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Calculated Shard: 18 (from laksa080)

2. Crawled Status Check

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curl -X POST \
  'http://laksa018.int.ahrefs:8124/' \
  -H 'Content-Type: text/plain' \
  -H 'X-ClickHouse-Database: crawler3' \
  -H 'Authorization: Basic YXBpOg==' \
  -d 'SELECT getAhrefsURLFromUnparsed(src_unparsed) AS found_url, ifNull(toUnixTimestamp(download_stamp), 0) AS crawl_time, ifNull(toUnixTimestamp(props_url_first_seen), 0) AS first_indexed_time, download_http_code AS http_code, src_unparsed AS src_unparsed, src_root_hash AS src_root_hash, history_drop_reason AS history_drop_reason, meta_title AS meta_title, meta_descriptions AS meta_descriptions, attrs_boilerpipe_text AS attrs_boilerpipe_text, attrs_markdown AS attrs_markdown, attrs_readable_markdown AS attrs_readable_markdown, meta_canonical AS meta_canonical FROM crawler3.page_info_local FINAL PREWHERE (src_root_hash, src_unparsed) IN ((getAhrefsRootHashFromUnparsed(getAhrefsUnparsedNoserviceFromURL(\'https://physics.stackexchange.com/questions/448935/restoring-force-vs-applied-force-for-hookes-law\')), getAhrefsUnparsedNoserviceFromURL(\'https://physics.stackexchange.com/questions/448935/restoring-force-vs-applied-force-for-hookes-law\'))) FORMAT JSONEachRow'

Response:

{"found_url":"https:\/\/physics.stackexchange.com\/questions\/448935\/restoring-force-vs-applied-force-for-hookes-law","crawl_time":1774730964,"first_indexed_time":0,"http_code":200,"src_unparsed":"com,stackexchange!physics,\/questions\/448935\/restoring-force-vs-applied-force-for-hookes-law s443","src_root_hash":"8045678284012640218","history_drop_reason":null,"meta_title":"newtonian mechanics - Restoring Force vs Applied Force for Hooke's Law - Physics Stack Exchange","meta_descriptions":[],"attrs_boilerpipe_text":"1\n$\\begingroup$\nAccording to Hooke's Law,\n$F=-kx$\nwhere\n$F$\nis the restoring force,\n$k$\nis the spring constant and\n$x$\nis the length of extension\/compression.\nWhen an applied force compresses a spring, a restoring force will act in the opposite direction.\nWhen a spring is compressed and is in equilibrium (not extending\/compressing), the restoring force should be equal to the applied force since the resultant force is 0 and the acceleration is 0. (Thus the spring is not extending\/compressing)\nHowever, the act of compressing a spring will require the applied force to be greater than the restoring force even at an\ninfinitesimally small time interval\n, am I wrong?\nIs it more correct to say that\n$F=kx$\nwhere\n$F$\nis the applied force or\n$F=-kx$\nwhere\n$F$\nis the restoring force? I think that\n$F=-kx$\nthe better option, but I'm pretty confused here.\nDoes it matter at all?\nQmechanic\n♦\n226k\n52 gold badges\n651 silver badges\n2.7k bronze badges\nasked\nDec 17, 2018 at 16:31\n$\\endgroup$\n5\n4\n$\\begingroup$\nWould x be proportional to the applied force or the restoring force?\nThe equation\n$F=-kx$\nis for the spring (restoring) force. If we apply a force equal to this (\n$F_{app}=kx)$\nthen the spring edge where the force is applied will not accelerate (it will be stationary or moving at a constant velocity depending on what the velocity is when we set our force to start being equal to\n$kx$\n). Note that this is a very specific case\nYou are right, in order to stretch\/compress a spring at rest we need to initially apply a force different than\n$kx$\n, but this is true for other scenarios too. For example, to lift a box we need to first apply a force larger than its weight to start it's velocity upwards. Then we can reduce our force to be equal to the weight to lift the box at a constant speed, or we can keep our force constant and the box will accelerate upwards. The same analogy can be made for the spring at each position\n$x$\n, since the force depends on\n$x$\n.\nSo overall, the spring (restoring) force is always given by\n$F=-kx$\nwhere\n$x$\nis the displacement from equilibrium. The applied force is really anything you want it to be, and depending on what it is the spring will respond accordingly based on the initial conditions. For example, we can hang a mass from a spring so that the mass remains stationary, or we can stretch and release that same mass so that oscillations occur. Notice how in either scenario the applied force is a constant\n$mg$\n, but the spring behavior is different due to the initial conditions of the system\nanswered\nDec 18, 2018 at 3:48\n$\\endgroup$\n0\n0\n$\\begingroup$\nI think that for your question, the sign of the force is not something that matters. What really matters is that an unbalance between the applied force and the elastic restoring force is effectively needed in order for the applied force to accelerate the spring.\nIf you think of this problem as static then the it takes infinite time for the applied force to compress the spring. This is the only case in which you can consider the applied and the restoring elastic force equal.\nanswered\nDec 18, 2018 at 2:25\n$\\endgroup$\n0\n$\\begingroup$\nWhen we say\n$F=kx$\n, we are essentially saying that based on\n$kx$\nof the spring, we can determine the applied force.  The same goes for an accelerating object.  The\n$ma$\nof the object will be equivalent to the applied force.\nIn response, the spring exerts a restoring force back to its equilibrium position whether it is stretched or compressed.  Since we are taking the direction of applied force to be positive, the direction of the restoring force will thus be negative, given by\n$F=-kx$\n.\nanswered\nDec 18, 2018 at 7:21\n$\\endgroup$\nStart asking to get answers\nFind the answer to your question by asking.