{"id":223949,"date":"2025-06-02T13:53:25","date_gmt":"2025-06-02T13:53:25","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=223949"},"modified":"2025-06-02T13:53:27","modified_gmt":"2025-06-02T13:53:27","slug":"the-weight-of-a-1200-kg-car-is-supported-equally-by-four-tires-which-are-inflated-to-the-same-gauge-pressure-what-gauge-pressure-is-required-so-the-area-of-contact-of-each-tire-with-the-road-is-0-001","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/02\/the-weight-of-a-1200-kg-car-is-supported-equally-by-four-tires-which-are-inflated-to-the-same-gauge-pressure-what-gauge-pressure-is-required-so-the-area-of-contact-of-each-tire-with-the-road-is-0-001\/","title":{"rendered":"The weight of a 1200 kg car is supported equally by four tires which are inflated to the same gauge pressure what gauge pressure is required so the area of contact of each tire with the road is 0.001 m^2 to the"},"content":{"rendered":"\n

The weight of a 1200 kg car is supported equally by four tires which are inflated to the same gauge pressure what gauge pressure is required so the area of contact of each tire with the road is 0.001 m^2 to the<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

Required gauge pressure<\/strong> Pgauge\u2005\u200a\u2248\u2005\u200a2.9\u00d7106 Pa\u2005\u200a=\u2005\u200a2\u2009940 kPa\u2005\u200a=\u2005\u200a4.3\u00d7102 psiP_{\\text{gauge}} \\;\\approx\\; 2.9 \\times 10^{6}\\ \\text{Pa} \\;=\\; 2\\,940\\ \\text{kPa} \\;=\\; 4.3 \\times 10^{2}\\ \\text{psi}<\/p>\n\n\n\n

\n\n\n\n

Why that number?<\/h3>\n\n\n\n
    \n
  1. Weight on each tyre<\/strong>
    The 1200 kg car\u2019s weight is shared equally by the four tyres, so each supports one-quarter of the total mass: mper\u00a0tyre=1200\u00a0kg4=300\u00a0kg.m_{\\text{per tyre}} = \\frac{1200\\ \\text{kg}}{4} = 300\\ \\text{kg}. Converting this mass to a force (weight) with g=9.8\u00a0m\u00a0s\u22122g = 9.8\\ \\text{m s}^{-2}: F=mg=300\u00d79.8\u2005\u200a\u2248\u2005\u200a2.94\u00d7103\u00a0N.F = mg = 300 \\times 9.8 \\;\\approx\\; 2.94 \\times 10^{3}\\ \\text{N}.<\/li>\n\n\n\n
  2. Pressure\u2013area relation<\/strong>
    Pressure is force per unit area, P=F\/AP = F\/A. The specified contact patch for each tyre is only A=0.001\u00a0m2A = 0.001\\ \\text{m}^{2} (a square 3.2 cm on a side). Substituting: Pinside=FA=2.94\u00d7103\u00a0N0.001\u00a0m2=2.94\u00d7106\u00a0Pa.P_{\\text{inside}} = \\frac{F}{A} = \\frac{2.94 \\times 10^{3}\\ \\text{N}}{0.001\\ \\text{m}^{2}} = 2.94 \\times 10^{6}\\ \\text{Pa}.<\/li>\n\n\n\n
  3. Gauge vs. absolute<\/strong>
    \u201cGauge\u201d pressure is the amount above<\/em> atmospheric pressure. Because atmospheric pressure (\u2248 1.01 \u00d7 10\u2075 Pa) is less than 4 % of the value we have just calculated, adding or subtracting it changes the result only in the third significant figure. Thus the gauge pressure effectively equals the computed 2.94 MPa.<\/li>\n<\/ol>\n\n\n\n
    \n\n\n\n

    Interpreting the result<\/h3>\n\n\n\n
      \n
    • Unusually high:<\/strong> 2.9 MPa is roughly 14 times higher<\/strong> than the 200 kPa (30 psi) found in everyday passenger-car tyres. To achieve a 0.001 m\u00b2 contact patch the tyre would have to be inflated to racing-car or small-aircraft levels\u2014far beyond safe limits for road tyres.<\/li>\n\n\n\n
    • Why the mismatch?<\/strong> In reality a typical tyre\u2019s contact area is 0.025\u20130.030 m\u00b2, not 0.001 m\u00b2. That larger footprint lowers the required gauge pressure into the familiar 180\u2013250 kPa range.<\/li>\n\n\n\n
    • Lesson:<\/strong> For any load-bearing surface, halving the contact area doubles<\/em> the required pressure; reducing the area by a factor of 25 (as here) increases pressure twenty-fivefold. Engineers manipulate contact area (wider tyres, twin wheels, caterpillar tracks) precisely to keep working pressures within safe, efficient ranges.<\/li>\n<\/ul>\n\n\n\n

      Hence, to support the car on a 0.001 m\u00b2 patch per tyre, a gauge pressure of about 2.9 MPa (430 psi)<\/strong> would be necessary\u2014a value that highlights how crucial adequate contact area is for practical tyre design.<\/p>\n\n\n\n

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

      The weight of a 1200 kg car is supported equally by four tires which are inflated to the same gauge pressure what gauge pressure is required so the area of contact of each tire with the road is 0.001 m^2 to the The Correct Answer and Explanation is: Required gauge pressure Pgauge\u2005\u200a\u2248\u2005\u200a2.9\u00d7106 Pa\u2005\u200a=\u2005\u200a2\u2009940 kPa\u2005\u200a=\u2005\u200a4.3\u00d7102 psiP_{\\text{gauge}} \\;\\approx\\; 2.9 \\times […]<\/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-223949","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/223949","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=223949"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/223949\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=223949"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=223949"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=223949"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}