{"id":250,"date":"2023-06-23T07:32:36","date_gmt":"2023-06-23T07:32:36","guid":{"rendered":"https:\/\/physigeek.com\/id\/gaya-gesek-atau-gaya-gesek\/"},"modified":"2023-06-23T07:32:36","modified_gmt":"2023-06-23T07:32:36","slug":"gaya-gesek-atau-gaya-gesek","status":"publish","type":"post","link":"https:\/\/physigeek.com\/id\/gaya-gesek-atau-gaya-gesek\/","title":{"rendered":"Gaya gesekan (atau gaya gesekan)"},"content":{"rendered":"<p>Artikel ini menjelaskan apa itu gaya gesekan (atau gaya gesekan) dalam fisika dan cara menghitungnya. Oleh karena itu, Anda akan menemukan sifat-sifat gaya gesekan, dua jenis gaya gesekan yang ada, dan, sebagai tambahan, latihan khusus untuk dipraktikkan. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"%C2%BFQue-es-la-fuerza-de-rozamiento\"><\/span> Apa itu gaya gesekan?<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> <strong>Gaya gesekan<\/strong> , disebut juga <strong>gaya gesek<\/strong> , adalah gaya kontak yang terjadi ketika suatu benda berusaha menggerakkan permukaan benda lain.<\/p>\n<p> Lebih tepatnya, gaya gesek adalah gaya yang dikerjakan dalam arah yang sejajar dan berlawanan dengan gerak. <\/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-de-friction-ou-force-de-friction.png\" alt=\"\" class=\"wp-image-4319\" width=\"293\" height=\"292\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/force-de-friction-ou-force-de-friction-300x300.png 300w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/force-de-friction-ou-force-de-friction-150x150.png 150w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/force-de-friction-ou-force-de-friction.png 625w\" sizes=\"(max-width: 300px) 100vw, 300px\"><\/figure>\n<p> Ada dua jenis gaya gesekan: gaya gesekan statis dan gaya gesekan dinamis. Tergantung pada kasusnya, satu atau beberapa tindakan lainnya. Di bawah ini kita akan melihat perbedaan di antara keduanya.<\/p>\n<p> Secara umum gaya gesek dilambangkan dengan simbol F <sub>R.<\/sub> <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Caracteristicas-de-la-fuerza-de-rozamiento\"><\/span> Karakteristik gaya gesekan<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Setelah kita mengetahui pengertian gaya gesek (atau gaya gesek), mari kita lihat apa saja ciri-ciri gaya jenis ini:<\/p>\n<ul style=\"color:#4fd12f; font-weight: bold;\">\n<li style=\"margin-bottom:15px\"> <span style=\"color:#101010;font-weight: normal;\">Gaya gesek merupakan gaya kontak, artinya hanya bekerja jika dua permukaan bersentuhan.<\/span><\/li>\n<li style=\"margin-bottom:15px\"> <span style=\"color:#101010;font-weight: normal;\">Selain itu, gaya gesekan hanya muncul ketika suatu benda bergerak atau berupaya untuk menumpangkan benda lain.<\/span><\/li>\n<li style=\"margin-bottom:15px\"> <span style=\"color:#101010;font-weight: normal;\">Arah gaya gesek sejajar dengan arah gerak.<\/span><\/li>\n<li style=\"margin-bottom:15px\"> <span style=\"color:#101010;font-weight: normal;\">Arah gaya gesekan berlawanan dengan gerak.<\/span><\/li>\n<li style=\"margin-bottom:15px\"> <span style=\"color:#101010;font-weight: normal;\">Gaya gesekan tidak bergantung pada kecepatan geser benda.<\/span><\/li>\n<li style=\"margin-bottom:15px\"> <span style=\"color:#101010;font-weight: normal;\">Gaya gesekan tidak bergantung pada ukuran permukaan yang bersentuhan.<\/span><\/li>\n<li style=\"margin-bottom:15px\"> <span style=\"color:#101010;font-weight: normal;\">Namun gaya gesekan bergantung pada bahan yang bersentuhan, lapisan akhir, dan suhu.<\/span><\/li>\n<li> <span style=\"color:#101010;font-weight: normal;\">Gaya gesekan berbanding lurus dengan gaya normal.<\/span> <\/li>\n<\/ul>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Formula-de-la-fuerza-de-rozamiento\"><\/span> Rumus gaya gesekan<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Gaya gesekan sama dengan koefisien gesekan dikalikan gaya normal. Oleh karena itu, untuk menghitung gaya gesekan, pertama-tama harus dicari gaya normalnya kemudian dikalikan dengan koefisien gesekan antara kedua permukaan kontak.