\nAsk question\nExplore related questions\nSee similar questions with these tags.","attrs_markdown":"[Skip to main content](https:\/\/physics.stackexchange.com\/questions\/448935\/restoring-force-vs-applied-force-for-hookes-law#content)\n\n#### Stack Exchange Network\nStack Exchange network consists of 183 Q\\&A communities including [Stack Overflow](https:\/\/stackoverflow.com\/), the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.\n\n[Visit Stack Exchange](https:\/\/stackexchange.com\/)\n\n1. - [Tour Start here for a quick overview of the site](https:\/\/physics.stackexchange.com\/tour)\n   - [Help Center Detailed answers to any questions you might have](https:\/\/physics.stackexchange.com\/help)\n   - [Meta Discuss the workings and policies of this site](https:\/\/physics.meta.stackexchange.com\/)\n   - [About Us Learn more about Stack Overflow the company, and our products](https:\/\/stackoverflow.co\/)\n2. ### [current community](https:\/\/physics.stackexchange.com\/)\n   - [Physics](https:\/\/physics.stackexchange.com\/)\n     [help](https:\/\/physics.stackexchange.com\/help) [chat](https:\/\/chat.stackexchange.com\/?tab=site&host=physics.stackexchange.com)\n   - [Physics Meta](https:\/\/physics.meta.stackexchange.com\/)\n   ### your communities\n   [Sign up](https:\/\/physics.stackexchange.com\/users\/signup?ssrc=site_switcher&returnurl=https%3A%2F%2Fphysics.stackexchange.com%2Fquestions%2F448935%2Frestoring-force-vs-applied-force-for-hookes-law) or [log in](https:\/\/physics.stackexchange.com\/users\/login?ssrc=site_switcher&returnurl=https%3A%2F%2Fphysics.stackexchange.com%2Fquestions%2F448935%2Frestoring-force-vs-applied-force-for-hookes-law) to customize your list.\n   \n   ### [more stack exchange communities](https:\/\/stackexchange.com\/sites)\n   [company blog](https:\/\/stackoverflow.blog\/)\n3. 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[Learn more](https:\/\/stackoverflow.co\/internal\/?utm_medium=referral&utm_source=physics-community&utm_campaign=side-bar&utm_content=explore-teams-compact)\n\n**Stack Internal**\n\nKnowledge at work\n\nBring the best of human thought and AI automation together at your work.\n\n[Explore Stack Internal](https:\/\/stackoverflow.co\/internal\/?utm_medium=referral&utm_source=physics-community&utm_campaign=side-bar&utm_content=explore-teams-compact-popover)\n\n# [Restoring Force vs Applied Force for Hooke's Law](https:\/\/physics.stackexchange.com\/questions\/448935\/restoring-force-vs-applied-force-for-hookes-law)\n[Ask Question](https:\/\/physics.stackexchange.com\/questions\/ask)\n\nAsked\n\n7 years, 3 months ago\n\nModified [7 years, 3 months ago](https:\/\/physics.stackexchange.com\/questions\/448935\/restoring-force-vs-applied-force-for-hookes-law?lastactivity \"2018-12-18 07:21:29Z\")\n\nViewed 4k times\n\n1\n\n\\$\\\\begingroup\\$\n\nAccording to Hooke's Law, \\$F=-kx\\$ where \\$F\\$ is the restoring force, \\$k\\$ is the spring constant and \\$x\\$ is the length of extension\/compression.\n\nWhen an applied force compresses a spring, a restoring force will act in the opposite direction.  \n When a spring is compressed and is in equilibrium (not extending\/compressing), the restoring force should be equal to the applied force since the resultant force is 0 and the acceleration is 0. (Thus the spring is not extending\/compressing)\n\nHowever, the act of compressing a spring will require the applied force to be greater than the restoring force even at an **infinitesimally small time interval**, am I wrong?\n\nIs it more correct to say that \\$F=kx\\$ where \\$F\\$ is the applied force or \\$F=-kx\\$ where \\$F\\$ is the restoring force? I think that \\$F=-kx\\$ the better option, but I'm pretty confused here. **Does it matter at all?**\n\n- [newtonian-mechanics](https:\/\/physics.stackexchange.com\/questions\/tagged\/newtonian-mechanics \"show questions tagged 'newtonian-mechanics'\")\n- [forces](https:\/\/physics.stackexchange.com\/questions\/tagged\/forces \"show questions tagged 'forces'\")\n- [spring](https:\/\/physics.stackexchange.com\/questions\/tagged\/spring \"show questions tagged 'spring'\")\n\n[Share](https:\/\/physics.stackexchange.com\/q\/448935 \"Short permalink to this question\")\n\nCite\n\n[Improve this question](https:\/\/physics.stackexchange.com\/posts\/448935\/edit)\n\nFollow\n\n[edited Dec 17, 2018 at 16:49](https:\/\/physics.stackexchange.com\/posts\/448935\/revisions \"show all edits to this post\")\n\n[![Qmechanic's user avatar](https:\/\/www.gravatar.com\/avatar\/86f4f64980b4736b1f6e5e3056d08e15?s=64&d=identicon&r=PG)](https:\/\/physics.stackexchange.com\/users\/2451\/qmechanic)\n\n[Qmechanic](https:\/\/physics.stackexchange.com\/users\/2451\/qmechanic)♦\n\n226k5252 gold badges651651 silver badges2\\.7k2\\.7k bronze badges\n\nasked Dec 17, 2018 at 16:31\n\n[![helpme's user avatar](https:\/\/www.gravatar.com\/avatar\/f19a3ac761df525d20390cf126e6219e?s=64&d=identicon&r=PG&f=y&so-version=2)](https:\/\/physics.stackexchange.com\/users\/212622\/helpme)\n\n[helpme](https:\/\/physics.stackexchange.com\/users\/212622\/helpme)\n\n39377 silver badges1818 bronze badges\n\n\\$\\\\endgroup\\$\n\n5\n\n- \\$\\\\begingroup\\$ Compress the spring such that the end of the spring moves at a constant speed, and the two forces will at all times be equal to each other (except for right at the beginning and right at the end, but these times can be made arbitrarily small, theoretically). \\$\\\\endgroup\\$\n  \n  march\n  \n  – [march](https:\/\/physics.stackexchange.com\/users\/83835\/march \"9,893 reputation\")\n  \n  2018-12-17 16:50:52 +00:00\n  \n  Commented Dec 17, 2018 at 