{"id":246,"date":"2023-06-23T09:26:11","date_gmt":"2023-06-23T09:26:11","guid":{"rendered":"https:\/\/physigeek.com\/id\/kekuatan-normal\/"},"modified":"2023-06-23T09:26:11","modified_gmt":"2023-06-23T09:26:11","slug":"kekuatan-normal","status":"publish","type":"post","link":"https:\/\/physigeek.com\/id\/kekuatan-normal\/","title":{"rendered":"Kekuatan normal"},"content":{"rendered":"<p>Artikel ini menjelaskan apa itu gaya normal dan cara menentukannya tergantung pada jenis masalahnya. Dengan demikian, Anda akan menemukan karakteristik gaya normal dan, sebagai tambahan, Anda akan dapat mempraktikkan gaya jenis ini dengan latihan yang diselesaikan langkah demi langkah. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"%C2%BFQue-es-la-fuerza-normal\"><\/span>Apa itu gaya normal?<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Dalam fisika, <strong>gaya normal<\/strong> adalah gaya yang dilakukan oleh suatu permukaan pada suatu benda yang bertumpu padanya. Oleh karena itu, arah gaya normal tegak lurus permukaan dan arah gaya normal keluar, yaitu permukaan menerapkan gaya normal terhadap benda.<\/p>\n<p> Secara umum, gaya normal berfungsi untuk melawan <a href=\"https:\/\/physigeek.com\/id\/berat-fisik\/\">gaya berat<\/a> , yaitu tarikan gravitasi bumi pada benda bermassa. Namun, ketika benda bertumpu pada permukaan miring, nilai gaya normal mungkin tidak cukup. Di bawah ini kita akan melihat cara menghitung gaya normal pada bidang miring.<\/p>\n<p> Secara singkat <strong><u style=\"text-decoration-color:#4fd12f\">ciri-ciri gaya normal<\/u><\/strong> adalah:<\/p>\n<ul style=\"color:#4fd12f; font-weight: bold;\">\n<li style=\"margin-bottom:12px\"> <span style=\"color:#101010;font-weight: normal;\">Gaya normal merupakan gaya kontak, yaitu gaya yang hanya dapat diterapkan jika dua permukaan bersentuhan.<\/span><\/li>\n<li style=\"margin-bottom:12px\"> <span style=\"color:#101010;font-weight: normal;\">Arah gaya normal tegak lurus terhadap permukaan tempat benda berada.<\/span><\/li>\n<li style=\"margin-bottom:12px\"> <span style=\"color:#101010;font-weight: normal;\">Arah gaya normal selalu ke luar, karena permukaanlah yang menerapkan gaya normal pada benda.<\/span><\/li>\n<li style=\"margin-bottom:12px\"> <span style=\"color:#101010;font-weight: normal;\">Secara umum, besarnya gaya normal setara dengan proyeksi gaya yang dihasilkan pada permukaan penyangga.<\/span><\/li>\n<li> <span style=\"color:#101010;font-weight: normal;\">Biasanya, gaya normal biasanya dilambangkan dengan simbol N atau F <sub>N.<\/sub><\/span> <\/li>\n<\/ul>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Como-calcular-la-fuerza-normal\"><\/span> Cara menghitung gaya normal<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Secara umum, untuk <strong>menghitung gaya normal,<\/strong> kita harus menerapkan persamaan kesetimbangan, yang menetapkan bahwa suatu benda berada dalam kesetimbangan ketika jumlah gaya vertikal dan jumlah gaya horizontal sama dengan nol.<\/p>\n<p> Dengan menerapkan kondisi kesetimbangan pada soal, kita akan dapat menyelesaikan gaya normal dari persamaan yang diusulkan dan oleh karena itu menentukan nilai gaya normal. <\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-a9333695ba02f6e089d628fe3622a2e5_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{array}{c}\\displaystyle\\sum \\vv{F_x}=0\\\\[2ex]\\displaystyle\\sum \\vv{F_y}=0\\end{array}\" title=\"Rendered by QuickLaTeX.com\" height=\"65\" width=\"81\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Ejemplo-del-calculo-de-la-fuerza-normal\"><\/span> Contoh perhitungan gaya normal<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Setelah kita mengetahui definisi gaya normal, mari kita lihat contoh konkrit penghitungan gaya normal.<\/p>\n<ul>\n<li> Sebuah benda bermassa 8 kg berada dalam keadaan diam di tanah datar. Berapakah nilai gaya normal yang dilakukan tanah pada benda tersebut?<\/li>\n<\/ul>\n<p> Pada soal ini, benda berada dalam keadaan diam pada permukaan datar, sehingga gaya yang bekerja padanya hanyalah gaya berat dan gaya normal. <\/p>\n<figure class=\"wp-block-image aligncenter size-full is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/force-et-poids-normaux.png\" alt=\"kekuatan dan berat normal\" class=\"wp-image-4215\" width=\"273\" height=\"297\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/force-et-poids-normaux-275x300.png 275w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/force-et-poids-normaux.png 570w\" sizes=\"(max-width: 275px) 100vw, 275px\"><\/figure>\n<p> Jadi, agar suatu benda berada dalam kesetimbangan pada permukaan datar, gaya normal (N) dan gaya berat (P) harus sama. Oleh karena itu, garis normal dan berat mempunyai arah yang sama, modulus yang sama, tetapi arahnya berlawanan.