<\/p>\n<p> Oleh karena itu, <strong>rumus gaya gesekan<\/strong> (atau gaya gesekan) adalah sebagai 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-a865b1cd2e263b944debf58666ec1269_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 gesek atau gesekan yang dinyatakan dalam newton. <\/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 yang tidak mempunyai satuan.<\/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 gaya normal yang dinyatakan dalam newton. <\/span><\/li>\n<\/ul>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Fuerza-de-rozamiento-estatico-y-dinamico\"><\/span> Gaya gesekan statis dan dinamis<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Besarnya gaya gesek bergantung pada apakah benda dalam keadaan diam atau bergerak. Misalnya saja Anda pasti mencoba menyeret suatu benda yang sangat berat dan pada awalnya sulit untuk digerakkan, namun begitu Anda berhasil sedikit menggerakkan badan tersebut, maka akan lebih mudah untuk terus menyeret benda tersebut.<\/p>\n<p> Memang secara umum gaya gesek pada saat benda diam lebih besar dibandingkan pada saat benda bergerak.<\/p>\n<p> Jadi, kita membedakan dua jenis gaya gesekan (atau gaya gesekan):<\/p>\n<ul style=\"color:#4fd12f; font-weight: bold;\">\n<li style=\"margin-bottom:15px\"> <span style=\"color:#101010;font-weight: normal;\"><strong>Gaya gesekan statis<\/strong> : Ini adalah gaya gesekan yang bekerja ketika benda belum bergerak.<\/span><\/li>\n<li> <span style=\"color:#101010;font-weight: normal;\"><strong>Gaya gesekan dinamis (atau kinetik)<\/strong> : ini adalah gaya gesekan yang bekerja ketika benda sudah mulai bergerak.<\/span><\/li>\n<\/ul>\n<p> Demikian pula, koefisien gesekan statis juga dibedakan dari koefisien gesekan dinamis, yang masing-masing digunakan untuk menentukan gaya gesekan statis dan gaya gesekan dinamis.<\/p>\n<p> Terakhir, besarnya gaya gesek berubah-ubah seperti terlihat pada grafik berikut: <\/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\/graphique-force-statique-et-dynamique-friction.png\" alt=\"gaya gesekan statis dan dinamis\" class=\"wp-image-4356\" width=\"381\" height=\"261\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/graphique-force-statique-et-dynamique-friction-300x205.png 300w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/graphique-force-statique-et-dynamique-friction.png 725w\" sizes=\"(max-width: 300px) 100vw, 300px\"><\/figure>\n<p> Gaya gesekan statis sama dengan gaya yang diterapkan untuk mencoba menggerakkan benda tetapi arahnya berlawanan. Nilai maksimumnya merupakan hasil kali antara koefisien gesekan statis dan gaya normal. Ketika gaya yang diberikan melebihi nilai ini, benda mulai bergerak.<\/p>\n<p> Jadi, ketika benda sudah bergerak, gaya gesekan dinamis mempunyai nilai konstan yang setara dengan hasil kali antara koefisien gesekan dinamis dan gaya normal, berapa pun nilai gaya yang diterapkan. Selain itu, nilai ini sedikit lebih rendah dari nilai maksimum gaya gesek statis. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Ejercicios-resueltos-de-la-fuerza-de-rozamiento\"><\/span> Latihan soal gaya gesek terselesaikan<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h3 class=\"wp-block-heading\"> Latihan 1<\/h3>\n<p> Benda tersebut dimaksudkan untuk memindahkan sebuah balok bermassa m=12 kg pada permukaan datar dan balok tersebut mulai bergerak ketika gaya sebesar 35 N diberikan. Berapakah koefisien gesekan statik antara tanah dan balok? Data: g=10 m\/s <sup>2<\/sup> . <\/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-resolu-coefficient-de-frottement-statique.png\" alt=\"memecahkan masalah koefisien gesekan statis\" class=\"wp-image-4302\" width=\"285\" height=\"125\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/probleme-resolu-coefficient-de-frottement-statique-300x132.png 300w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/probleme-resolu-coefficient-de-frottement-statique.png 650w\" 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\"> Pertama, kita buat grafik semua gaya yang bekerja pada balok: <\/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-force-de-friction-statique.png\" alt=\"menyelesaikan latihan tentang koefisien gesekan statis atau koefisien gesekan statis\" class=\"wp-image-4303\" width=\"269\" height=\"359\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-resolu-force-de-friction-statique-225x300.png 