16:50\n- \\$\\\\begingroup\\$ If the applied force was not equal to the springs restoring force at any time there would be acceleration. The spring can be compressed and have no net force on it \\$\\\\endgroup\\$\n  \n  Triatticus\n  \n  – [Triatticus](https:\/\/physics.stackexchange.com\/users\/23615\/triatticus \"247 reputation\")\n  \n  2018-12-17 16:51:32 +00:00\n  \n  Commented Dec 17, 2018 at 16:51\n- \\$\\\\begingroup\\$ If the spring is compressed at an accelerating rate, the applied force and restoring force are not equal. Would x be proportional to the applied force or the restoring force? \\$\\\\endgroup\\$\n  \n  helpme\n  \n  – [helpme](https:\/\/physics.stackexchange.com\/users\/212622\/helpme \"393 reputation\")\n  \n  2018-12-17 18:19:09 +00:00\n  \n  Commented Dec 17, 2018 at 18:19\n- \\$\\\\begingroup\\$ Newton's 3rd law requires that the force and opposing force be equal at all times. \\$\\\\endgroup\\$\n  \n  David White\n  \n  – [David White](https:\/\/physics.stackexchange.com\/users\/86822\/david-white \"12,662 reputation\")\n  \n  2018-12-17 20:35:42 +00:00\n  \n  Commented Dec 17, 2018 at 20:35\n- \\$\\\\begingroup\\$ The 3rd law partner forces act on different bodies \\$\\\\endgroup\\$\n  \n  helpme\n  \n  – [helpme](https:\/\/physics.stackexchange.com\/users\/212622\/helpme \"393 reputation\")\n  \n  2018-12-17 20:39:33 +00:00\n  \n  Commented Dec 17, 2018 at 20:39\n\n[Add a comment](https:\/\/physics.stackexchange.com\/questions\/448935\/restoring-force-vs-applied-force-for-hookes-law \"Use comments to ask for more information or suggest improvements. Avoid answering questions in comments.\")   \\|\n\n## 3 Answers 3\nSorted by:\n\n[Reset to default](https:\/\/physics.stackexchange.com\/questions\/448935\/restoring-force-vs-applied-force-for-hookes-law?answertab=scoredesc#tab-top)\n\n4\n\n\\$\\\\begingroup\\$\n\n> Would x be proportional to the applied force or the restoring force?\n\nThe equation \\$F=-kx\\$ is for the spring (restoring) force. If we apply a force equal to this (\\$F\\_{app}=kx)\\$ then the spring edge where the force is applied will not accelerate (it will be stationary or moving at a constant velocity depending on what the velocity is when we set our force to start being equal to \\$kx\\$). Note that this is a very specific case\n\nYou are right, in order to stretch\/compress a spring at rest we need to initially apply a force different than \\$kx\\$, but this is true for other scenarios too. For example, to lift a box we need to first apply a force larger than its weight to start it's velocity upwards. Then we can reduce our force to be equal to the weight to lift the box at a constant speed, or we can keep our force constant and the box will accelerate upwards. The same analogy can be made for the spring at each position \\$x\\$, since the force depends on \\$x\\$.\n\nSo overall, the spring (restoring) force is always given by \\$F=-kx\\$ where \\$x\\$ is the displacement from equilibrium. The applied force is really anything you want it to be, and depending on what it is the spring will respond accordingly based on the initial conditions. For example, we can hang a mass from a spring so that the mass remains stationary, or we can stretch and release that same mass so that oscillations occur. Notice how in either scenario the applied force is a constant \\$mg\\$, but the spring behavior is different due to the initial conditions of the system\n\n[Share](https:\/\/physics.stackexchange.com\/a\/449034 \"Short permalink to this answer\")\n\nCite\n\n[Improve this answer](https:\/\/physics.stackexchange.com\/posts\/449034\/edit)\n\nFollow\n\n[edited Dec 18, 2018 at 4:26](https:\/\/physics.stackexchange.com\/posts\/449034\/revisions \"show all edits to this post\")\n\nanswered Dec 18, 2018 at 3:48\n\n[![BioPhysicist's user avatar](https:\/\/i.sstatic.net\/uGTTK.png?s=64)](https:\/\/physics.stackexchange.com\/users\/179151\/biophysicist)\n\n[BioPhysicist](https:\/\/physics.stackexchange.com\/users\/179151\/biophysicist)\n\n59\\.5k1919 gold badges119119 silver badges199199 bronze badges\n\n\\$\\\\endgroup\\$\n\n0\n\n[Add a comment](https:\/\/physics.stackexchange.com\/questions\/448935\/restoring-force-vs-applied-force-for-hookes-law \"Use comments to ask for more information or suggest improvements. Avoid comments like “+1” or “thanks”.\")   \\|\n\n0\n\n\\$\\\\begingroup\\$\n\nI think that for your question, the sign of the force is not something that matters. What really matters is that an unbalance between the applied force and the elastic restoring force is effectively needed in order for the applied force to accelerate the spring.\n\nIf you think of this problem as static then the it takes infinite time for the applied force to compress the spring. This is the only case in which you can consider the applied and the restoring elastic force equal.\n\n[Share](https:\/\/physics.stackexchange.com\/a\/449020 \"Short permalink to this answer\")\n\nCite\n\n[Improve this answer](https:\/\/physics.stackexchange.com\/posts\/449020\/edit)\n\nFollow\n\nanswered Dec 18, 2018 at 2:25\n\n[![nodarkside's user avatar](https:\/\/i.sstatic.net\/J1QBM.jpg?s=64)](https:\/\/physics.stackexchange.com\/users\/186231\/nodarkside)\n\n[nodarkside](https:\/\/physics.stackexchange.com\/users\/186231\/nodarkside)\n\n35522 silver badges99 bronze badges\n\n\\$\\\\endgroup\\$\n\n[Add a comment](https:\/\/physics.stackexchange.com\/questions\/448935\/restoring-force-vs-applied-force-for-hookes-law \"Use comments to ask for more information or suggest improvements. Avoid comments like “+1” or “thanks”.