{"id":39,"date":"2023-06-26T23:01:22","date_gmt":"2023-06-26T23:01:22","guid":{"rendered":"https:\/\/physigeek.com\/tr\/newtonun-ilk-eylemsizlik-yasasi-yasasi\/"},"modified":"2023-06-26T23:01:22","modified_gmt":"2023-06-26T23:01:22","slug":"newtonun-ilk-eylemsizlik-yasasi-yasasi","status":"publish","type":"post","link":"https:\/\/physigeek.com\/tr\/newtonun-ilk-eylemsizlik-yasasi-yasasi\/","title":{"rendered":"Newton'un birinci yasas\u0131 (eylemsizlik yasas\u0131)"},"content":{"rendered":"

Bu makale, eylemsizlik yasas\u0131 olarak da bilinen Newton’un birinci yasas\u0131n\u0131n ne s\u00f6yledi\u011fini a\u00e7\u0131klamaktad\u0131r. Newton’un birinci yasas\u0131n\u0131n a\u00e7\u0131klamas\u0131na ek olarak, bu yasan\u0131n \u00f6rneklerini ve matematiksel form\u00fcl\u00fcn\u00fc g\u00f6receksiniz. Son olarak Newton’un birinci yasas\u0131n\u0131n ad\u0131m ad\u0131m \u00e7\u00f6z\u00fcml\u00fc al\u0131\u015ft\u0131rmalar\u0131yla pratik yapabileceksiniz. <\/p>\n

<\/span> Newton’un birinci yasas\u0131 nedir?<\/span><\/h2>\n

Newton’un eylemsizlik yasas\u0131 olarak da adland\u0131r\u0131lan birinci yasa a\u00e7\u0131klamas\u0131<\/strong> \u015funlar\u0131 belirtir: <\/p>\n

\n

Bir cisim \u00fczerine herhangi bir d\u0131\u015f kuvvet etki etmedi\u011fi s\u00fcrece hareketsiz kal\u0131r veya sabit h\u0131zda kal\u0131r. Ba\u015fka bir deyi\u015fle, bir cismin hareket veya dinlenme durumunu de\u011fi\u015ftirmek i\u00e7in ona bir kuvvet uygulanmas\u0131 gerekir.<\/p>\n<\/div>\n

\u00d6rne\u011fin yerde duran bir cisim, \u00fczerine bir kuvvet etki edene kadar hareket etmeyecektir.<\/p>\n

Bu nedenle, Newton’un birinci yasas\u0131, e\u011fer bir cisim d\u00fczg\u00fcn do\u011frusal hareketle hareket ediyorsa, bunun \u00fczerine hi\u00e7bir d\u0131\u015f kuvvetin etki etmedi\u011fi veya t\u00fcm sistemin bile\u015fke kuvvetinin s\u0131f\u0131r oldu\u011fu anlam\u0131na gelir.<\/p>\n

Toplamda Newton’un \u00fc\u00e7 yasas\u0131 vard\u0131r; bunlardan birine eylemsizlik ilkesi de denir, ikinci yasa veya dinami\u011fin temel ilkesi ve \u00fc\u00e7\u00fcnc\u00fc yasa veya etki ve tepki ilkesi de denir.<\/p>\n

Mant\u0131ksal olarak, \u00fc\u00e7 yasaya fizik\u00e7i Isaac Newton’un ad\u0131 verilmi\u015ftir \u00e7\u00fcnk\u00fc bunlar\u0131 Do\u011fa Felsefesinin Matematiksel \u0130lkeleri<\/em> adl\u0131 \u00e7al\u0131\u015fmas\u0131nda a\u00e7\u0131klayan ilk ki\u015fi odur. Bu yay\u0131n fizi\u011fin temel direklerinden biri olarak kabul edilir. <\/p>\n

<\/span> Newton’un Birinci Yasas\u0131na \u00d6rnekler<\/span><\/h2>\n

Newton’un birinci yasas\u0131n\u0131n (veya eylemsizlik yasas\u0131n\u0131n) tan\u0131m\u0131n\u0131 g\u00f6z \u00f6n\u00fcnde bulundurarak, a\u015fa\u011f\u0131da bu kural\u0131n birka\u00e7 \u00f6rne\u011fini analiz edece\u011fiz.<\/p>\n