<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-318d3aaff48777c13e5ac24cb775f6b0_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"N=P\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"54\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> Jadi, untuk menentukan nilai gaya normal, cukup menghitung berat benda, yang setara dengan massanya dikalikan percepatan gravitasi: <\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-a7355a420808c17875e97713c4bef5ec_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"N=P=m\\cdot g=8 \\cdot 9,81 = 78,48 \\ N\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"283\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Fuerza-normal-en-un-plano-inclinado\"><\/span> gaya normal pada bidang miring<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Pada bagian ini, kita akan memperoleh rumus gaya normal pada bidang miring, karena nilainya berubah bergantung pada apakah permukaannya datar atau miring.<\/p>\n<p> Jadi, gaya-gaya yang bekerja pada benda yang bertumpu pada bidang miring adalah sebagai berikut: <\/p>\n<figure class=\"wp-block-image aligncenter size-large is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/force-normale-sur-un-plan-incline.png\" alt=\"gaya normal pada bidang miring\" class=\"wp-image-4220\" width=\"308\" height=\"417\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/force-normale-sur-un-plan-incline-222x300.png 222w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/force-normale-sur-un-plan-incline-757x1024.png 757w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/force-normale-sur-un-plan-incline-768x1038.png 768w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/force-normale-sur-un-plan-incline.png 770w\" sizes=\"(max-width: 222px) 100vw, 222px\"><\/figure>\n<p> Perhatikan gambar di atas: Jika bidang dimiringkan, akan lebih mudah jika menggunakan arah sejajar bidang (sumbu 1) dan arah tegak lurus bidang (sumbu 2) sebagai sumbunya. Dengan cara ini lebih mudah untuk menyatakan persamaan keseimbangan.<\/p>\n<p> Untuk menghitung <strong>gaya normal pada bidang miring,<\/strong> perlu diterapkan kondisi kesetimbangan pada sumbu tegak lurus bidang miring, karena dapat dijamin bahwa benda berada dalam kesetimbangan pada sumbu tersebut tetapi tidak pada sumbu yang sejajar bidang miring. .<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-badff1735c827c6562a4e074ea4b6bd2_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\sum \\vv{F_2}=0\" title=\"Rendered by QuickLaTeX.com\" height=\"25\" width=\"80\" style=\"vertical-align: -8px;\"><\/p>\n<\/p>\n<p> Jadi gaya normal pada bidang miring setara dengan komponen berat sumbu yang tegak lurus bidang:<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-ea3f790cf878ca23f77405f73a20e7c6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"N=P_2\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"58\" style=\"vertical-align: -3px;\"><\/p>\n<\/p>\n<p> Komponen berat sumbu yang tegak lurus bidang sama dengan rumus berat dikalikan kosinus sudut kemiringan bidang:<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-efeff10487f285abba9d74ee3eba6b45_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_2=P\\cdot \\cos(\\alpha)\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"117\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-9fe1bc1d3a7fbacecb2ce1ccc1dadc67_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_2=m\\cdot g\\cdot \\cos(\\alpha)\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"141\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p> Singkatnya, <strong>rumus gaya normal pada bidang miring<\/strong> menyatakan bahwa gaya normal sama dengan massa benda dikali gravitasi dikali kosinus sudut kemiringan bidang: <\/p>\n<figure class=\"wp-block-image aligncenter size-full is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/formule-de-la-force-normale-dans-un-plan-incline.png\" alt=\"rumus gaya normal pada bidang miring\" class=\"wp-image-4232\" width=\"269\" height=\"92\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/formule-de-la-force-normale-dans-un-plan-incline-300x102.png 300w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/formule-de-la-force-normale-dans-un-plan-incline.png 576w\" sizes=\"(max-width: 300px) 100vw, 300px\"><\/figure>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Fuerza-normal-y-fuerza-de-rozamiento\"><\/span> gaya normal dan gaya gesek<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Pada bagian ini kita akan melihat hubungan antara gaya normal dan gaya gesekan, karena keduanya merupakan dua jenis gaya yang dihubungkan secara matematis. Namun pertama-tama, Anda perlu mengetahui apa itu gaya gesekan.