\")   \\|\n\n0\n\n\\$\\\\begingroup\\$\n\nWhen we say \\$F=kx\\$, we are essentially saying that based on \\$kx\\$ of the spring, we can determine the applied force. The same goes for an accelerating object. The \\$ma\\$ of the object will be equivalent to the applied force.\n\nIn response, the spring exerts a restoring force back to its equilibrium position whether it is stretched or compressed. Since we are taking the direction of applied force to be positive, the direction of the restoring force will thus be negative, given by \\$F=-kx\\$.\n\n[Share](https:\/\/physics.stackexchange.com\/a\/449051 \"Short permalink to this answer\")\n\nCite\n\n[Improve this answer](https:\/\/physics.stackexchange.com\/posts\/449051\/edit)\n\nFollow\n\nanswered Dec 18, 2018 at 7:21\n\n[![QuIcKmAtHs's user avatar](https:\/\/www.gravatar.com\/avatar\/074040db5e59bc608cfcebf9b259ebc7?s=64&d=identicon&r=PG)](https:\/\/physics.stackexchange.com\/users\/179737\/quickmaths)\n\n[QuIcKmAtHs](https:\/\/physics.stackexchange.com\/users\/179737\/quickmaths)\n\n3,79344 gold badges2121 silver badges4040 bronze badges\n\n\\$\\\\endgroup\\$\n\n[Add a comment](https:\/\/physics.stackexchange.com\/questions\/448935\/restoring-force-vs-applied-force-for-hookes-law \"Use comments to ask for more information or suggest improvements. Avoid comments like “+1” or “thanks”.\")   \\|\n\nStart asking to get answers\n\nFind the answer to your question by asking.\n\n[Ask question](https:\/\/physics.stackexchange.com\/questions\/ask)\n\nExplore related questions\n\n- [newtonian-mechanics](https:\/\/physics.stackexchange.com\/questions\/tagged\/newtonian-mechanics \"show questions tagged 'newtonian-mechanics'\")\n- [forces](https:\/\/physics.stackexchange.com\/questions\/tagged\/forces \"show questions tagged 'forces'\")\n- [spring](https:\/\/physics.stackexchange.com\/questions\/tagged\/spring \"show questions tagged 'spring'\")\n\nSee similar questions with these tags.\n\n#### Related\n[7](https:\/\/physics.stackexchange.com\/questions\/54924\/is-restoring-force-a-particular-type-of-force \"Question score (upvotes - downvotes)\")\n\n[Is 'restoring force' a particular type of force?](https:\/\/physics.stackexchange.com\/questions\/54924\/is-restoring-force-a-particular-type-of-force)\n\n[2](https:\/\/physics.stackexchange.com\/questions\/86344\/is-mass-proportional-to-the-displacement-from-equilibrium-in-hookes-law \"Question score (upvotes - downvotes)\")\n\n[Is mass proportional to the displacement from equilibrium in Hooke's law?](https:\/\/physics.stackexchange.com\/questions\/86344\/is-mass-proportional-to-the-displacement-from-equilibrium-in-hookes-law)\n\n[0](https:\/\/physics.stackexchange.com\/questions\/149111\/can-hookes-law-go-along-with-newtons-3rd-law-how-is-momentum-conserved-in-ca \"Question score (upvotes - downvotes)\")\n\n[Can Hooke's law go along with Newton's 3rd law & how is momentum conserved in case of spring?](https:\/\/physics.stackexchange.com\/questions\/149111\/can-hookes-law-go-along-with-newtons-3rd-law-how-is-momentum-conserved-in-ca)\n\n[1](https:\/\/physics.stackexchange.com\/questions\/167458\/on-the-definition-of-elastic-restoring-force-in-a-spring \"Question score (upvotes - downvotes)\")\n\n[On the definition of elastic restoring force in a spring](https:\/\/physics.stackexchange.com\/questions\/167458\/on-the-definition-of-elastic-restoring-force-in-a-spring)\n\n[3](https:\/\/physics.stackexchange.com\/questions\/208038\/how-to-apply-newtons-third-law-when-a-massless-spring-is-compressed-by-a-moving \"Question score (upvotes - downvotes)\")\n\n[How to apply Newton's Third law when a massless spring is compressed by a moving block?](https:\/\/physics.stackexchange.com\/questions\/208038\/how-to-apply-newtons-third-law-when-a-massless-spring-is-compressed-by-a-moving)\n\n[1](https:\/\/physics.stackexchange.com\/questions\/246604\/why-is-the-restoring-force-directly-proportional-to-extension \"Question score (upvotes - downvotes)\")\n\n[Why is the restoring force directly proportional to extension?](https:\/\/physics.stackexchange.com\/questions\/246604\/why-is-the-restoring-force-directly-proportional-to-extension)\n\n[0](https:\/\/physics.stackexchange.com\/questions\/358446\/what-do-force-extension-graphs-for-springs-show \"Question score (upvotes - downvotes)\")\n\n[What do force extension graphs for springs show?](https:\/\/physics.stackexchange.com\/questions\/358446\/what-do-force-extension-graphs-for-springs-show)\n\n#### [Hot Network Questions](https:\/\/stackexchange.com\/questions?tab=hot)\n- [\\-ие always replaceable by -ье?](https:\/\/russian.stackexchange.com\/questions\/29787\/%D0%B8%D0%B5-always-replaceable-by-%D1%8C%D0%B5)\n- [Parallels in sequences?](https:\/\/music.stackexchange.com\/questions\/143417\/parallels-in-sequences)\n- [Ignoring the tail of the letter \"g\" in setting the baseline of text in nodes of a tikz diagram](https:\/\/tex.stackexchange.com\/questions\/761310\/ignoring-the-tail-of-the-letter-g-in-setting-the-baseline-of-text-in-nodes-of)\n- [feature generating strategy](https:\/\/datascience.stackexchange.com\/questions\/137866\/feature-generating-strategy)\n- [What does \"world\" mean in \"God was reconciling the world to himself\"? 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(Thus the spring is not extending\/compressing)\n\nHowever, the act of compressing a spring will require the applied force to be greater than the restoring force even at an **infinitesimally small time interval**, am I wrong?\n\nIs it more correct to say that \\$F=kx\\$ where \\$F\\$ is the applied force or \\$F=-kx\\$ where \\$F\\$ is the restoring force? I think that \\$F=-kx\\$ the better option, but I'm pretty confused here. **Does it matter at all?