    \n
  1. Newton’un birinci yasas\u0131n\u0131n a\u00e7\u0131k bir \u00f6rne\u011fi, yerde duran bir kanepedir. Kanepe \u00fczerine herhangi bir kuvvet uygulanmad\u0131\u011f\u0131 takdirde hareket etmeyecek ve sabit kalacakt\u0131r. Ancak kanepe yeterince b\u00fcy\u00fck bir kuvvetle itilirse kanepe h\u0131z kazanacak ve dolay\u0131s\u0131yla hareket durumu de\u011fi\u015fecektir.<\/span><\/li>\n
  2. Newton’un birinci yasas\u0131n\u0131n bir ba\u015fka \u00f6rne\u011fi, uzayda sabit bir h\u0131zla hareket eden bir uzay sondas\u0131n\u0131n \u00f6rne\u011fidir. Gezegenlerin \u00e7ekim etkisi a\u015f\u0131ld\u0131\u011f\u0131nda uzayda s\u00fcrt\u00fcnme veya ba\u015fka bir kuvvet kalmaz. Bu nedenle, bir uzay sondas\u0131 uzayda sabit bir h\u0131zla hareket eder \u00e7\u00fcnk\u00fc ona etki eden bir kuvvet yoktur.<\/span><\/li>\n
  3. Sabit h\u0131zla hareket eden bir araba ayn\u0131 zamanda Newton’un birinci yasas\u0131na (ya da eylemsizlik yasas\u0131na) bir \u00f6rnektir, \u00e7\u00fcnk\u00fc ivmelenmeden hareket etmek i\u00e7in ortaya \u00e7\u0131kan kuvvetin s\u0131f\u0131r olmas\u0131 gerekir. Araba ileri do\u011fru hareket etti\u011finde, harekete kar\u015f\u0131 bir s\u00fcrt\u00fcnme kuvveti ona etki eder, dolay\u0131s\u0131yla sabit bir h\u0131zda hareket etmek i\u00e7in araban\u0131n motorunun ayn\u0131 b\u00fcy\u00fckl\u00fckte ve y\u00f6nde ancak ters y\u00f6nde bir kuvvet uygulamas\u0131 gerekir. Bu \u015fekilde iki kuvvet birbirine kar\u015f\u0131 \u00e7\u0131kar ve araba ayn\u0131 h\u0131zda hareket eder.<\/span> <\/li>\n<\/ol>\n

    <\/span> Newton yasas\u0131n\u0131n ilk form\u00fcl\u00fc<\/span><\/h2>\n

    Newton’un birinci yasas\u0131 kavram\u0131n\u0131 derinle\u015ftirmek i\u00e7in bu b\u00f6l\u00fcmde s\u00f6z konusu yasan\u0131n ifade edilebilece\u011fi form\u00fcl\u00fc g\u00f6rece\u011fiz.<\/p>\n

    Matematiksel olarak Newton’un birinci yasas\u0131n\u0131n form\u00fcl\u00fc,<\/strong> bir sistemin kuvvetlerinin toplam\u0131 s\u0131f\u0131rsa, s\u00f6z konusu sistemin ivmesinin de s\u0131f\u0131r olaca\u011f\u0131n\u0131 belirtir. Bunun tersi de do\u011frudur.<\/p>\n

    \n

    \"\\displaystyle<\/p>\n<\/p>\n

    Benzer \u015fekilde, e\u011fer kuvvetlerin toplam\u0131 s\u0131f\u0131rsa, bu momentumun (veya do\u011frusal momentumun) sabit oldu\u011fu anlam\u0131na gelir.<\/p>\n

    \n

    \"\\displaystyle<\/p>\n<\/p>\n

    Her hal\u00fckarda bu ifadeler cebir yoluyla yasay\u0131 ifade etmekten ba\u015fka bir i\u015fe yaramaz. \u00d6nemli olan Newton’un birinci yasas\u0131n\u0131n anlam\u0131n\u0131 anlaman\u0131z ve bunun do\u011fru olmas\u0131 i\u00e7in t\u00fcm kuvvetlerin toplam\u0131n\u0131n s\u0131f\u0131r olmas\u0131 gerekti\u011fidir. <\/p>\n

    <\/span> Newton’un birinci yasas\u0131n\u0131n \u00e7\u00f6z\u00fclm\u00fc\u015f al\u0131\u015ft\u0131rmalar\u0131<\/span><\/h2>\n

    1. Egzersiz<\/h3>\n

    Bir asans\u00f6r\u00fcn 7 kg’l\u0131k bir nesneyi kald\u0131rmak i\u00e7in ne kadar kuvvet uygulamas\u0131 gerekir? <\/p>\n