225w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-resolu-force-de-friction-statique.png 670w\" sizes=\"(max-width: 225px) 100vw, 225px\"><\/figure>\n<p class=\"has-text-align-left\"> Dalam situasi batas kesetimbangan, dua persamaan berikut terpenuhi:<\/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 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-b2d0fc9325264d9c3dceae21b529d2c5_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"F_R=F\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"60\" style=\"vertical-align: -3px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Jadi gaya gesekan akan setara dengan gaya horizontal yang diterapkan pada 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-f433d79addb4c8cfc17c71cf797f4905_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"F_R=F=35 \\ N\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"123\" style=\"vertical-align: -3px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Sebaliknya, kita dapat menghitung nilai gaya normal dengan menggunakan rumus gaya berat:<\/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-fdd5a10090733132a78410e57a059c2d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{array}{l}N=P\\\\[3ex] N=m\\cdot g\\\\[3ex] N=12\\cdot 10 \\\\[3ex] N=120 \\ N\\end{array }\" title=\"Rendered by QuickLaTeX.com\" height=\"147\" width=\"88\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Terakhir, setelah kita mengetahui nilai gaya gesekan dan gaya normal, kita menerapkan rumus koefisien gesekan statis untuk menentukan 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-20cdd7c8dcf9b6af55e1fe95eedf8da2_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\mu_e=\\cfrac{F_R}{N}=\\cfrac{35}{120}=0.29\" title=\"Rendered by QuickLaTeX.com\" height=\"39\" width=\"174\" style=\"vertical-align: -12px;\"><\/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> Sebuah benda bermassa m=6 kg kita letakkan di puncak bidang miring 45\u00ba. Jika benda meluncur pada bidang miring dengan percepatan 4 m\/s <sup>2<\/sup> , berapakah koefisien gesekan dinamis antara permukaan bidang miring dengan permukaan benda? Data: g=10 m\/s <sup>2<\/sup> . <\/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-resolu-coefficient-de-frottement-dynamique.png\" alt=\"masalah koefisien gesekan atau gesekan dinamis\" class=\"wp-image-4281\" width=\"203\" height=\"205\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/probleme-resolu-coefficient-de-frottement-dynamique-298x300.png 298w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/probleme-resolu-coefficient-de-frottement-dynamique-150x150.png 150w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/probleme-resolu-coefficient-de-frottement-dynamique.png 479w\" sizes=\"(max-width: 298px) 100vw, 298px\"><\/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\"> Hal pertama yang perlu kita lakukan untuk menyelesaikan masalah fisika mengenai dinamika adalah menggambar diagram benda bebas. Jadi, gaya-gaya yang bekerja pada sistem tersebut 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-force-de-friction-dynamique.png\" alt=\"menyelesaikan latihan koefisien gesekan atau gesekan dinamis\" class=\"wp-image-4282\" width=\"248\" height=\"301\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-resolu-force-de-friction-dynamique-247x300.png 247w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-resolu-force-de-friction-dynamique.png 572w\" sizes=\"(max-width: 247px) 100vw, 247px\"><\/figure>\n<p class=\"has-text-align-left\"> Pada arah sumbu 1 (sejajar bidang miring) benda mengalami percepatan, namun pada arah sumbu 2 (tegak lurus bidang miring) benda diam. Dari informasi ini, kami mengusulkan persamaan gaya-gaya sistem:<\/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-d87a1ef6aaa3476891df5da8334cbc49_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_1-F_R=m\\cdot a\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"124\" style=\"vertical-align: -3px;\"><\/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-6bdf90ed250934bf6cffbb110bc792a4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_2-N=0\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"90\" style=\"vertical-align: -3px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Jadi, kita dapat menghitung 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-59341555fe3d5fe315ceb1864547873b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{array}{l}N=P_2\\\\[3ex]N=m\\cdot g\\cdot \\text{cos}(\\alpha) \\\\[3ex] N=6 \\cdot 10 \\cdot \\ text{cos}(45\u00ba)\\\\[3ex]N=42,43 \\ N\\end{array}\" title=\"Rendered by QuickLaTeX.com\" height=\"151\" width=\"185\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Sebaliknya, kita menghitung nilai gaya gesekan (atau gaya gesekan) dari persamaan pertama yang disajikan:<\/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-d8f2aff2a81d98ddcea04b1988282fda_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{array}{l}P_1-F_R=m\\cdot a\\\\[3ex]F_R=P_1-m\\cdot a\\\\[3ex]F_R=m\\cdot g\\cdot \\text{sin} (\\alpha)-m\\cdot a\\\\[3ex]F_R=6\\cdot 10\\cdot \\text{sin}(45\u00ba)-6\\cdot 4\\\\[3ex]F_R=18.43 \\ N\\end{ array} \" title=\"Rendered by QuickLaTeX.com\" height=\"195\" width=\"204\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Dan setelah kita mengetahui nilai gaya normal dan gaya gesekan, kita dapat menentukan koefisien gesekan dinamis menggunakan rumus yang sesuai: <\/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-31af78ef6e04fa66121d64aa3570f5a6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\mu_d=\\cfrac{F_R}{N}=\\cfrac{18.43}{43.43}=\\bm{0.42}\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"187\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n<h3 class=\"wp-block-heading\">Latihan 3<\/h3>\n<p> Sebuah kereta luncur bermassa 70 kg meluncur menuruni kemiringan 30\u00ba dengan kecepatan awal 2 m\/s. Jika koefisien gesekan dinamis antara kereta luncur dan salju adalah 0,2, hitunglah kecepatan yang diperoleh kereta luncur tersebut setelah menempuh jarak 20 meter. Data: g=10 m\/s <sup>2<\/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\"> Pertama-tama, kita membuat diagram benda bebas kereta luncur: <\/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-plan-incline.png\" alt=\"ditentukan pelaksanaan gaya gesekan pada bidang miring\" class=\"wp-image-4345\" width=\"305\" height=\"355\" srcset=\"https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-resolu-plan-incline-258x300.png 258w, https:\/\/physigeek.com\/wp-content\/uploads\/2023\/09\/exercice-resolu-plan-incline.png 706w\" sizes=\"(max-width: 258px) 100vw, 258px\"><\/figure>\n<p class=\"has-text-align-left\"> Kereta luncur mempunyai percepatan searah sumbu 1 (sejajar bidang miring) namun tetap diam pada arah sumbu 2 (tegak lurus bidang miring), sehingga persamaan gayanya 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-d87a1ef6aaa3476891df5da8334cbc49_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_1-F_R=m\\cdot a\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"124\" style=\"vertical-align: -3px;\"><\/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-6bdf90ed250934bf6cffbb110bc792a4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_2-N=0\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"90\" style=\"vertical-align: -3px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Dari persamaan kedua kita dapat menghitung gaya normal yang bekerja pada kereta luncur<\/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-90b32b903f8be520ec73748b3de9b8b3_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{array}{l}N=P_2\\\\[3ex]N=m\\cdot g\\cdot \\text{cos}(\\alpha) \\\\[3ex] N=70 \\cdot 10 \\cdot \\ text{cos}(30\u00ba)\\\\[3ex]N=606,22 \\ N\\end{array}\" title=\"Rendered by QuickLaTeX.com\" height=\"151\" width=\"194\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Karena sekarang kita mengetahui nilai gaya normal dan koefisien gesekan dinamis, kita dapat menghitung gaya gesekan dengan menerapkan rumus yang sesuai:<\/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-e0a32cc7650b33325233258788c218d4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"F_R=\\mu\\cdot N=0,2 \\cdot 606,22=121,24 \\ N \" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"298\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Jadi, untuk menentukan kecepatan akhir, kita harus mencari percepatan kereta luncur terlebih dahulu, dan ini dapat dihitung