**\n\n[![Qmechanic's user avatar](https:\/\/www.gravatar.com\/avatar\/86f4f64980b4736b1f6e5e3056d08e15?s=64&d=identicon&r=PG)](https:\/\/physics.stackexchange.com\/users\/2451\/qmechanic)\n\n[Qmechanic](https:\/\/physics.stackexchange.com\/users\/2451\/qmechanic)♦\n\n226k52 gold badges651 silver badges2\\.7k bronze badges\n\nasked Dec 17, 2018 at 16:31\n\n[![helpme's user avatar](https:\/\/www.gravatar.com\/avatar\/f19a3ac761df525d20390cf126e6219e?s=64&d=identicon&r=PG&f=y&so-version=2)](https:\/\/physics.stackexchange.com\/users\/212622\/helpme)\n\n\\$\\\\endgroup\\$\n\n5\n\n4\n\n\\$\\\\begingroup\\$\n\n> Would x be proportional to the applied force or the restoring force?\n\nThe equation \\$F=-kx\\$ is for the spring (restoring) force. If we apply a force equal to this (\\$F\\_{app}=kx)\\$ then the spring edge where the force is applied will not accelerate (it will be stationary or moving at a constant velocity depending on what the velocity is when we set our force to start being equal to \\$kx\\$). Note that this is a very specific case\n\nYou are right, in order to stretch\/compress a spring at rest we need to initially apply a force different than \\$kx\\$, but this is true for other scenarios too. For example, to lift a box we need to first apply a force larger than its weight to start it's velocity upwards. Then we can reduce our force to be equal to the weight to lift the box at a constant speed, or we can keep our force constant and the box will accelerate upwards. The same analogy can be made for the spring at each position \\$x\\$, since the force depends on \\$x\\$.\n\nSo overall, the spring (restoring) force is always given by \\$F=-kx\\$ where \\$x\\$ is the displacement from equilibrium. The applied force is really anything you want it to be, and depending on what it is the spring will respond accordingly based on the initial conditions. For example, we can hang a mass from a spring so that the mass remains stationary, or we can stretch and release that same mass so that oscillations occur. Notice how in either scenario the applied force is a constant \\$mg\\$, but the spring behavior is different due to the initial conditions of the system\n\nanswered Dec 18, 2018 at 3:48\n\n[![BioPhysicist's user avatar](https:\/\/i.sstatic.net\/uGTTK.png?s=64)](https:\/\/physics.stackexchange.com\/users\/179151\/biophysicist)\n\n\\$\\\\endgroup\\$\n\n0\n\n0\n\n\\$\\\\begingroup\\$\n\nI think that for your question, the sign of the force is not something that matters. What really matters is that an unbalance between the applied force and the elastic restoring force is effectively needed in order for the applied force to accelerate the spring.\n\nIf you think of this problem as static then the it takes infinite time for the applied force to compress the spring. This is the only case in which you can consider the applied and the restoring elastic force equal.\n\nanswered Dec 18, 2018 at 2:25\n\n[![nodarkside's user avatar](https:\/\/i.sstatic.net\/J1QBM.jpg?s=64)](https:\/\/physics.stackexchange.com\/users\/186231\/nodarkside)\n\n\\$\\\\endgroup\\$\n\n0\n\n\\$\\\\begingroup\\$\n\nWhen we say \\$F=kx\\$, we are essentially saying that based on \\$kx\\$ of the spring, we can determine the applied force. The same goes for an accelerating object. The \\$ma\\$ of the object will be equivalent to the applied force.\n\nIn response, the spring exerts a restoring force back to its equilibrium position whether it is stretched or compressed. Since we are taking the direction of applied force to be positive, the direction of the restoring force will thus be negative, given by \\$F=-kx\\$.\n\nanswered Dec 18, 2018 at 7:21\n\n[![QuIcKmAtHs's user avatar](https:\/\/www.gravatar.com\/avatar\/074040db5e59bc608cfcebf9b259ebc7?s=64&d=identicon&r=PG)](https:\/\/physics.stackexchange.com\/users\/179737\/quickmaths)\n\n\\$\\\\endgroup\\$\n\nStart asking to get answers\n\nFind the answer to your question by asking.\n\n[Ask question](https:\/\/physics.stackexchange.com\/questions\/ask)\n\nExplore related questions\n\nSee similar questions with these tags.","meta_canonical":null}

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Boilerpipe Text	1 $\begingroup$ According to Hooke's Law, $F=-kx$ where $F$ is the restoring force, $k$ is the spring constant and $x$ is the length of extension/compression. When an applied force compresses a spring, a restoring force will act in the opposite direction. When a spring is compressed and is in equilibrium (not extending/compressing), the restoring force should be equal to the applied force since the resultant force is 0 and the acceleration is 0. (Thus the spring is not extending/compressing) However, the act of compressing a spring will require the applied force to be greater than the restoring force even at an infinitesimally small time interval , am I wrong? Is it more correct to say that $F=kx$ where $F$ is the applied force or $F=-kx$ where $F$ is the restoring force? I think that $F=-kx$ the better option, but I'm pretty confused here. Does it matter at all? Qmechanic ♦ 226k 52 gold badges 651 silver badges 2.7k bronze badges asked Dec 17, 2018 at 16:31 $\endgroup$ 5 4 $\begingroup$ Would x be proportional to the applied force or the restoring force? The equation $F=-kx$ is for the spring (restoring) force. If we apply a force equal to this ( $F_{app}=kx)$ then the spring edge where the force is applied will not accelerate (it will be stationary or moving at a constant velocity depending on what the velocity is when we set our force to start being equal to $kx$ ). Note that this is a very specific case You are right, in order to