    \n
    \u00e7\u00f6z\u00fcm\u00fc g\u00f6r<\/strong><\/div>\n<\/div>\n

    Bu sorunu \u00e7\u00f6zmek i\u00e7in yapmam\u0131z gereken ilk \u015fey, D\u00fcnya’n\u0131n cisme uygulad\u0131\u011f\u0131 yer\u00e7ekimi kuvvetini hesaplamakt\u0131r. Bunu yapmak i\u00e7in a\u011f\u0131rl\u0131k kuvveti form\u00fcl\u00fcn\u00fc kullan\u0131r\u0131z:<\/p>\n

    \n

    \"P=m\\cdot<\/p>\n<\/p>\n

    Yani Newton’un birinci kanununa g\u00f6re asans\u00f6r yukar\u0131ya do\u011fru 68,67 N’luk bir dikey kuvvet uygularsa ortaya \u00e7\u0131kan kuvvet s\u0131f\u0131r olaca\u011f\u0131ndan cisim sabit kalacakt\u0131r. Bu nedenle asans\u00f6r\u00fcn y\u00fckselmeye ba\u015flayabilmesi i\u00e7in 68,67 N’den daha b\u00fcy\u00fck bir kuvvet uygulamas\u0131 gerekir.<\/p>\n

    <\/div>\n

    Al\u0131\u015ft\u0131rma 2<\/h3>\n

    Bir asans\u00f6r k\u00fctlesi 100 kg olan bir cismi kald\u0131r\u0131yor. Herhangi bir anda harekete kar\u015f\u0131 gelen s\u00fcrt\u00fcnme kuvveti 300 N ve kablonun yukar\u0131ya do\u011fru uygulad\u0131\u011f\u0131 kuvvet 1100 N’dir. Asans\u00f6r h\u0131zlan\u0131yor mu, yava\u015fl\u0131yor mu yoksa sabit h\u0131zla m\u0131 hareket ediyor? <\/p>\n

    \n
    \u00e7\u00f6z\u00fcm\u00fc g\u00f6r<\/strong><\/div>\n<\/div>\n

    \u00d6ncelikle D\u00fcnya’n\u0131n v\u00fccuda uygulad\u0131\u011f\u0131 \u00e7ekim kuvvetini a\u011f\u0131rl\u0131k form\u00fcl\u00fcyle hesapl\u0131yoruz:<\/p>\n

    \n

    \"P=m\\cdot<\/p>\n<\/p>\n

    Yani asans\u00f6r\u00fc a\u015fa\u011f\u0131ya \u00e7eken kuvvetlerin toplam\u0131:<\/p>\n

    \n

    \"F_{\\text{down}}=300+981=1281<\/p>\n<\/p>\n

    Asans\u00f6r\u00fc yukar\u0131ya iten tek kuvvet ise halat kuvvetidir.<\/p>\n

    \n

    \"F_{\\text{up}}=1100<\/p>\n<\/p>\n

    Yani a\u015fa\u011f\u0131 do\u011fru olan kuvvetlerin toplam\u0131 yukar\u0131 do\u011fru olan kuvvetlerden daha b\u00fcy\u00fck oldu\u011fundan asans\u00f6r bu noktada yava\u015flar. <\/p>\n

    \n

    \"F_{\\text{down}}F_{\\text{up}} \\ \\longrightarrow \\ \\text{freins d’ascenseur}” title=”Rendered by QuickLaTeX.com” height=”19″ width=”282″ style=”vertical-align: -6px;”><\/p>\n<\/p>\n

    <\/div>\n

    Al\u0131\u015ft\u0131rma 3<\/h3>\n

    K\u00fctlesi 60 kg olan bir kutu yerle 30 derecelik a\u00e7\u0131 yapan bir halatla s\u00fcr\u00fckleniyor. Kutuyu 10 m\/s sabit h\u0131zla hareket ettirmek i\u00e7in ipe 120 N’luk bir kuvvet uygulanmas\u0131 gerekiyorsa, kutu ile yer aras\u0131ndaki kinetik s\u00fcrt\u00fcnme katsay\u0131s\u0131n\u0131n yakla\u015f\u0131k de\u011feri nedir? <\/p>\n

    \n
    \"Newton<\/figure>\n<\/div>\n
    \n
    \u00e7\u00f6z\u00fcm\u00fc g\u00f6r<\/strong><\/div>\n<\/div>\n