<\/p>\n<p> Gaya gesek (atau gaya gesek) adalah gaya yang terjadi pada saat mencoba menggerakkan suatu benda pada permukaan yang tidak licin. Oleh karena itu, gaya gesekan adalah gaya yang melawan gerak suatu benda.<\/p>\n<p> Gaya gesekan dihitung dari gaya normal. Lebih tepatnya, <strong>gaya gesekan sama dengan koefisien gesekan permukaan dikalikan gaya normal.<\/strong><\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-b8e2dc6a1180d664163aeb969b289073_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"F_R=\\mu \\cdot N\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"86\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p style=\"margin-bottom:5px\"> Emas: <\/p>\n<ul style=\"color:#4fd12f; font-weight: bold;\">\n<li style=\"margin-bottom:8px\"><span style=\"color:#101010;font-weight: normal;\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-5b005ac29604de5f2904d2da7ade0238_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"F_R\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"22\" style=\"vertical-align: -3px;\"><\/p>\n<p> adalah gaya gesekan. <\/span><\/li>\n<li style=\"margin-bottom:8px\"><span style=\"color:#101010;font-weight: normal;\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-05d9eae892416bd34247a25207f8b718_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\mu\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"11\" style=\"vertical-align: -4px;\"><\/p>\n<p> adalah koefisien gesekan.<\/span><\/li>\n<li><span style=\"color:#101010;font-weight: normal;\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-7354bae77b50b7d1faed3e8ea7a3511a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"N\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"16\" style=\"vertical-align: 0px;\"><\/p>\n<p> adalah resistensi normal. <\/span><\/li>\n<\/ul>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Ejercicios-resueltos-de-la-fuerza-normal\"><\/span> Menyelesaikan latihan kekuatan normal<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h3 class=\"wp-block-heading\"> Latihan 1<\/h3>\n<p> Sebuah benda bermassa 5 kg berada dalam keadaan diam di tanah datar. Jika kemudian benda lain bermassa 3 kg ditambahkan di atas benda pertama, berapakah gaya normal yang dilakukan tanah untuk menopang kedua benda tersebut? Data: g=9,81 m\/ <sup>s2<\/sup> . <\/p>\n<div class=\"wp-block-otfm-box-spoiler-start otfm-sp__wrapper otfm-sp__box js-otfm-sp-box__closed otfm-sp__FFF8E1\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\" data-otfm-spc=\"#FFF8E1\" style=\"text-align:center\">\n<div class=\"otfm-sp__title\"> <strong>Lihat solusinya<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> Karena tanah harus menopang kedua benda, gaya normalnya adalah jumlah gaya berat masing-masing benda. Oleh karena itu, pertama-tama kita akan menghitung berat masing-masing benda lalu menjumlahkannya.<\/p>\n<p class=\"has-text-align-left\"> Ingatlah bahwa gaya beban dihitung dengan mengalikan massa benda dengan gravitasi.<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-c0cdb663ec9f8fe79fbecd960b50fc39_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P=m\\cdot g\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"75\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Jadi, kami menghitung berat badan 5 kg:<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-a15ef1db0a6608fa8e6165ac0e12e925_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_1=5\\cdot 9.81=49.05\\N\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"160\" style=\"vertical-align: -3px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Kedua, kita tentukan berat benda kedua yang massanya 3 kg:<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-d57016899db324fc0f785e92341e9f2f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_2=3\\cdot 9.81=29.43\\N\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"161\" style=\"vertical-align: -3px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Jadi dengan menerapkan kondisi keseimbangan vertikal, kita memperoleh bahwa gaya normal setara dengan jumlah dua beban: <\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-c65761c8213d33892f422dd6b0a29121_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle\\sum \\vv{F_y}=0\" title=\"Rendered by QuickLaTeX.com\" height=\"25\" width=\"81\" style=\"vertical-align: -8px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-808b8980c7dbb5f2b1cdf14418fea88c_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"N=P_1+P_2\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"99\" style=\"vertical-align: -3px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Kesimpulannya, nilai gaya normal yang dilakukan oleh tanah adalah: <\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-c5819ab4ff951b1edbe6efa0a0111243_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"N=49,05+29,43=78,48 \\ N\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"237\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n<h3 class=\"wp-block-heading\">Latihan 2<\/h3>\n<p> Seperti terlihat pada gambar berikut, dua buah benda dihubungkan dengan tali dan katrol yang massanya dapat diabaikan. Jika benda 2 bermassa m <sub>2<\/sub> =7 kg dan kemiringan lerengnya 50\u00ba, hitunglah gaya normal yang dikerjakan oleh bidang miring pada benda bermassa m <sub>1<\/sub> sehingga seluruh sistem berada dalam keadaan setimbang. Abaikan kekuatan gesekan selama latihan. <\/p>\n<figure class=\"wp-block-image aligncenter size-full is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/probleme-dequilibre-des-forces.png\" alt=\"masalah keseimbangan translasi\" class=\"wp-image-295\" width=\"299\" height=\"240\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/probleme-dequilibre-des-forces-300x241.png 300w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/probleme-dequilibre-des-forces.png 718w\" sizes=\"(max-width: 300px) 100vw, 300px\"><\/figure>\n<div class=\"wp-block-otfm-box-spoiler-start otfm-sp__wrapper otfm-sp__box js-otfm-sp-box__closed otfm-sp__FFF8E1\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\" data-otfm-spc=\"#FFF8E1\" style=\"text-align:center\">\n<div class=\"otfm-sp__title\"> <strong>Lihat solusinya<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> Benda 1 berada pada bidang miring, jadi hal pertama yang harus dilakukan adalah memvektorisasikan gaya beratnya agar mempunyai gaya-gaya pada sumbu lereng: <\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-c05811c44aa2d58295c811d612a54eee_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_{1x}=P_1\\cdot \\text{sin}(\\alpha)\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"128\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-1a0b77602980cc17cce9b3baef744df8_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_{1y}=P_1\\cdot \\text{cos}(\\alpha)\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"130\" style=\"vertical-align: -6px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Jadi, himpunan gaya yang bekerja pada keseluruhan sistem adalah: <\/p>\n<figure class=\"wp-block-image aligncenter size-full is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-resolu-equilibre-des-forces.png\" alt=\"latihan keseimbangan translasi terselesaikan\" class=\"wp-image-296\" width=\"338\" height=\"272\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-resolu-equilibre-des-forces-300x241.png 300w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-resolu-equilibre-des-forces.png 718w\" sizes=\"(max-width: 300px) 100vw, 300px\"><\/figure>\n<p class=\"has-text-align-left\"> Rumusan masalah menyatakan bahwa sistem gaya-gaya berada dalam keadaan setimbang, sehingga kedua benda harus berada dalam keadaan setimbang. Dari informasi ini kita dapat mengajukan persamaan kesetimbangan kedua benda: <\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-b62bbb21cbec2be0bba7f8a839b12ba9_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"1\\ \\rightarrow \\ \\begin{cases}P_{1x}=T\\\\[2ex]P_{1y}=N\\end{cases} \\qquad\\qquad 2 \\ \\rightarrow \\ T=P_2[\/latex ] Par cons\u00e9quent, la composante vectorielle du poids du corps 1 inclin\u00e9 dans le sens de la pente doit \u00eatre \u00e9gale au poids de l'objet 2. [latex]P_{1x}=P_2\" title=\"Rendered by QuickLaTeX.com\" height=\"83\" width=\"1404\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-4e1b75b6ba5d7bbe88d23e014eb011c5_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_1\\cdot \\text{sin}(\\alpha)=P_2\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"120\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Dari persamaan sebelumnya, kita dapat menghitung massa benda 1: <\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-06a53a846ad5bc034f69fa05488404c4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"m_1\\cdot g \\cdot \\text{sin}(\\alpha) =m_2 \\cdot g\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"174\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-802fde26f3388538d766a709d60cf48b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"m_1 \\cdot \\text{sin}(\\alpha) =m_2\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"130\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-16ae359d38a8a11d1b1db4988b8eeaf1_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"m_1 \\cdot \\text{sin}(50\\text{\u00ba}) =7\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"130\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-4249c6e274233595f50eedc1da64f56f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"m_1 =\\cfrac{7}{\\text{sin}(50\\text{\u00ba})}\" title=\"Rendered by QuickLaTeX.com\" height=\"44\" width=\"111\" style=\"vertical-align: -17px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-6e80f0daabb2167ec2f6622b08001a97_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"m_1=9,14 \\ kg\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"106\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Sebaliknya, jika kita melihat diagram gaya sistem, kita mengamati bahwa gaya normal harus sama dengan komponen vektor berat benda 1 yang tegak lurus bidang miring. <\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-82b47c80ab7ef66a41fc4d4425032831_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_{1y}=N\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"66\" style=\"vertical-align: -6px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-56ad7b690b37b3f53ca20597e165860b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_1\\cdot \\text{cos}(\\alpha)=N\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"120\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Jadi, dari persamaan ini kita dapat mencari nilai gaya normal: <\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-f258dccd08d6573f74a2261b2192a92f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{array}{l}N=P_1\\cdot \\text{cos}(\\alpha)\\\\[3ex]N=m_1 \\cdot g\\cdot \\text{cos}(\\alpha)\\\\[ 3ex]N=9,14 \\cdot 9,81 \\cdot \\text{cos}(50\\text{\u00ba})\\\\[3ex]N=\\bm{57,63 \\ N}\\end{array}[\/ latex]\n\n<div class=&quot;wp-block-otfm-box-spoiler-end otfm-sp_end&quot;><\/div>\n<h3 class=&quot;wp-block-heading&quot;> Exercice 3<\/h3>\n<p> Nous pla\u00e7ons un corps de masse m=2 kg au sommet d&#8217;une rampe avec un angle d&#8217;inclinaison de 30\u00ba. Quel est le coefficient de frottement entre la rampe et le corps si celui-ci est maintenu en \u00e9quilibre ? Donn\u00e9es : g=9,81 m\/s <sup>2<\/sup> <\/p>\n<figure class=&quot;wp-block-image aligncenter size-full is-resized&quot;><img decoding=&quot;async&quot; loading=&quot;lazy&quot; src=&quot;https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/probleme-de-force-normale-et-de-force-de-friction.png&quot; alt=&quot;&quot; class=&quot;wp-image-4253&quot; width=&quot;285&quot; height=&quot;176&quot; srcset=&quot;https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/probleme-de-force-normale-et-de-force-de-friction-300x185.png 300w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/probleme-de-force-normale-et-de-force-de-friction.png 702w&quot; sizes=&quot;(max-width: 300px) 100vw, 300px&quot;><\/figure>\n<div class=&quot;wp-block-otfm-box-spoiler-start otfm-sp__wrapper otfm-sp__box js-otfm-sp-box__closed otfm-sp__FFF8E1&quot; role=&quot;button&quot; tabindex=&quot;0&quot; aria-expanded=&quot;false&quot; data-otfm-spc=&quot;#FFF8E1&quot; style=&quot;text-align:center&quot;>\n<div class=&quot;otfm-sp__title&quot;> <strong>Voir la solution<\/strong><\/div>\n<\/div>\n<p> Comme dans tout probl\u00e8me de physique portant sur les forces, la premi\u00e8re chose \u00e0 faire est de dessiner le diagramme du corps libre du syst\u00e8me. Ainsi, toutes les forces qui agissent dans ce syst\u00e8me sont : <\/p>\n<figure class=&quot;wp-block-image aligncenter size-full is-resized&quot;><img