{"id":1808,"date":"2026-08-29T00:20:22","date_gmt":"2026-08-29T00:20:22","guid":{"rendered":"https:\/\/physicsfundamentalsinfo.com\/blog\/?p=1808"},"modified":"2026-08-29T21:39:10","modified_gmt":"2026-08-29T21:39:10","slug":"free-body-diagram","status":"publish","type":"post","link":"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/free-body-diagram\/","title":{"rendered":"Free Body Diagram: How to Draw One Step by Step"},"content":{"rendered":"\n<div class=\"pf-citation\"><div class=\"eyebrow\">Definition<\/div><p>\nA free body diagram is a simplified sketch that shows one chosen object as a dot or box, with every external force acting on it drawn as a labelled arrow. It strips away the surroundings, so the vector sum of those arrows gives the net force \u2014 which equals mass times acceleration.\n<\/p><\/div>\n\n<p>You already do this without noticing. Push a stalled car and you feel exactly three things that matter: your push, the ground dragging back, and the car&#8217;s stubborn weight pressing down. Your brain quietly discards the paint, the passengers, the shopping in the boot.<\/p> <p>A free body diagram is that instinct, made rigorous. Get it right and a page of confusing physics collapses into two short equations. Get it wrong \u2014 one stray arrow, one missing contact \u2014 and every number after it is wrong too.<\/p> <h2>What Is a Free Body Diagram?<\/h2> <p>A free body diagram is a drawing of a single object showing only the external forces acting on it, with each force drawn as an arrow starting at the object. Nothing else appears: no surfaces, no ropes, no neighbouring blocks.<\/p> <p>The word <em>free<\/em> is doing real work here. You are freeing the body from its surroundings, then replacing every removed contact with the force it was exerting. The floor vanishes and becomes an upward arrow. The rope vanishes and becomes a pull along its old direction.<\/p> <p>Why bother? Because forces obey vector arithmetic, and vector arithmetic needs a clean list. As soon as the picture contains a wall, a pulley and three blocks, you cannot tell which forces belong to which object.<\/p> <h3>One body, one diagram<\/h3> <p>This is the rule beginners break most often. If a problem has two blocks, it has two free body diagrams \u2014 never one crowded picture with arrows going everywhere.<\/p> <p>The payoff is that each diagram gives you its own equation. Two diagrams, two equations, two unknowns: the system solves.<\/p> <h2>The Free Body Diagram Formula<\/h2> <p>Every free body diagram exists to feed one equation \u2014 Newton&#8217;s second law, applied separately to each axis.<\/p>\n\n<div class=\"pf-formula\">\u03a3F = ma<\/div>\n\n<p>In practice you never use it in that vector form. You split it the moment you have chosen axes:<\/p>\n\n<div class=\"pf-formula\">\u03a3Fx = max and \u03a3Fy = may<\/div>\n\n<ul> <li><strong>\u03a3F<\/strong> \u2014 the net force, the vector sum of every arrow on the diagram, in newtons (N)<\/li> <li><strong>\u03a3Fx<\/strong>, <strong>\u03a3Fy<\/strong> \u2014 the net force along the x and y axes, in newtons (N)<\/li> <li><strong>m<\/strong> \u2014 the mass of the body you isolated, in kilograms (kg)<\/li> <li><strong>a<\/strong> \u2014 the acceleration of that body, in metres per second squared (m\/s\u00b2)<\/li> <li><strong>ax<\/strong>, <strong>ay<\/strong> \u2014 the components of that acceleration along each axis, in m\/s\u00b2<\/li> <\/ul> <p>Two details decide whether your answer is right. First, \u03a3F means <em>sum<\/em>: forces pointing along the negative axis enter with a minus sign. Second, m is the mass of the isolated body only \u2014 not the whole system, unless you deliberately drew the whole system as one body.<\/p> <p>Once the diagram is done and the components are listed, the arithmetic is mechanical, and you can check any single-force result against our <a href=\"https:\/\/physicsfundamentalsinfo.com\/calculators\/newtons-second-law\">Newton&#8217;s Second Law Calculator<\/a> before committing it to a full solution.<\/p> <h2>How to Draw a Free Body Diagram in 5 Steps<\/h2> <p>Draw a free body diagram by isolating one body, reducing it to a dot, adding one arrow for gravity and one for each physical contact, labelling every arrow, then choosing axes and resolving. These five steps work for every mechanics problem you will meet.<\/p> <h3>Step 1 \u2014 Pick one body and circle it<\/h3> <p>Decide what you are analysing before you draw anything. Literally draw a loop around it in the original picture \u2014 this single habit prevents most stray-arrow errors.<\/p> <h3>Step 2 \u2014 Redraw it as a dot<\/h3> <p>Shape and size rarely matter for force problems, so shrink the object to a point and put that point at the origin of your axes. All arrows will now start from this dot.<\/p> <h3>Step 3 \u2014 Add gravity, then walk the boundary<\/h3> <p>Weight always acts, so draw it first, straight down. Then trace around the object&#8217;s surface and ask at every point: <em>is something touching me here?<\/em> Each contact earns exactly one force \u2014 or two if the surface is rough, since a rough surface gives both a normal force and friction.<\/p> <h3>Step 4 \u2014 Draw and label every arrow<\/h3> <p>Each arrow starts at the dot and points the way the force actually acts. Length should roughly reflect size: if you know the object is not sinking through the floor, draw N about as long as W.<\/p> <p>Label with symbols, not numbers \u2014 N, W, T, f. Numbers come later, and symbols keep the algebra honest.<\/p> <h3>Step 5 \u2014 Choose axes, then resolve<\/h3> <p>Pick axes that put as many arrows as possible directly on an axis. Any arrow left at an angle gets split into components, and only then do you write \u03a3Fx and \u03a3Fy.<\/p> <svg viewBox=\"0 0 700 330\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" role=\"img\" aria-label=\"A crate pulled by a rope at 30 degrees, redrawn as a free body diagram with four force arrows acting on a single point\" style=\"width:100%;height:auto;max-width:700px;\"> <defs> <marker id=\"fbdArrInk\" markerWidth=\"10\" markerHeight=\"10\" refX=\"9\" refY=\"3.5\" orient=\"auto\"><polygon points=\"0 0, 10 3.5, 0 7\" fill=\"#0A1628\"\/><\/marker> <marker id=\"fbdArrWine\" markerWidth=\"10\" markerHeight=\"10\" refX=\"9\" refY=\"3.5\" orient=\"auto\"><polygon points=\"0 0, 10 3.5, 0 