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1. Shard Calculation

Query:

Response:

Calculated Shard: 18 (from laksa019)

2. Crawled Status Check

Query:

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/86895/hydrostatic-pressure\')), getAhrefsUnparsedNoserviceFromURL(\'https://physics.stackexchange.com/questions/86895/hydrostatic-pressure\'))) FORMAT JSONEachRow'

Response:

{"found_url":"https:\/\/physics.stackexchange.com\/questions\/86895\/hydrostatic-pressure","crawl_time":1772858089,"first_indexed_time":0,"http_code":200,"src_unparsed":"com,stackexchange!physics,\/questions\/86895\/hydrostatic-pressure s443","src_root_hash":"8045678284012640218","history_drop_reason":null,"meta_title":"fluid dynamics - Hydrostatic pressure? - Physics Stack Exchange","meta_descriptions":[],"attrs_boilerpipe_text":"5\n$\\begingroup$\nWhat what I understand, hydrostatic pressure is the \"weight\" of the water pushing against objects. But if this is true, why is hydrostatic pressure perpendicular to the surface it acts on instead of always going down?\nFor example, if you placed a book on a desk, the book's weight would push against the desk, but gravity is pulling it \"down\". But if you put another book beside this book the first book wouldn't apply any force on it. The weight exerted from the book always\ngoes down\n.\nSimilarly, why doesn't hydrostatic pressure always go down? I understand that if you put a plate on the seabed, the weight of the water would push down on it, but if you just had a vertical plate standing on the bed, why would force push it from the sides?\nTo sum up, why does hydrostatic pressure act perpendicular to the object instead of always down?\nasked\nNov 17, 2013 at 22:26\n$\\endgroup$\n1\n7\n$\\begingroup$\nBut if this is true, why is hydrostatic pressure perpendicular to the surface it acts on instead of always going down?\nBecause of the properties of a fluid.  A fluid will not tolerate directionality in force.  If you have a jug of liquid and poke a hole at a surface below the water level, the fluid is ready to flow out in any direction, left, up, down, wherever.  It will have the same push in any direction.\nYou confusion might be cleared up by trigonometry.  If a surface on the bottom of a tank is slanted, then the force\npushing down\nis the pressure times $\\cos{\\theta}$, if $\\theta$ is the slope.  Mentally deconstruct the force vector into two - a downward component and a sideways.  The downward component fits the model that you want pressure to fit.  It is equal to the weight for fluid above it.  The sideways force is also there, but it doesn't really change things.  Since it's sideways, it won't affect the weight of a jug of water you're holding.  All of the sideways force components will cancel each other out so that the weight felt by the bottom surface is exactly equal to the weight of the fluid in the container.\nanswered\nNov 17, 2013 at 23:01\n$\\endgroup$\n1\n0\n$\\begingroup$\nHydrostatic pressure concers pressure that happens on perfect fluids in equilibrium. A perfect fluid is slippery, devoid of viscosity: when it is in equilibrium, it cannot exert nor resist shear (tangential force). Therefore, on the walls of a vessel sustaining a perfect fluid at rest, the normal component is solely responsible for resisting the weight of the liquid.\nanswered\nAug 20, 2021 at 13:32\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\/86895\/hydrostatic-pressure#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%2F86895%2Fhydrostatic-pressure) or [log in](https:\/\/physics.stackexchange.com\/users\/login?ssrc=site_switcher&returnurl=https%3A%2F%2Fphysics.stackexchange.com%2Fquestions%2F86895%2Fhydrostatic-pressure) 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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Bring the best of human thought and AI automation together at your work.