{"id":222090,"date":"2025-05-31T03:40:45","date_gmt":"2025-05-31T03:40:45","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=222090"},"modified":"2025-05-31T03:40:47","modified_gmt":"2025-05-31T03:40:47","slug":"there-are-two-forces-on-the-2-00-kg-box-in-the-overhead-view-of-figure-but-only-one-is-shown","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/31\/there-are-two-forces-on-the-2-00-kg-box-in-the-overhead-view-of-figure-but-only-one-is-shown\/","title":{"rendered":"There are two forces on the 2.00 kg box in the overhead view of Figure, but only one is shown"},"content":{"rendered":"\n<p>There are two forces on the 2.00 kg box in the overhead view of Figure, but only one is shown. For F1 = 20.0 N, a = 12.0 m\/s\u00b2, and ? = 30.0\u00b0, find the second force. (a) In unit-vector notation and as (b) A magnitude and (c) An angle relative to the positive direction of the x axis.&nbsp;<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/files.transtutors.com\/questions\/transtutors001\/images\/transtutors001_9c8384da-d95e-4d4d-919f-c04ab65da78a.png\" alt=\"\"\/><\/figure>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is<\/mark><\/strong>:<\/p>\n\n\n\n<p>We are given the following:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Mass of the box: m=2.00\u2009kgm = 2.00\\, \\text{kg}<\/li>\n\n\n\n<li>Acceleration: a\u20d7\\vec{a}, magnitude a=12.0\u2009m\/s2a = 12.0\\, \\text{m\/s}^2, direction \u03b8=30.0\u2218\\theta = 30.0^\\circ below the +x-axis<\/li>\n\n\n\n<li>Force F\u20d71=20.0\u2009Ni^\\vec{F}_1 = 20.0\\, \\text{N} \\hat{i}<\/li>\n\n\n\n<li>We are to find the second force F\u20d72\\vec{F}_2<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Newton\u2019s Second Law:<\/h3>\n\n\n\n<p>F\u20d7net=ma\u20d7=F\u20d71+F\u20d72\\vec{F}_{\\text{net}} = m\\vec{a} = \\vec{F}_1 + \\vec{F}_2<\/p>\n\n\n\n<p>So we rearrange to solve for F\u20d72\\vec{F}_2: F\u20d72=ma\u20d7\u2212F\u20d71\\vec{F}_2 = m\\vec{a} &#8211; \\vec{F}_1<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Find ma\u20d7m\\vec{a} in unit-vector form<\/h3>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>a=12.0\u2009m\/s2a = 12.0\\, \\text{m\/s}^2<\/li>\n\n\n\n<li>\u03b8=30\u2218\\theta = 30^\\circ <strong>below<\/strong> the +x-axis<\/li>\n<\/ul>\n\n\n\n<p>Break into components: ax=acos\u2061(30\u2218)=12.0\u00d732=10.39\u2009m\/s2a_x = a \\cos(30^\\circ) = 12.0 \\times \\frac{\\sqrt{3}}{2} = 10.39\\, \\text{m\/s}^2 ay=\u2212asin\u2061(30\u2218)=\u221212.0\u00d712=\u22126.00\u2009m\/s2a_y = -a \\sin(30^\\circ) = -12.0 \\times \\frac{1}{2} = -6.00\\, \\text{m\/s}^2<\/p>\n\n\n\n<p>Now multiply by mass: ma\u20d7=2.00\u2009kg\u00d7(10.39i^\u22126.00j^)=(20.78i^\u221212.0j^)\u2009Nm\\vec{a} = 2.00\\, \\text{kg} \\times (10.39 \\hat{i} &#8211; 6.00 \\hat{j}) = (20.78 \\hat{i} &#8211; 12.0 \\hat{j})\\, \\text{N}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Subtract F\u20d71\\vec{F}_1 to get F\u20d72\\vec{F}_2<\/h3>\n\n\n\n<p>F\u20d72=(20.78i^\u221212.0j^)\u2212(20.0i^)=(0.78i^\u221212.0j^)\u2009N\\vec{F}_2 = (20.78 \\hat{i} &#8211; 12.0 \\hat{j}) &#8211; (20.0 \\hat{i}) = (0.78 \\hat{i} &#8211; 12.0 \\hat{j})\\, \\text{N}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answers:<\/h3>\n\n\n\n<p><strong>(a) Unit vector notation:<\/strong> F\u20d72=0.78i^\u221212.0j^\u2009N\\boxed{\\vec{F}_2 = 0.78 \\hat{i} &#8211; 12.0 \\hat{j}\\, \\text{N}}<\/p>\n\n\n\n<p><strong>(b) Magnitude of