decoding=&quot;async&quot; loading=&quot;lazy&quot; src=&quot;https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-resolu-force-normale-et-friction-force.png&quot; alt=&quot;exercice r\u00e9solu de la force normale et de la force de frottement&quot; class=&quot;wp-image-4254&quot; width=&quot;285&quot; height=&quot;333&quot; srcset=&quot;https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-resolu-force-normale-et-friction-force-256x300.png 256w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-resolu-force-normale-et-friction-force.png 702w&quot; sizes=&quot;(max-width: 256px) 100vw, 256px&quot;><\/figure>\n<p> Ainsi, pour que le syst\u00e8me soit en \u00e9quilibre, la somme des forces sur les axes 1 et 2 doit \u00eatre \u00e9gale \u00e0 z\u00e9ro. Par cons\u00e9quent, les \u00e9quations suivantes sont vraies : [latex]F_R=P_1&#8243; title=&#8221;Rendered by QuickLaTeX.com&#8221; height=&#8221;454&#8243; width=&#8221;7014&#8243; style=&#8221;vertical-align: 0px;&#8221;><\/p>\n<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-ea3f790cf878ca23f77405f73a20e7c6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"N=P_2\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"58\" style=\"vertical-align: -3px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Sekarang kita dapat menghitung nilai gaya normal dari persamaan kedua:<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-780db8c589b96d398e1400444a11db30_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{array}{l}N=P_2\\\\[3ex]N=P\\cdot \\text{cos}(\\alpha)\\\\[3ex]N=m \\cdot g\\cdot \\text{cos }(\\alpha)\\\\[3ex]N=2 \\cdot 9,81 \\cdot \\text{cos}(30\\text{\u00ba})\\\\[3ex]N=16,99 \\ N\\end{array} \" title=\"Rendered by QuickLaTeX.com\" height=\"196\" width=\"171\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Sebaliknya kita menentukan nilai gaya gesekan menggunakan persamaan pertama:<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-bef5af0f3a7e907aa90f08435f538cf7_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{array}{l}F_R=P_1\\\\[3ex]N=P\\cdot \\text{sin}(\\alpha)\\\\[3ex]F_R=m \\cdot g\\cdot \\text{sin }(\\alpha)\\\\[3ex]F_R=2 \\cdot 9,81 \\cdot \\text{sin}(30\\text{\u00ba})\\\\[3ex]F_R=9,81 \\ N\\end{array} \" title=\"Rendered by QuickLaTeX.com\" height=\"196\" width=\"175\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Demikian pula gaya gesekan dapat dihubungkan dengan gaya normal dan koefisien gesekan menggunakan rumus berikut:<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-b8e2dc6a1180d664163aeb969b289073_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"F_R=\\mu \\cdot N\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"86\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Jadi kita menghapus koefisien gesekan dari persamaan dan menghitung nilainya: <\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-2bee3710c7506bf8ff2456662a57f279_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\mu=\\cfrac{F_R}{N}\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"59\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-69da73a9c8ca8ef047563bcb0b957d4b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\mu=\\cfrac{9,81}{16,99}\" title=\"Rendered by QuickLaTeX.com\" height=\"42\" width=\"80\" style=\"vertical-align: -16px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\">\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/physigeek.com\/wp-content\/ql-cache\/quicklatex.com-87da99c1b6541f3ad374e4ebb3e9daf1_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\bm{\\mu=0.58}\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"66\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n","protected":false},"excerpt":{"rendered":"<p>Artikel ini menjelaskan apa itu gaya normal dan cara menentukannya tergantung pada jenis masalahnya. Dengan demikian, Anda akan menemukan karakteristik gaya normal dan, sebagai tambahan, Anda akan dapat mempraktikkan gaya jenis ini dengan latihan yang diselesaikan langkah demi langkah. Apa itu gaya normal? Dalam fisika, gaya normal adalah gaya yang dilakukan oleh suatu permukaan pada &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"\" href=\"https:\/\/physigeek.com\/id\/kekuatan-normal\/\"> <span class=\"screen-reader-text\">Kekuatan normal<\/span> Baca selengkapnya &quot;<\/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":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v21.4 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>\u25b7 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