\n   \n   [Try for free](https:\/\/stackoverflowteams.com\/teams\/create\/free\/?utm_medium=referral&utm_source=physics-community&utm_campaign=side-bar&utm_content=explore-teams) [Learn more](https:\/\/stackoverflow.co\/internal\/?utm_medium=referral&utm_source=physics-community&utm_campaign=side-bar&utm_content=explore-teams)\n3. [Stack Internal]()\n4. Bring the best of human thought and AI automation together at your work. [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# [Hydrostatic pressure?](https:\/\/physics.stackexchange.com\/questions\/86895\/hydrostatic-pressure)\n[Ask Question](https:\/\/physics.stackexchange.com\/questions\/ask)\n\nAsked\n\n12 years, 3 months ago\n\nModified [4 years, 6 months ago](https:\/\/physics.stackexchange.com\/questions\/86895\/hydrostatic-pressure?lastactivity \"2021-08-20 13:32:04Z\")\n\nViewed 5k times\n\n5\n\n\\$\\\\begingroup\\$\n\nWhat what I understand, hydrostatic pressure is the \"weight\" of the water pushing against objects. But if this is true, why is hydrostatic pressure perpendicular to the surface it acts on instead of always going down?\n\nFor example, if you placed a book on a desk, the book's weight would push against the desk, but gravity is pulling it \"down\". But if you put another book beside this book the first book wouldn't apply any force on it. The weight exerted from the book always *goes down*.\n\nSimilarly, why doesn't hydrostatic pressure always go down? I understand that if you put a plate on the seabed, the weight of the water would push down on it, but if you just had a vertical plate standing on the bed, why would force push it from the sides?\n\nTo sum up, why does hydrostatic pressure act perpendicular to the object instead of always down?\n\n- [fluid-dynamics](https:\/\/physics.stackexchange.com\/questions\/tagged\/fluid-dynamics \"show questions tagged 'fluid-dynamics'\")\n- [pressure](https:\/\/physics.stackexchange.com\/questions\/tagged\/pressure \"show questions tagged 'pressure'\")\n- [water](https:\/\/physics.stackexchange.com\/questions\/tagged\/water \"show questions tagged 'water'\")\n\n[Share](https:\/\/physics.stackexchange.com\/q\/86895 \"Short permalink to this question\")\n\nCite\n\n[Improve this question](https:\/\/physics.stackexchange.com\/posts\/86895\/edit)\n\nFollow\n\nasked Nov 17, 2013 at 22:26\n\n[![dfg's user avatar](https:\/\/www.gravatar.com\/avatar\/c32f1df155b1e911ca216fa0426c1b08?s=64&d=identicon&r=PG)](https:\/\/physics.stackexchange.com\/users\/28118\/dfg)\n\n[dfg](https:\/\/physics.stackexchange.com\/users\/28118\/dfg)\n\n2,04977 gold badges3939 silver badges5757 bronze badges\n\n\\$\\\\endgroup\\$\n\n1\n\n- 1\n  \\$\\\\begingroup\\$ For the same reason hydraulic hoses work, with bends in them. Pressure in fluids can go around corners. If it couldn't, any sort of tubing would be useless, from hydroelectric sluice gates, to bent straws. \\$\\\\endgroup\\$\n  \n  Mike Dunlavey\n  \n  – [Mike Dunlavey](https:\/\/physics.stackexchange.com\/users\/5223\/mike-dunlavey \"17,428 reputation\")\n  \n  2013-11-18 21:18:46 +00:00\n  \n  Commented Nov 18, 2013 at 21:18\n\n[Add a comment](https:\/\/physics.stackexchange.com\/questions\/86895\/hydrostatic-pressure \"Use comments to ask for more information or suggest improvements. Avoid answering questions in comments.\")   \\|\n\n## 2 Answers 2\nSorted by:\n\n[Reset to default](https:\/\/physics.stackexchange.com\/questions\/86895\/hydrostatic-pressure?answertab=scoredesc#tab-top)\n\n7\n\n\\$\\\\begingroup\\$\n\n> But if this is true, why is hydrostatic pressure perpendicular to the surface it acts on instead of always going down?\n\nBecause of the properties of a fluid. A fluid will not tolerate directionality in force. If you have a jug of liquid and poke a hole at a surface below the water level, the fluid is ready to flow out in any direction, left, up, down, wherever. It will have the same push in any direction.\n\nYou confusion might be cleared up by trigonometry. If a surface on the bottom of a tank is slanted, then the force *pushing down* is the pressure times \\$\\\\cos{\\\\theta}\\$, if \\$\\\\theta\\$ is the slope. Mentally deconstruct the force vector into two - a downward component and a sideways. The downward component fits the model that you want pressure to fit. It is equal to the weight for fluid above it. The sideways force is also there, but it doesn't really change things. Since it's sideways, it won't affect the weight of a jug of water you're holding. All of the sideways force components will cancel each other out so that the weight felt by the bottom surface is exactly equal to the weight of the fluid in the container.