7\" fill=\"#7A1F2B\"\/><\/marker> <marker id=\"fbdArrGold\" markerWidth=\"10\" markerHeight=\"10\" refX=\"9\" refY=\"3.5\" orient=\"auto\"><polygon points=\"0 0, 10 3.5, 0 7\" fill=\"#C8932A\"\/><\/marker> <\/defs> <rect x=\"0\" y=\"0\" width=\"700\" height=\"330\" fill=\"#F5F2EA\"\/> <rect x=\"0\" y=\"0\" width=\"700\" height=\"330\" fill=\"none\" stroke=\"#D9CFB8\" stroke-width=\"2\"\/> <text x=\"175\" y=\"34\" font-family=\"Georgia,serif\" font-size=\"16\" fill=\"#7A1F2B\" text-anchor=\"middle\" font-weight=\"bold\">1. The real situation<\/text> <text x=\"520\" y=\"34\" font-family=\"Georgia,serif\" font-size=\"16\" fill=\"#7A1F2B\" text-anchor=\"middle\" font-weight=\"bold\">2. The free body diagram<\/text> <line x1=\"45\" y1=\"245\" x2=\"300\" y2=\"245\" stroke=\"#0A1628\" stroke-width=\"3\"\/> <line x1=\"55\" y1=\"245\" x2=\"45\" y2=\"257\" stroke=\"#0A1628\" stroke-width=\"1.5\"\/> <line x1=\"85\" y1=\"245\" x2=\"75\" y2=\"257\" stroke=\"#0A1628\" stroke-width=\"1.5\"\/> <line x1=\"115\" y1=\"245\" x2=\"105\" y2=\"257\" stroke=\"#0A1628\" stroke-width=\"1.5\"\/> <line x1=\"145\" y1=\"245\" x2=\"135\" y2=\"257\" stroke=\"#0A1628\" stroke-width=\"1.5\"\/> <line x1=\"175\" y1=\"245\" x2=\"165\" y2=\"257\" stroke=\"#0A1628\" stroke-width=\"1.5\"\/> <line x1=\"205\" y1=\"245\" x2=\"195\" y2=\"257\" stroke=\"#0A1628\" stroke-width=\"1.5\"\/> <line x1=\"235\" y1=\"245\" x2=\"225\" y2=\"257\" stroke=\"#0A1628\" stroke-width=\"1.5\"\/> <line x1=\"265\" y1=\"245\" x2=\"255\" y2=\"257\" stroke=\"#0A1628\" stroke-width=\"1.5\"\/> <line x1=\"295\" y1=\"245\" x2=\"285\" y2=\"257\" stroke=\"#0A1628\" stroke-width=\"1.5\"\/> <rect x=\"105\" y=\"185\" width=\"95\" height=\"60\" fill=\"#142139\" stroke=\"#0A1628\" stroke-width=\"2\" rx=\"3\"\/> <text x=\"152\" y=\"222\" font-family=\"Georgia,serif\" font-size=\"15\" fill=\"#FAF6EE\" text-anchor=\"middle\">crate<\/text> <line x1=\"200\" y1=\"185\" x2=\"278\" y2=\"140\" stroke=\"#C8932A\" stroke-width=\"4\" marker-end=\"url(#fbdArrGold)\"\/> <text x=\"255\" y=\"128\" font-family=\"Georgia,serif\" font-size=\"16\" fill=\"#C8932A\" font-weight=\"bold\">rope<\/text> <path d=\"M 245 185 A 45 45 0 0 0 239 162.5\" fill=\"none\" stroke=\"#0A1628\" stroke-width=\"1.6\"\/> <text x=\"250\" y=\"176\" font-family=\"Georgia,serif\" font-size=\"14\" fill=\"#0A1628\">30\u00b0<\/text> <text x=\"175\" y=\"292\" font-family=\"Georgia,serif\" font-size=\"13\" fill=\"#1F2E47\" text-anchor=\"middle\" font-style=\"italic\">A rough floor, a crate, a rope<\/text> <line x1=\"318\" y1=\"200\" x2=\"368\" y2=\"200\" stroke=\"#7A1F2B\" stroke-width=\"3\" marker-end=\"url(#fbdArrWine)\"\/> <line x1=\"410\" y1=\"200\" x2=\"632\" y2=\"200\" stroke=\"#C5D0DC\" stroke-width=\"1.5\" stroke-dasharray=\"5,4\"\/> <line x1=\"520\" y1=\"88\" x2=\"520\" y2=\"292\" stroke=\"#C5D0DC\" stroke-width=\"1.5\" stroke-dasharray=\"5,4\"\/> <text x=\"640\" y=\"205\" font-family=\"Georgia,serif\" font-size=\"14\" fill=\"#1F2E47\">x<\/text> <text x=\"527\" y=\"86\" font-family=\"Georgia,serif\" font-size=\"14\" fill=\"#1F2E47\">y<\/text> <line x1=\"520\" y1=\"200\" x2=\"520\" y2=\"112\" stroke=\"#0A1628\" stroke-width=\"4\" marker-end=\"url(#fbdArrInk)\"\/> <text