F\u20d72\\vec{F}_2:<\/strong> \u2223F\u20d72\u2223=0.782+12.02=0.61+144\u224812.03\u2009N|\\vec{F}_2| = \\sqrt{0.78^2 + 12.0^2} = \\sqrt{0.61 + 144} \\approx \\boxed{12.03\\, \\text{N}}<\/p>\n\n\n\n<p><strong>(c) Direction relative to +x-axis:<\/strong> tan\u2061\u03b8=\u2223\u221212.0\u22230.78\u21d2\u03b8=tan\u2061\u22121(12.00.78)\u224886.3\u2218\\tan\\theta = \\frac{|-12.0|}{0.78} \\Rightarrow \\theta = \\tan^{-1}\\left(\\frac{12.0}{0.78}\\right) \\approx 86.3^\\circ<\/p>\n\n\n\n<p>Since it&#8217;s below the +x-axis: \u03b8=86.3\u2218&nbsp;below&nbsp;the&nbsp;+x-axis\\boxed{\\theta = 86.3^\\circ\\ \\text{below the +x-axis}}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation<\/h3>\n\n\n\n<p>This problem involves determining an unknown force acting on a 2.00 kg box, given its acceleration and one known force. According to Newton\u2019s second law, the net force acting on an object is equal to the product of its mass and acceleration (F\u20d7net=ma\u20d7\\vec{F}_{\\text{net}} = m\\vec{a}). The box is subject to two forces: one known force, F\u20d71\\vec{F}_1, acting directly along the +x-axis, and a second unknown force, F\u20d72\\vec{F}_2, which we are solving for.<\/p>\n\n\n\n<p>We start by calculating the net force required to accelerate the 2.00 kg box at 12.0\u2009m\/s212.0\\, \\text{m\/s}^2 at a 30\u221830^\\circ angle below the x-axis. To do this, we decompose the acceleration vector into its x and y components using trigonometric functions. The x-component is found using cosine, and the y-component using sine\u2014keeping in mind the direction (negative y because it&#8217;s below the x-axis).<\/p>\n\n\n\n<p>Multiplying these acceleration components by the mass yields the net force in vector form. From this net force, we subtract the given force F\u20d71\\vec{F}_1 (which has only an x-component) to isolate the second force F\u20d72\\vec{F}_2. This gives us the x and y components of F\u20d72\\vec{F}_2.<\/p>\n\n\n\n<p>Finally, we compute the magnitude of F\u20d72\\vec{F}_2 using the Pythagorean theorem and determine the direction using the arctangent of the ratio of the y-component to the x-component. Because the force points downward, the angle is measured below the x-axis.<\/p>\n\n\n\n<p>This approach combines Newton\u2019s laws with vector analysis to find a missing force in a dynamic system.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/05\/learnexams-banner4-165.jpeg\" alt=\"\" class=\"wp-image-222091\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>There are two forces on the 2.00 kg box in the overhead view of Figure, but only one is shown. For F1 = 20.0 N, a = 12.0 m\/s\u00b2, and ? = 30.0\u00b0, find the second force. (a) In unit-vector notation and as (b) A magnitude and (c) An angle relative to the positive direction [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","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":"","ast-disable-related-posts":"","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":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-222090","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/222090","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=222090"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/222090\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=222090"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=222090"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=222090"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}