\n\n[Share](https:\/\/physics.stackexchange.com\/a\/86899 \"Short permalink to this answer\")\n\nCite\n\n[Improve this answer](https:\/\/physics.stackexchange.com\/posts\/86899\/edit)\n\nFollow\n\nanswered Nov 17, 2013 at 23:01\n\n[![Alan Rominger's user avatar](https:\/\/www.gravatar.com\/avatar\/1e2e8e9b67646db490fc27d32e015805?s=64&d=identicon&r=PG)](https:\/\/physics.stackexchange.com\/users\/1255\/alan-rominger)\n\n[Alan Rominger](https:\/\/physics.stackexchange.com\/users\/1255\/alan-rominger)\n\n21\\.3k1010 gold badges6363 silver badges146146 bronze badges\n\n\\$\\\\endgroup\\$\n\n1\n\n- \\$\\\\begingroup\\$ But why if we place an object (e.g. a box) inside a liquid it would also experience a force from the bottom side? I can't understand this. Wouldn't we assume that the force comes from the weight of the liquid above the object? \\$\\\\endgroup\\$\n  \n  Antonios Sarikas\n  \n  – [Antonios Sarikas](https:\/\/physics.stackexchange.com\/users\/208825\/antonios-sarikas \"1,738 reputation\")\n  \n  2020-06-26 09:52:24 +00:00\n  \n  Commented Jun 26, 2020 at 9:52\n\n[Add a comment](https:\/\/physics.stackexchange.com\/questions\/86895\/hydrostatic-pressure \"Use comments to ask for more information or suggest improvements. Avoid comments like “+1” or “thanks”.\")   \\|\n\n0\n\n\\$\\\\begingroup\\$\n\nHydrostatic pressure concers pressure that happens on perfect fluids in equilibrium. A perfect fluid is slippery, devoid of viscosity: when it is in equilibrium, it cannot exert nor resist shear (tangential force). Therefore, on the walls of a vessel sustaining a perfect fluid at rest, the normal component is solely responsible for resisting the weight of the liquid.\n\n[Share](https:\/\/physics.stackexchange.com\/a\/660510 \"Short permalink to this answer\")\n\nCite\n\n[Improve this answer](https:\/\/physics.stackexchange.com\/posts\/660510\/edit)\n\nFollow\n\nanswered Aug 20, 2021 at 13:32\n\n[![rdabra's user avatar](https:\/\/www.gravatar.com\/avatar\/294eb826ae336b2871d002c481fcea14?s=64&d=identicon&r=PG&f=y&so-version=2)](https:\/\/physics.stackexchange.com\/users\/232698\/rdabra)\n\n[rdabra](https:\/\/physics.stackexchange.com\/users\/232698\/rdabra)\n\n10322 bronze badges\n\n\\$\\\\endgroup\\$\n\n[Add a comment](https:\/\/physics.stackexchange.com\/questions\/86895\/hydrostatic-pressure \"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- [fluid-dynamics](https:\/\/physics.stackexchange.com\/questions\/tagged\/fluid-dynamics \"show questions tagged 'fluid-dynamics'\")\n- [pressure](https:\/\/physics.stackexchange.com\/questions\/tagged\/pressure \"show questions tagged 'pressure'\")\n- [water](https:\/\/physics.stackexchange.com\/questions\/tagged\/water \"show questions tagged 'water'\")\n\nSee similar questions with these tags.\n\n- The Overflow Blog\n- [What’s new at Stack Overflow: March 2026](https:\/\/stackoverflow.blog\/2026\/03\/02\/what-s-new-at-stack-overflow-march-2026\/)\n- Featured on Meta\n- [Logo updates to Stack Overflow's visual identity](https:\/\/meta.stackexchange.com\/questions\/417394\/logo-updates-to-stack-overflows-visual-identity)\n\n#### Linked\n[3](https:\/\/physics.stackexchange.com\/questions\/500341\/why-the-force-exerted-by-a-fluid-on-an-object-submerged-in-it-is-always-perpendi \"Question score (upvotes - downvotes)\")\n\n[Why the force exerted by a fluid on an object submerged in it is always perpendicular to it's surface?](https:\/\/physics.stackexchange.com\/questions\/500341\/why-the-force-exerted-by-a-fluid-on-an-object-submerged-in-it-is-always-perpendi?noredirect=1)\n\n#### Related\n[1](https:\/\/physics.stackexchange.com\/questions\/232838\/gravity-acceleration-and-hydrostatic-pressure \"Question score (upvotes - downvotes)\")\n\n[Gravity acceleration and hydrostatic pressure](https:\/\/physics.stackexchange.com\/questions\/232838\/gravity-acceleration-and-hydrostatic-pressure)\n\n[0](https:\/\/physics.stackexchange.com\/questions\/359136\/understanding-liquid-pressure-formula-p-h-rho-g \"Question score (upvotes - downvotes)\")\n\n[Understanding liquid pressure formula, \\$P =h \\\\rho g\\$?](https:\/\/physics.stackexchange.com\/questions\/359136\/understanding-liquid-pressure-formula-p-h-rho-g)\n\n[0](https:\/\/physics.stackexchange.com\/questions\/367214\/is-atmospheric-pressure-constant-a-question-about-the-mercury-cylinder-experime \"Question score (upvotes - downvotes)\")\n\n[Is atmospheric pressure constant? 