dari persamaan gaya pertama yang disajikan: <\/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-d87a1ef6aaa3476891df5da8334cbc49_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P_1-F_R=m\\cdot a\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"124\" style=\"vertical-align: -3px;\"><\/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-fa13e0490f51e32ac03b455043f6f32d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"a=\\cfrac{P_1-F_R}{m}\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"99\" 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-a6274d836af5618f7ef99e7f179c3902_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"a=\\cfrac{m\\cdot g\\cdot \\text{sin}(\\alpha)-F_R}{m}\" title=\"Rendered by QuickLaTeX.com\" height=\"40\" width=\"177\" 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-a3a9db70462cd187d50b851ede83983f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"a=\\cfrac{70\\cdot 10\\cdot \\text{sin}(30\u00ba)-121.24}{70}\" title=\"Rendered by QuickLaTeX.com\" height=\"40\" width=\"221\" 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-ba0d7325efa059351cc3d9aef838a9e2_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"a=3,27 \\ \\cfrac{m}{s^2}\" title=\"Rendered by QuickLaTeX.com\" height=\"34\" width=\"92\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Setelah kita mengetahui percepatan kereta luncur, kita menghitung waktu yang diperlukan untuk menempuh jarak 20 meter dengan persamaan gerak lurus dengan percepatan tetap: <\/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-69f632cd171007df0f5bd6f0fa458a5c_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"x=v_0\\cdot t +\\cfrac{1}{2}\\cdot a \\cdot t^2\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"150\" 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-b97ac72bf22d70273fece0cce195f4ca_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"20=2\\cdot t +\\cfrac{1}{2}\\cdot 3.27 \\cdot t^2\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"172\" 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-11a99cb686bf68cbcca594d0d60f801b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"0=1,64t^2+2t-20\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"158\" style=\"vertical-align: -4px;\"><\/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-7aa7e01e70b4199d597d05e261c970df_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle t=\\cfrac{-2\\pm \\sqrt{2^2-4\\cdot 1.64\\cdot (-20)}}{2\\cdot 1.64}=\\cfrac{-2\\ pm 11.63}{ 3.28}=\\begin{cases}2.94\\\\[2ex]-4.15 \\ \\color{red}\\bm{\\times}\\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"65\" width=\"507\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Logikanya, kita mengesampingkan solusi negatif karena waktu adalah besaran fisika yang tidak boleh negatif.<\/p>\n<p class=\"has-text-align-left\"> Terakhir, kita menghitung kecepatan akhir menggunakan rumus percepatan konstan: <\/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-5ace98bfb166f5b813f593760fcfa048_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"a=\\cfrac{v_f-v_0}{t_f-t_0}\\quad \\longrightarrow \\quad v_f=a\\cdot (t_f-t_0)+v_0\" title=\"Rendered by QuickLaTeX.com\" height=\"40\" width=\"330\" style=\"vertical-align: -18px;\"><\/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-e7b1fffc78a1f3be82ead78bf2635dc0_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"v_f=3,27\\cdot (2,94-0)+2=11,61 \\ \\cfrac{m}{s}\" title=\"Rendered by QuickLaTeX.com\" height=\"34\" width=\"288\" style=\"vertical-align: -12px;\"><\/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 gesekan (atau gaya gesekan) dalam fisika dan cara menghitungnya. Oleh karena itu, Anda akan menemukan sifat-sifat gaya gesekan, dua jenis gaya gesekan yang ada, dan, sebagai tambahan, latihan khusus untuk dipraktikkan. Apa itu gaya gesekan? Gaya gesekan , disebut juga gaya gesek , adalah gaya kontak yang terjadi ketika &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"\" href=\"https:\/\/physigeek.com\/id\/gaya-gesek-atau-gaya-gesek\/\"> <span class=\"screen-reader-text\">Gaya gesekan (atau gaya gesekan)<\/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 Gaya 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