x=\"533\" y=\"120\" font-family=\"Georgia,serif\" font-size=\"17\" fill=\"#0A1628\" font-weight=\"bold\">N<\/text> <line x1=\"520\" y1=\"200\" x2=\"520\" y2=\"284\" stroke=\"#7A1F2B\" stroke-width=\"4\" marker-end=\"url(#fbdArrWine)\"\/> <text x=\"533\" y=\"278\" font-family=\"Georgia,serif\" font-size=\"17\" fill=\"#7A1F2B\" font-weight=\"bold\">W = mg<\/text> <line x1=\"520\" y1=\"200\" x2=\"597\" y2=\"156\" stroke=\"#C8932A\" stroke-width=\"4\" marker-end=\"url(#fbdArrGold)\"\/> <text x=\"601\" y=\"150\" font-family=\"Georgia,serif\" font-size=\"17\" fill=\"#C8932A\" font-weight=\"bold\">T<\/text> <line x1=\"520\" y1=\"200\" x2=\"438\" y2=\"200\" stroke=\"#0A1628\" stroke-width=\"4\" marker-end=\"url(#fbdArrInk)\"\/> <text x=\"428\" y=\"192\" font-family=\"Georgia,serif\" font-size=\"17\" fill=\"#0A1628\" font-weight=\"bold\">f<\/text> <circle cx=\"520\" cy=\"200\" r=\"8\" fill=\"#0A1628\"\/> <text x=\"520\" y=\"315\" font-family=\"Georgia,serif\" font-size=\"13\" fill=\"#1F2E47\" text-anchor=\"middle\" font-style=\"italic\">The crate becomes a dot. Only forces ON it survive.<\/text> <\/svg> <p style=\"text-align:center;font-size:13px;font-style:italic;color:#1F2E47;\">The five steps in action: a real crate on a rough floor becomes four labelled arrows on a point.<\/p>\n\n<div class=\"pf-sim-slot\"><div class=\"pf-sim-slot-header\"><span class=\"icon-dot\"><\/span><span class=\"label\">Free Body Diagram Lab<\/span><\/div><div class=\"pf-sim-slot-body\"><style>.pf-sim-frame{width:100%;border:none;height:600px}@media(max-width:760px){.pf-sim-frame{height:1000px}}<\/style><iframe src=\"\/labs\/free-body-diagram.html?embed=1\" class=\"pf-sim-frame\" loading=\"lazy\"><\/iframe><\/div><\/div>\n\n<h2>The 4 Forces That Never Belong on a Free Body Diagram<\/h2> <p>Four forces show up on student diagrams again and again, and none of them is real in this context. Learning to spot them is faster than learning any formula.<\/p> <h3>1. The &#8220;force of motion&#8221;<\/h3> <p>A ball rolling across a floor has no forward arrow. It is moving because nothing has stopped it yet \u2014 motion needs no cause, only <em>changes<\/em> in motion do. Velocity is not a force and never earns an arrow.<\/p> <svg viewBox=\"0 0 700 300\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" role=\"img\" aria-label=\"Two free body diagrams of a ball rolling on a floor: the incorrect one adds a phantom forward force of motion, the correct one shows only weight, normal force and friction\" style=\"width:100%;height:auto;max-width:700px;\"> <defs> <marker id=\"phArrInk\" markerWidth=\"10\" markerHeight=\"10\" refX=\"9\" refY=\"3.5\" orient=\"auto\"><polygon points=\"0 0, 10 3.5, 0 7\" fill=\"#0A1628\"\/><\/marker> <marker id=\"phArrWine\" markerWidth=\"10\" markerHeight=\"10\" refX=\"9\" refY=\"3.5\" orient=\"auto\"><polygon points=\"0 0, 10 3.5, 0 7\" fill=\"#7A1F2B\"\/><\/marker> <marker id=\"phArrMist\" markerWidth=\"10\" markerHeight=\"10\" refX=\"9\" refY=\"3.5\" orient=\"auto\"><polygon points=\"0 0, 10 3.5, 0 7\" fill=\"#8A8FA0\"\/><\/marker> <\/defs> <rect x=\"0\" y=\"0\" width=\"700\" height=\"300\" fill=\"#F5F2EA\"\/> <rect