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I understand, hydrostatic pressure is the \"weight\" of the water pushing against objects. But if this is true, why is hydrostatic pressure perpendicular to the surface it acts on instead of always going down?\n\nFor example, if you placed a book on a desk, the book's weight would push against the desk, but gravity is pulling it \"down\". But if you put another book beside this book the first book wouldn't apply any force on it. The weight exerted from the book always *goes down*.\n\nSimilarly, why doesn't hydrostatic pressure always go down? I understand that if you put a plate on the seabed, the weight of the water would push down on it, but if you just had a vertical plate standing on the bed, why would force push it from the sides?\n\nTo sum up, why does hydrostatic pressure act perpendicular to the object instead of always down?\n\nasked Nov 17, 2013 at 22:26\n\n[![dfg's user avatar](https:\/\/www.gravatar.com\/avatar\/c32f1df155b1e911ca216fa0426c1b08?s=64&d=identicon&r=PG)](https:\/\/physics.stackexchange.com\/users\/28118\/dfg)\n\n\\$\\\\endgroup\\$\n\n1\n\n7\n\n\\$\\\\begingroup\\$\n\n> But if this is true, why is hydrostatic pressure perpendicular to the surface it acts on instead of always going down?\n\nBecause of the properties of a fluid. A fluid will not tolerate directionality in force. If you have a jug of liquid and poke a hole at a surface below the water level, the fluid is ready to flow out in any direction, left, up, down, wherever. It will have the same push in any direction.\n\nYou confusion might be cleared up by trigonometry. If a surface on the bottom of a tank is slanted, then the force *pushing down* is the pressure times \\$\\\\cos{\\\\theta}\\$, if \\$\\\\theta\\$ is the slope. Mentally deconstruct the force vector into two - a downward component and a sideways. The downward component fits the model that you want pressure to fit. It is equal to the weight for fluid above it. The sideways force is also there, but it doesn't really change things. Since it's sideways, it won't affect the weight of a jug of water you're holding. All of the sideways force components will cancel each other out so that the weight felt by the bottom surface is exactly equal to the weight of the fluid in the container.\n\nanswered Nov 17, 2013 at 23:01\n\n[![Alan Rominger's user avatar](https:\/\/www.gravatar.com\/avatar\/1e2e8e9b67646db490fc27d32e015805?s=64&d=identicon&r=PG)](https:\/\/physics.stackexchange.com\/users\/1255\/alan-rominger)\n\n\\$\\\\endgroup\\$\n\n1\n\n0\n\n\\$\\\\begingroup\\$\n\nHydrostatic pressure concers pressure that happens on perfect fluids in equilibrium. A perfect fluid is slippery, devoid of viscosity: when it is in equilibrium, it cannot exert nor resist shear (tangential force). Therefore, on the walls of a vessel sustaining a perfect fluid at rest, the normal component is solely responsible for resisting the weight of the liquid.\n\nanswered Aug 20, 2021 at 13:32\n\n[![rdabra's user avatar](https:\/\/www.gravatar.com\/avatar\/294eb826ae336b2871d002c481fcea14?s=64&d=identicon&r=PG&f=y&so-version=2)](https:\/\/physics.stackexchange.com\/users\/232698\/rdabra)\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	5 $\begingroup$ What what I understand, hydrostatic pressure is the "weight" of the water pushing against objects. But if this is true, why is hydrostatic pressure perpendicular to the surface it acts on instead of always going down? For example, if you placed a book on a desk, the book's weight would push against the desk, but gravity is pulling it "down". But if you put another book beside this book the first book wouldn't apply any force on it. The weight exerted from the book always goes down . Similarly, why doesn't hydrostatic pressure always go down? I understand that if you put a plate on the seabed, the weight of the water would push down on it, but if you just had a vertical plate standing on the bed, why would force push it from the sides? To sum up, why does hydrostatic pressure act perpendicular to the object instead of always down? asked Nov 17, 2013 at 22:26 $\endgroup$ 1 7 $\begingroup$ But if this is true, why is hydrostatic pressure perpendicular to the surface it acts on instead of always going down? Because of the properties of a fluid. A fluid will not tolerate directionality in force. If you have a jug of liquid and poke a hole at a surface below the water level, the