on the cable. Of course there will be forces inside the cable keeping it still. Hmm, I will try to reformulate it.
– mathreadler
16 hours ago
If the cable is homogenous ($rho$ constant) then I think you are right that it will be a caternary but I don't think it will be in a general case.
– mathreadler
16 hours ago
If you follow the wikipedia link, you can find the non-homogeneous case as well. Of course, the solution you seek might not have an analytical form (you can calculate it numerically).
– Andrei
16 hours ago
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3061712%2fcan-we-solve-the-functions-describing-the-bend-of-a-cable-at-rest-fixed-at-two-p%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
First of all, you cannot have equilibrium with only one force. You also need the tension in the cable.On any piece of the cable you have three forces: the tension to the left, the tension to the right, and the gravity.
What you are describing is the caternary problem.
answered 17 hours ago
Andrei
11.3k21026
Yes sorry, I mean the only force from outside on the cable. Of course there will be forces inside the cable keeping it still. Hmm, I will try to reformulate it.
– mathreadler
16 hours ago
If the cable is homogenous ($rho$ constant) then I think you are right that it will be a caternary but I don't think it will be in a general case.
– mathreadler
16 hours ago
If you follow the wikipedia link, you can find the non-homogeneous case as well. Of course, the solution you seek might not have an analytical form (you can calculate it numerically).
– Andrei
16 hours ago
add a comment |
First of all, you cannot have equilibrium with only one force. You also need the tension in the cable.On any piece of the cable you have three forces: the tension to the left, the tension to the right, and the gravity.
What you are describing is the caternary problem.
answered 17 hours ago
Andrei
11.3k21026
Yes sorry, I mean the only force from outside on the cable. Of course there will be forces inside the cable keeping it still. Hmm, I will try to reformulate it.
– mathreadler
16 hours ago
If the cable is homogenous ($rho$ constant) then I think you are right that it will be a caternary but I don't think it will be in a general case.
– mathreadler
16 hours ago
If you follow the wikipedia link, you can find the non-homogeneous case as well. Of course, the solution you seek might not have an analytical form (you can calculate it numerically).
– Andrei
16 hours ago
add a comment |
First of all, you cannot have equilibrium with only one force. You also need the tension in the cable.On any piece of the cable you have three forces: the tension to the left, the tension to the right, and the gravity.
What you are describing is the caternary problem.
answered 17 hours ago
Andrei
11.3k21026
First of all, you cannot have equilibrium with only one force. You also need the tension in the cable.On any piece of the cable you have three forces: the tension to the left, the tension to the right, and the gravity.
What you are describing is the caternary problem.
answered 17 hours ago
Andrei
11.3k21026
answered 17 hours ago
Andrei
11.3k21026
answered 17 hours ago
Andrei
11.3k21026
answered 17 hours ago
Andrei
11.3k21026
11.3k21026
Yes sorry, I mean the only force from outside on the cable. Of course there will be forces inside the cable keeping it still. Hmm, I will try to reformulate it.
– mathreadler
16 hours ago
If the cable is homogenous ($rho$ constant) then I think you are right that it will be a caternary but I don't think it will be in a general case.
– mathreadler
16 hours ago
If you follow the wikipedia link, you can find the non-homogeneous case as well. Of course, the solution you seek might not have an analytical form (you can calculate it numerically).
– Andrei
16 hours ago
add a comment |
Yes sorry, I mean the only force from outside on the cable. Of course there will be forces inside the cable keeping it still. Hmm, I will try to reformulate it.
– mathreadler
16 hours ago
If the cable is homogenous ($rho$ constant) then I think you are right that it will be a caternary but I don't think it will be in a general case.
– mathreadler
16 hours ago
If you follow the wikipedia link, you can find the non-homogeneous case as well. Of course, the solution you seek might not have an analytical form (you can calculate it numerically).
– Andrei
16 hours ago
Yes sorry, I mean the only force from outside on the cable. Of course there will be forces inside the cable keeping it still. Hmm, I will try to reformulate it.
– mathreadler
16 hours ago
Yes sorry, I mean the only force from outside on the cable. Of course there will be forces inside the cable keeping it still. Hmm, I will try to reformulate it.
– mathreadler
16 hours ago
If the cable is homogenous ($rho$ constant) then I think you are right that it will be a caternary but I don't think it will be in a general case.
– mathreadler
16 hours ago
If the cable is homogenous ($rho$ constant) then I think you are right that it will be a caternary but I don't think it will be in a general case.
– mathreadler
16 hours ago
If you follow the wikipedia link, you can find the non-homogeneous case as well. Of course, the solution you seek might not have an analytical form (you can calculate it numerically).
– Andrei
16 hours ago
If you follow the wikipedia link, you can find the non-homogeneous case as well. Of course, the solution you seek might not have an analytical form (you can calculate it numerically).
– Andrei
16 hours ago
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3061712%2fcan-we-solve-the-functions-describing-the-bend-of-a-cable-at-rest-fixed-at-two-p%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