    Uygulanan kuvvetin e\u011fim a\u00e7\u0131s\u0131n\u0131 bildi\u011fimizden, trigonometrik oranlar\u0131 kullanarak onu dikey kuvvet ve yatay kuvvet olarak ay\u0131rabiliriz:<\/p>\n

    \n

    \"F_x=120\\cdot<\/p>\n<\/p>\n

    \n

    \"F_y=120\\cdot<\/p>\n<\/p>\n

    Di\u011fer taraftan D\u00fcnya’n\u0131n kutuya uygulad\u0131\u011f\u0131 a\u011f\u0131rl\u0131\u011f\u0131n kuvvetini de hesapl\u0131yoruz:<\/p>\n

    \n

    \"P=m\\cdot<\/p>\n<\/p>\n

    Buna g\u00f6re sistemin serbest cisim diyagram\u0131 \u015fu \u015fekildedir: <\/p>\n

    \n
    \"Newton'un<\/figure>\n<\/div>\n

    G\u00f6sterilen F x<\/sub> ve F y<\/sub> kuvvetlerinin sadece 120 N’luk kuvvetin ayr\u0131\u015f\u0131m\u0131 oldu\u011funu, yani ayn\u0131 anda etki etmediklerini, bunun yerine iki kuvvetin 120 N’luk kuvvetin yerini ald\u0131\u011f\u0131n\u0131 unutmay\u0131n.<\/p>\n

    Kutu sabit bir h\u0131zla hareket etti\u011finden bu onun dengede oldu\u011fu anlam\u0131na gelir, dolay\u0131s\u0131yla egzersizi \u00e7\u00f6zmek i\u00e7in denge ko\u015fullar\u0131n\u0131 uygulayabiliriz. Normal kuvveti bulmak i\u00e7in \u00f6ncelikle dikey denge denklemini kuruyoruz: <\/p>\n

    \n

    \"\\somme<\/p>\n<\/p>\n

    \n

    \"N+F_y-P=0\"<\/p>\n<\/p>\n

    \n

    \"N=P-F_y\"<\/p>\n<\/p>\n

    \n

    \"N=588,6-60=528,6<\/p>\n<\/p>\n

    Ve son olarak s\u00fcrt\u00fcnme katsay\u0131s\u0131n\u0131 belirlemek i\u00e7in yatay denge denklemine dikkat edelim: <\/p>\n

    \n

    \"\\somme<\/p>\n<\/p>\n

    \n

    \"F_x-F_{\\mu}=0\"<\/p>\n<\/p>\n

    \n

    \"F_x-\\mu<\/p>\n<\/p>\n

    \n

    \"103,92-\\mu<\/p>\n<\/p>\n

    \n

    \"103,92=\\mu<\/p>\n<\/p>\n

    \n

    \"\\mu=\\cfrac{103,92}{528,6}=0,20\"<\/p>\n<\/p>\n

    <\/div>\n","protected":false},"excerpt":{"rendered":"

    Bu makale, eylemsizlik yasas\u0131 olarak da bilinen Newton’un birinci yasas\u0131n\u0131n ne s\u00f6yledi\u011fini a\u00e7\u0131klamaktad\u0131r. Newton’un birinci yasas\u0131n\u0131n a\u00e7\u0131klamas\u0131na ek olarak, bu yasan\u0131n \u00f6rneklerini ve matematiksel form\u00fcl\u00fcn\u00fc g\u00f6receksiniz. Son olarak Newton’un birinci yasas\u0131n\u0131n ad\u0131m ad\u0131m \u00e7\u00f6z\u00fcml\u00fc al\u0131\u015ft\u0131rmalar\u0131yla pratik yapabileceksiniz. Newton’un birinci yasas\u0131 nedir? Newton’un eylemsizlik yasas\u0131 olarak da adland\u0131r\u0131lan birinci yasa a\u00e7\u0131klamas\u0131 \u015funlar\u0131 belirtir: Bir cisim \u00fczerine …<\/p>\n

    Newton'un birinci yasas\u0131 (eylemsizlik yasas\u0131)<\/span> Devam\u0131 »<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"","footnotes":""},"categories":[5],"tags":[],"class_list":["post-39","post","type-post","status-publish","format-standard","hentry","category-dinamik"],"yoast_head":"\n\u25b7 Newton'un birinci yasas\u0131: ifade, \u00f6rnekler ve al\u0131\u015ft\u0131rmalar<\/title>\n<meta name=\"description\" content=\"Newton'un birinci yasas\u0131 (ya da eylemsizlik yasas\u0131) ne diyor? 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