fluid is ready to flow out in any direction, left, up, down, wherever. It will have the same push in any direction. You confusion might be cleared up by trigonometry. If a surface on the bottom of a tank is slanted, then the force pushing down is the pressure times $\cos{\theta}$, if $\theta$ is the slope. Mentally deconstruct the force vector into two - a downward component and a sideways. The downward component fits the model that you want pressure to fit. It is equal to the weight for fluid above it. The sideways force is also there, but it doesn't really change things. Since it's sideways, it won't affect the weight of a jug of water you're holding. All of the sideways force components will cancel each other out so that the weight felt by the bottom surface is exactly equal to the weight of the fluid in the container. answered Nov 17, 2013 at 23:01 $\endgroup$ 1 0 $\begingroup$ Hydrostatic pressure concers pressure that happens on perfect fluids in equilibrium. A perfect fluid is slippery, devoid of viscosity: when it is in equilibrium, it cannot exert nor resist shear (tangential force). Therefore, on the walls of a vessel sustaining a perfect fluid at rest, the normal component is solely responsible for resisting the weight of the liquid. answered Aug 20, 2021 at 13:32 $\endgroup$ Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions See similar questions with these tags.
Markdown	[Skip to main content](https://physics.stackexchange.com/questions/86895/hydrostatic-pressure#content) #### Stack Exchange Network Stack 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. 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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) # [Hydrostatic pressure?](https://physics.stackexchange.com/questions/86895/hydrostatic-pressure) [Ask Question](https://physics.stackexchange.com/questions/ask) Asked 12 years, 3 months ago Modified [4 years, 6 months ago](https://physics.stackexchange.com/questions/86895/hydrostatic-pressure?lastactivity "2021-08-20 13:32:04Z") Viewed 5k times 5 \$\\begingroup\$ What what I understand, hydrostatic pressure is the "weight" of the water pushing against objects. But if this is true, why is hydrostatic pressure perpendicular to the surface it acts on instead of always going down? For example, if you placed a book on a desk, the book's weight would push against the desk, but gravity is pulling it "down". But if you put another book beside this book the first book wouldn't apply any force on it. The weight exerted from the book always goes down. Similarly, why doesn't hydrostatic pressure always go down? I understand that if you put a plate on the seabed, the weight of the water would push down on it, but if you just had a vertical plate standing on the bed, why would force push it from the sides? To sum up, why does hydrostatic pressure act perpendicular to the object instead of always down? - [fluid-dynamics](https://physics.stackexchange.com/questions/tagged/fluid-dynamics "show questions tagged 'fluid-dynamics'") - [pressure](https://physics.stackexchange.com/questions/tagged/pressure "show questions tagged 'pressure'") - [water](https://physics.stackexchange.com/questions/tagged/water "show questions tagged 'water'") [Share](https://physics.stackexchange.com/q/86895 "Short permalink to this question") Cite [Improve this question](https://physics.stackexchange.com/posts/86895/edit) Follow asked Nov 17, 2013 at 22:26 [![dfg's user avatar](https://www.gravatar.com/avatar/c32f1df155b1e911ca216fa0426c1b08?s=64&d=identicon&r=PG)](https://physics.stackexchange.com/users/28118/dfg) [dfg](https://physics.stackexchange.com/users/28118/dfg) 2,04977 gold badges3939 silver badges5757 bronze badges \$\\endgroup\$ 1 - 1 \$\\begingroup\$ For the same reason hydraulic hoses work, with bends in them. Pressure in fluids can go around corners. If it couldn't, any sort of tubing would be useless, from hydroelectric sluice gates, to bent straws. \$\\endgroup\$ Mike Dunlavey – [Mike Dunlavey](https://physics.stackexchange.com/users/5223/mike-dunlavey "17,428 reputation") 2013-11-18 21:18:46 +00:00 Commented Nov 18, 2013 at 21:18 [Add a comment](https://physics.stackexchange.com/questions/86895/hydrostatic-pressure "Use comments to ask for more information or suggest improvements. Avoid answering questions in comments.") \\| ## 2 Answers 