stretch/compress a spring at rest we need to initially apply a force different than $kx$ , but this is true for other scenarios too. For example, to lift a box we need to first apply a force larger than its weight to start it's velocity upwards. Then we can reduce our force to be equal to the weight to lift the box at a constant speed, or we can keep our force constant and the box will accelerate upwards. The same analogy can be made for the spring at each position $x$ , since the force depends on $x$ . So overall, the spring (restoring) force is always given by $F=-kx$ where $x$ is the displacement from equilibrium. The applied force is really anything you want it to be, and depending on what it is the spring will respond accordingly based on the initial conditions. For example, we can hang a mass from a spring so that the mass remains stationary, or we can stretch and release that same mass so that oscillations occur. Notice how in either scenario the applied force is a constant $mg$ , but the spring behavior is different due to the initial conditions of the system answered Dec 18, 2018 at 3:48 $\endgroup$ 0 0 $\begingroup$ I think that for your question, the sign of the force is not something that matters. What really matters is that an unbalance between the applied force and the elastic restoring force is effectively needed in order for the applied force to accelerate the spring. If you think of this problem as static then the it takes infinite time for the applied force to compress the spring. This is the only case in which you can consider the applied and the restoring elastic force equal. answered Dec 18, 2018 at 2:25 $\endgroup$ 0 $\begingroup$ When we say $F=kx$ , we are essentially saying that based on $kx$ of the spring, we can determine the applied force. The same goes for an accelerating object. The $ma$ of the object will be equivalent to the applied force. In response, the spring exerts a restoring force back to its equilibrium position whether it is stretched or compressed. Since we are taking the direction of applied force to be positive, the direction of the restoring force will thus be negative, given by $F=-kx$ . answered Dec 18, 2018 at 7:21 $\endgroup$ Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions See similar questions with these tags.
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[Explore Stack Internal](https://stackoverflow.co/internal/?utm_medium=referral&utm_source=physics-community&utm_campaign=side-bar&utm_content=explore-teams-compact-popover) # [Restoring Force vs Applied Force for Hooke's Law](https://physics.stackexchange.com/questions/448935/restoring-force-vs-applied-force-for-hookes-law) [Ask Question](https://physics.stackexchange.com/questions/ask) Asked 7 years, 3 months ago Modified [7 years, 3 months ago](https://physics.stackexchange.com/questions/448935/restoring-force-vs-applied-force-for-hookes-law?lastactivity "2018-12-18 07:21:29Z") Viewed 4k times 1 \$\\begingroup\$ According to Hooke's Law, \$F=-kx\$ where \$F\$ is the restoring force, \$k\$ is the spring constant and \$x\$ is the length of extension/compression. When an applied force compresses a spring, a restoring force will act in the opposite direction. When a spring is compressed and is in equilibrium (not extending/compressing), the restoring force should be equal to the applied force since the resultant force is 0 and the acceleration is 0. (Thus the spring is not extending/compressing) However, the act of compressing a spring will require the applied force to be greater than the restoring force even at an infinitesimally small time interval, am I wrong? Is it more correct to say that \$F=kx\$ where \$F\$ is the applied force or \$F=-kx\$ where \$F\$ is the restoring force? I think that \$F=-kx\$ the better option, but I'm pretty confused here. Does it matter at all? - [newtonian-mechanics](https://physics.stackexchange.com/questions/tagged/newtonian-mechanics "show questions tagged 'newtonian-mechanics'") - [forces](https://physics.stackexchange.com/questions/tagged/forces "show questions tagged 'forces'") - [spring](https://physics.stackexchange.com/questions/tagged/spring "show questions tagged 'spring'") [Share](https://physics.stackexchange.com/q/448935 "Short permalink to this question") Cite [Improve this question](https://physics.stackexchange.com/posts/448935/edit) Follow [edited Dec 17, 2018 at 16:49](https://physics.stackexchange.com/posts/448935/revisions "show all edits to this post") [![Qmechanic's user avatar](https://www.gravatar.com/avatar/86f4f64980b4736b1f6e5e3056d08e15?s=64&d=identicon&r=PG)](https://physics.stackexchange.com/users/2451/qmechanic) [Qmechanic](https://physics.stackexchange.com/users/2451/qmechanic)♦ 226k5252 gold badges651651 silver badges2\.7k2\.7k bronze badges asked Dec 17, 2018 at 16:31 [![helpme's user avatar](https://www.gravatar.com/avatar/f19a3ac761df525d20390cf126e6219e?s=64&d=identicon&r=PG&f=y&so-version=2)](https://physics.stackexchange.com/users/212622/helpme) [helpme](https://physics.stackexchange.com/users/212622/helpme) 39377 silver badges1818 bronze badges \$\\endgroup\$ 5 - \$\\begingroup\$ Compress the spring such that the end of the spring moves at a constant speed, and the two forces will at all times be equal to each other (except for right at the beginning and right at the end, but these times can be made arbitrarily small, theoretically). \$\\endgroup\$ march – [march](https://physics.stackexchange.com/users/83835/march "9,893 reputation") 2018-12-17 16:50:52 +00:00 Commented Dec 17, 2018 at 16:50 - \$\\begingroup\$ If the applied force was not equal to the springs restoring force at any time