2 Sorted by: [Reset to default](https://physics.stackexchange.com/questions/86895/hydrostatic-pressure?answertab=scoredesc#tab-top) 7 \$\\begingroup\$ > But if this is true, why is hydrostatic pressure perpendicular to the surface it acts on instead of always going down? Because of the properties of a fluid. A fluid will not tolerate directionality in force. If you have a jug of liquid and poke a hole at a surface below the water level, the fluid is ready to flow out in any direction, left, up, down, wherever. It will have the same push in any direction. You confusion might be cleared up by trigonometry. If a surface on the bottom of a tank is slanted, then the force pushing down is the pressure times \$\\cos{\\theta}\$, if \$\\theta\$ is the slope. Mentally deconstruct the force vector into two - a downward component and a sideways. The downward component fits the model that you want pressure to fit. It is equal to the weight for fluid above it. The sideways force is also there, but it doesn't really change things. Since it's sideways, it won't affect the weight of a jug of water you're holding. All of the sideways force components will cancel each other out so that the weight felt by the bottom surface is exactly equal to the weight of the fluid in the container. [Share](https://physics.stackexchange.com/a/86899 "Short permalink to this answer") Cite [Improve this answer](https://physics.stackexchange.com/posts/86899/edit) Follow answered Nov 17, 2013 at 23:01 [![Alan Rominger's user avatar](https://www.gravatar.com/avatar/1e2e8e9b67646db490fc27d32e015805?s=64&d=identicon&r=PG)](https://physics.stackexchange.com/users/1255/alan-rominger) [Alan Rominger](https://physics.stackexchange.com/users/1255/alan-rominger) 21\.3k1010 gold badges6363 silver badges146146 bronze badges \$\\endgroup\$ 1 - \$\\begingroup\$ But why if we place an object (e.g. a box) inside a liquid it would also experience a force from the bottom side? I can't understand this. Wouldn't we assume that the force comes from the weight of the liquid above the object? \$\\endgroup\$ Antonios Sarikas – [Antonios Sarikas](https://physics.stackexchange.com/users/208825/antonios-sarikas "1,738 reputation") 2020-06-26 09:52:24 +00:00 Commented Jun 26, 2020 at 9:52 [Add a comment](https://physics.stackexchange.com/questions/86895/hydrostatic-pressure "Use comments to ask for more information or suggest improvements. Avoid comments like “+1” or “thanks”.") \\| 0 \$\\begingroup\$ Hydrostatic pressure concers pressure that happens on perfect fluids in equilibrium. A perfect fluid is slippery, devoid of viscosity: when it is in equilibrium, it cannot exert nor resist shear (tangential force). Therefore, on the walls of a vessel sustaining a perfect fluid at rest, the normal component is solely responsible for resisting the weight of the liquid. [Share](https://physics.stackexchange.com/a/660510 "Short permalink to this answer") Cite [Improve this answer](https://physics.stackexchange.com/posts/660510/edit) Follow answered Aug 20, 2021 at 13:32 [![rdabra's user avatar](https://www.gravatar.com/avatar/294eb826ae336b2871d002c481fcea14?s=64&d=identicon&r=PG&f=y&so-version=2)](https://physics.stackexchange.com/users/232698/rdabra) [rdabra](https://physics.stackexchange.com/users/232698/rdabra) 10322 bronze badges \$\\endgroup\$ [Add a comment](https://physics.stackexchange.com/questions/86895/hydrostatic-pressure "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. [Ask question](https://physics.stackexchange.com/questions/ask) Explore related questions - [fluid-dynamics](https://physics.stackexchange.com/questions/tagged/fluid-dynamics "show questions tagged 'fluid-dynamics'") - [pressure](https://physics.stackexchange.com/questions/tagged/pressure "show questions tagged 'pressure'") - [water](https://physics.stackexchange.com/questions/tagged/water "show questions tagged 'water'") See similar questions with these tags. - The Overflow Blog - [What’s new at Stack Overflow: March 2026](https://stackoverflow.blog/2026/03/02/what-s-new-at-stack-overflow-march-2026/) - Featured on Meta - [Logo updates to Stack Overflow's visual identity](https://meta.stackexchange.com/questions/417394/logo-updates-to-stack-overflows-visual-identity) #### Linked [3](https://physics.stackexchange.com/questions/500341/why-the-force-exerted-by-a-fluid-on-an-object-submerged-in-it-is-always-perpendi "Question score (upvotes - downvotes)") [Why the force exerted by a fluid on an object submerged in it is always perpendicular to it's surface?](https://physics.stackexchange.com/questions/500341/why-the-force-exerted-by-a-fluid-on-an-object-submerged-in-it-is-always-perpendi?noredirect=1) #### Related [1](https://physics.stackexchange.com/questions/232838/gravity-acceleration-and-hydrostatic-pressure "Question score (upvotes - downvotes)") [Gravity acceleration and hydrostatic pressure](https://physics.stackexchange.com/questions/232838/gravity-acceleration-and-hydrostatic-pressure) [0](https://physics.stackexchange.com/questions/359136/understanding-liquid-pressure-formula-p-h-rho-g "Question score (upvotes - downvotes)") [Understanding liquid pressure formula, \$P =h \\rho g\$?](https://physics.stackexchange.com/questions/359136/understanding-liquid-pressure-formula-p-h-rho-g) [0](https://physics.stackexchange.com/questions/367214/is-atmospheric-pressure-constant-a-question-about-the-mercury-cylinder-experime "Question score (upvotes - downvotes)") [Is atmospheric pressure constant? 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Readable Markdown	5 \$\\begingroup\$ What what I understand, hydrostatic pressure is the "weight" of the water pushing against objects. But if this is true, why is hydrostatic pressure perpendicular to the surface it acts on instead of always going down? For example, if you placed a book on a desk, the book's weight would push against the desk, but gravity is pulling it "down". But if you put another book beside this book the first book wouldn't apply any force on it. The weight exerted from the book always goes down. Similarly, why doesn't hydrostatic pressure always go down? I understand that if you put a plate on the seabed, the weight of the water would push down on it, but if you just had a vertical plate standing on the bed, why would force push it from the sides? To sum up, why does hydrostatic pressure act perpendicular to the object instead of always down? asked Nov 17, 2013 at 22:26 [![dfg's user avatar](https://www.gravatar.com/avatar/c32f1df155b1e911ca216fa0426c1b08?s=64&d=identicon&r=PG)](https://physics.stackexchange.com/users/28118/dfg) \$\\endgroup\$ 1 7 \$\\begingroup\$ > But if this is true, why is hydrostatic pressure perpendicular to the surface it acts on instead of always going down? Because of the properties of a fluid. A fluid will not tolerate directionality in force. If you have a jug of liquid and poke a hole at a surface below the water level, the fluid is ready to flow out in any direction, left, up, down, wherever. It will have the same push in any direction. You confusion might be cleared up by trigonometry. If a surface on the bottom of a tank is slanted, then the force pushing down is the pressure times \$\\cos{\\theta}\$, if \$\\theta\$ is the slope. Mentally deconstruct the force vector into two - a downward component and a sideways. The downward component fits the model that you want pressure to fit. It is equal to the weight for fluid above it. The sideways force is also there, but it doesn't really change things. Since it's sideways, it won't affect the weight of a jug of water you're holding. All of the sideways force components will cancel each other out so that the weight felt by the bottom surface is exactly equal to the weight of the fluid in the container. answered Nov 17, 2013 at 23:01 [![Alan Rominger's user avatar](https://www.gravatar.com/avatar/1e2e8e9b67646db490fc27d32e015805?s=64&d=identicon&r=PG)](https://physics.stackexchange.com/users/1255/alan-rominger) \$\\endgroup\$ 1 0 \$\\begingroup\$ Hydrostatic pressure concers pressure that happens on perfect fluids in equilibrium. A perfect fluid is slippery, devoid of viscosity: when it is in equilibrium, it cannot exert nor resist shear (tangential force). Therefore, on the walls of a vessel sustaining a perfect fluid at rest, the normal component is solely responsible for resisting the weight of the liquid. answered Aug 20, 2021 at 13:32 [![rdabra's user avatar](https://www.gravatar.com/avatar/294eb826ae336b2871d002c481fcea14?s=64&d=identicon&r=PG&f=y&so-version=2)](https://physics.stackexchange.com/users/232698/rdabra) \$\\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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