there would be acceleration. The spring can be compressed and have no net force on it \$\\endgroup\$ Triatticus – [Triatticus](https://physics.stackexchange.com/users/23615/triatticus "247 reputation") 2018-12-17 16:51:32 +00:00 Commented Dec 17, 2018 at 16:51 - \$\\begingroup\$ If the spring is compressed at an accelerating rate, the applied force and restoring force are not equal. Would x be proportional to the applied force or the restoring force? \$\\endgroup\$ helpme – [helpme](https://physics.stackexchange.com/users/212622/helpme "393 reputation") 2018-12-17 18:19:09 +00:00 Commented Dec 17, 2018 at 18:19 - \$\\begingroup\$ Newton's 3rd law requires that the force and opposing force be equal at all times. \$\\endgroup\$ David White – [David White](https://physics.stackexchange.com/users/86822/david-white "12,662 reputation") 2018-12-17 20:35:42 +00:00 Commented Dec 17, 2018 at 20:35 - \$\\begingroup\$ The 3rd law partner forces act on different bodies \$\\endgroup\$ helpme – [helpme](https://physics.stackexchange.com/users/212622/helpme "393 reputation") 2018-12-17 20:39:33 +00:00 Commented Dec 17, 2018 at 20:39 [Add a comment](https://physics.stackexchange.com/questions/448935/restoring-force-vs-applied-force-for-hookes-law "Use comments to ask for more information or suggest improvements. Avoid answering questions in comments.") \\| ## 3 Answers 3 Sorted by: [Reset to default](https://physics.stackexchange.com/questions/448935/restoring-force-vs-applied-force-for-hookes-law?answertab=scoredesc#tab-top) 4 \$\\begingroup\$ > Would x be proportional to the applied force or the restoring force? The equation \$F=-kx\$ is for the spring (restoring) force. If we apply a force equal to this (\$F\_{app}=kx)\$ then the spring edge where the force is applied will not accelerate (it will be stationary or moving at a constant velocity depending on what the velocity is when we set our force to start being equal to \$kx\$). Note that this is a very specific case You are right, in order to stretch/compress a spring at rest we need to initially apply a force different than \$kx\$, but this is true for other scenarios too. For example, to lift a box we need to first apply a force larger than its weight to start it's velocity upwards. Then we can reduce our force to be equal to the weight to lift the box at a constant speed, or we can keep our force constant and the box will accelerate upwards. The same analogy can be made for the spring at each position \$x\$, since the force depends on \$x\$. So overall, the spring (restoring) force is always given by \$F=-kx\$ where \$x\$ is the displacement from equilibrium. The applied force is really anything you want it to be, and depending on what it is the spring will respond accordingly based on the initial conditions. For example, we can hang a mass from a spring so that the mass remains stationary, or we can stretch and release that same mass so that oscillations occur. Notice how in either scenario the applied force is a constant \$mg\$, but the spring behavior is different due to the initial conditions of the system [Share](https://physics.stackexchange.com/a/449034 "Short permalink to this answer") Cite [Improve this answer](https://physics.stackexchange.com/posts/449034/edit) Follow [edited Dec 18, 2018 at 4:26](https://physics.stackexchange.com/posts/449034/revisions "show all edits to this post") answered Dec 18, 2018 at 3:48 [![BioPhysicist's user avatar](https://i.sstatic.net/uGTTK.png?s=64)](https://physics.stackexchange.com/users/179151/biophysicist) [BioPhysicist](https://physics.stackexchange.com/users/179151/biophysicist) 59\.5k1919 gold badges119119 silver badges199199 bronze badges \$\\endgroup\$ 0 [Add a comment](https://physics.stackexchange.com/questions/448935/restoring-force-vs-applied-force-for-hookes-law "Use comments to ask for more information or suggest improvements. Avoid comments like “+1” or “thanks”.") \\| 0 \$\\begingroup\$ I think that for your question, the sign of the force is not something that matters. What really matters is that an unbalance between the applied force and the elastic restoring force is effectively needed in order for the applied force to accelerate the spring. If you think of this problem as static then the it takes infinite time for the applied force to compress the spring. This is the only case in which you can consider the applied and the restoring elastic force equal. [Share](https://physics.stackexchange.com/a/449020 "Short permalink to this answer") Cite [Improve this answer](https://physics.stackexchange.com/posts/449020/edit) Follow answered Dec 18, 2018 at 2:25 [![nodarkside's user avatar](https://i.sstatic.net/J1QBM.jpg?s=64)](https://physics.stackexchange.com/users/186231/nodarkside) [nodarkside](https://physics.stackexchange.com/users/186231/nodarkside) 35522 silver badges99 bronze badges \$\\endgroup\$ [Add a comment](https://physics.stackexchange.com/questions/448935/restoring-force-vs-applied-force-for-hookes-law "Use comments to ask for more information or suggest improvements. Avoid comments like “+1” or “thanks”.") \\| 0 \$\\begingroup\$ When we say \$F=kx\$, we are essentially saying that based on \$kx\$ of the spring, we can determine the applied force. The same goes for an accelerating object. The \$ma\$ of the object will be equivalent to the applied force. In response, the spring exerts a restoring force back to its equilibrium position whether it is stretched or compressed. Since we are taking the direction of applied force to be positive, the direction of the restoring force will thus be negative, given by \$F=-kx\$. [Share](https://physics.stackexchange.com/a/449051 "Short permalink to this answer") Cite [Improve this answer](https://physics.stackexchange.com/posts/449051/edit) Follow answered Dec 18, 2018 at 7:21 [![QuIcKmAtHs's user avatar](https://www.gravatar.com/avatar/074040db5e59bc608cfcebf9b259ebc7?s=64&d=identicon&r=PG)](https://physics.stackexchange.com/users/179737/quickmaths) [QuIcKmAtHs](https://physics.stackexchange.com/users/179737/quickmaths) 3,79344 gold badges2121 silver badges4040 bronze badges \$\\endgroup\$ [Add a comment](https://physics.stackexchange.com/questions/448935/restoring-force-vs-applied-force-for-hookes-law "Use comments to ask for more information or suggest improvements. Avoid comments like “+1” or “thanks”.") \\| Start asking to get answers Find the answer to your question by asking. 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Readable Markdown	1 \$\\begingroup\$ According to Hooke's Law, \$F=-kx\$ where \$F\$ is the restoring force, \$k\$ is the spring constant and \$x\$ is the length of extension/compression. When an applied force compresses a spring, a restoring force will act in the opposite direction. When a spring is compressed and is in equilibrium (not extending/compressing), the restoring force should be equal to the applied force since the resultant force is 0 and the acceleration is 0. (Thus the spring is not extending/compressing) However, the act of compressing a spring will require the applied force to be greater than the restoring force even at an infinitesimally small time interval, am I wrong? Is it more correct to say that \$F=kx\$ where \$F\$ is the applied force or \$F=-kx\$ where \$F\$ is the restoring force? I think that \$F=-kx\$ the better option, but I'm pretty confused here. Does it matter at all? [![Qmechanic's user avatar](https://www.gravatar.com/avatar/86f4f64980b4736b1f6e5e3056d08e15?s=64&d=identicon&r=PG)](https://physics.stackexchange.com/users/2451/qmechanic) [Qmechanic](https://physics.stackexchange.com/users/2451/qmechanic)♦ 226k52 gold badges651 silver badges2\.7k bronze badges asked Dec 17, 2018 at 16:31 [![helpme's user avatar](https://www.gravatar.com/avatar/f19a3ac761df525d20390cf126e6219e?s=64&d=identicon&r=PG&f=y&so-version=2)](https://physics.stackexchange.com/users/212622/helpme) \$\\endgroup\$ 5 4 \$\\begingroup\$ > Would x be proportional to the applied force or the restoring force? The equation \$F=-kx\$ is for the spring (restoring) force. If we apply a force equal to this (\$F\_{app}=kx)\$ then the spring edge where the force is applied will not accelerate (it will be stationary or moving at a constant velocity depending on what the velocity is when we set our force to start being equal to \$kx\$). Note that this is a very specific case You are right, in order to stretch/compress a spring at rest we need to initially apply a force different than \$kx\$, but this is true for other scenarios too. For example, to lift a box we need to first apply a force larger than its weight to start it's velocity upwards. Then we can reduce our force to be equal to the weight to lift the box at a constant speed, or we can keep our force constant and the box will accelerate upwards. The same analogy can be made for the spring at each position \$x\$, since the force depends on \$x\$. So overall, the spring (restoring) force is always given by \$F=-kx\$ where \$x\$ is the displacement from equilibrium. The applied force is really anything you want it to be, and depending on what it is the spring will respond accordingly based on the initial conditions. For example, we can hang a mass from a spring so that the mass remains stationary, or we can stretch and release that same mass so that oscillations occur. Notice how in either scenario the applied force is a constant \$mg\$, but the spring behavior is different due to the initial conditions of the system answered Dec 18, 2018 at 3:48 [![BioPhysicist's user avatar](https://i.sstatic.net/uGTTK.png?s=64)](https://physics.stackexchange.com/users/179151/biophysicist) \$\\endgroup\$ 0 0 \$\\begingroup\$ I think that for your question, the sign of the force is not something that matters. What really matters is that an unbalance between the applied force and the elastic restoring force is effectively needed in order for the applied force to accelerate the spring. If you think of this problem as static then the it takes infinite time for the applied force to compress the spring. This is the only case in which you can consider the applied and the restoring elastic force equal. answered Dec 18, 2018 at 2:25 [![nodarkside's user avatar](https://i.sstatic.net/J1QBM.jpg?s=64)](https://physics.stackexchange.com/users/186231/nodarkside) \$\\endgroup\$ 0 \$\\begingroup\$ When we say \$F=kx\$, we are essentially saying that based on \$kx\$ of the spring, we can determine the applied force. The same goes for an accelerating object. The \$ma\$ of the object will be equivalent to the applied force. In response, the spring exerts a restoring force back to its equilibrium position whether it is stretched or compressed. Since we are taking the direction of applied force to be positive, the direction of the restoring force will thus be negative, given by \$F=-kx\$. answered Dec 18, 2018 at 7:21 [![QuIcKmAtHs's user avatar](https://www.gravatar.com/avatar/074040db5e59bc608cfcebf9b259ebc7?s=64&d=identicon&r=PG)](https://physics.stackexchange.com/users/179737/quickmaths) \$\\endgroup\$ Start asking to get answers Find the answer to your question by asking. [Ask question](https://physics.stackexchange.com/questions/ask) Explore related questions See similar questions with these tags.
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