x=\"0\" y=\"0\" width=\"700\" height=\"300\" fill=\"none\" stroke=\"#D9CFB8\" stroke-width=\"2\"\/> <text x=\"350\" y=\"30\" font-family=\"Georgia,serif\" font-size=\"16\" fill=\"#7A1F2B\" text-anchor=\"middle\" font-weight=\"bold\">A ball coasting to the right, already moving, nothing pushing it<\/text> <line x1=\"350\" y1=\"48\" x2=\"350\" y2=\"272\" stroke=\"#D9CFB8\" stroke-width=\"2\"\/> <line x1=\"128\" y1=\"61\" x2=\"142\" y2=\"75\" stroke=\"#7A1F2B\" stroke-width=\"3\"\/><line x1=\"142\" y1=\"61\" x2=\"128\" y2=\"75\" stroke=\"#7A1F2B\" stroke-width=\"3\"\/><text x=\"152\" y=\"74\" font-family=\"Georgia,serif\" font-size=\"15\" fill=\"#7A1F2B\" font-weight=\"bold\">Wrong<\/text> <line x1=\"60\" y1=\"170\" x2=\"290\" y2=\"170\" stroke=\"#C5D0DC\" stroke-width=\"1.5\" stroke-dasharray=\"5,4\"\/> <line x1=\"175\" y1=\"170\" x2=\"175\" y2=\"98\" stroke=\"#0A1628\" stroke-width=\"4\" marker-end=\"url(#phArrInk)\"\/> <text x=\"186\" y=\"106\" font-family=\"Georgia,serif\" font-size=\"16\" fill=\"#0A1628\" font-weight=\"bold\">N<\/text> <line x1=\"175\" y1=\"170\" x2=\"175\" y2=\"242\" stroke=\"#7A1F2B\" stroke-width=\"4\" marker-end=\"url(#phArrWine)\"\/> <text x=\"186\" y=\"238\" font-family=\"Georgia,serif\" font-size=\"16\" fill=\"#7A1F2B\" font-weight=\"bold\">W<\/text> <line x1=\"175\" y1=\"170\" x2=\"113\" y2=\"170\" stroke=\"#0A1628\" stroke-width=\"4\" marker-end=\"url(#phArrInk)\"\/> <text x=\"104\" y=\"162\" font-family=\"Georgia,serif\" font-size=\"16\" fill=\"#0A1628\" font-weight=\"bold\">f<\/text> <line x1=\"175\" y1=\"170\" x2=\"253\" y2=\"170\" stroke=\"#8A8FA0\" stroke-width=\"4\" stroke-dasharray=\"7,5\" marker-end=\"url(#phArrMist)\"\/> <text x=\"196\" y=\"200\" font-family=\"Georgia,serif\" font-size=\"15\" fill=\"#8A8FA0\" font-weight=\"bold\">F(motion)<\/text> <circle cx=\"175\" cy=\"170\" r=\"8\" fill=\"#0A1628\"\/> <text x=\"175\" y=\"266\" font-family=\"Georgia,serif\" font-size=\"13\" fill=\"#7A1F2B\" text-anchor=\"middle\" font-style=\"italic\">Nothing is touching it from behind.<\/text> <polyline points=\"480,68 487,75 499,60\" fill=\"none\" stroke=\"#0A1628\" stroke-width=\"3\" stroke-linecap=\"round\" stroke-linejoin=\"round\"\/><text x=\"508\" y=\"74\" font-family=\"Georgia,serif\" font-size=\"15\" fill=\"#0A1628\" font-weight=\"bold\">Right<\/text> <line x1=\"410\" y1=\"170\" x2=\"640\" y2=\"170\" stroke=\"#C5D0DC\" stroke-width=\"1.5\" stroke-dasharray=\"5,4\"\/> <line x1=\"525\" y1=\"170\" x2=\"525\" y2=\"98\" stroke=\"#0A1628\" stroke-width=\"4\" marker-end=\"url(#phArrInk)\"\/> <text x=\"536\" y=\"106\" font-family=\"Georgia,serif\" font-size=\"16\" fill=\"#0A1628\" font-weight=\"bold\">N<\/text> <line x1=\"525\" y1=\"170\" x2=\"525\" y2=\"242\" stroke=\"#7A1F2B\" stroke-width=\"4\" marker-end=\"url(#phArrWine)\"\/> <text x=\"536\" y=\"238\" font-family=\"Georgia,serif\" font-size=\"16\" fill=\"#7A1F2B\" font-weight=\"bold\">W<\/text> <line x1=\"525\" y1=\"170\" x2=\"463\" y2=\"170\" stroke=\"#0A1628\" stroke-width=\"4\" marker-end=\"url(#phArrInk)\"\/> <text x=\"454\" y=\"162\" font-family=\"Georgia,serif\" font-size=\"16\" fill=\"#0A1628\" font-weight=\"bold\">f<\/text> <circle cx=\"525\" cy=\"170\" r=\"8\" fill=\"#0A1628\"\/> <text x=\"586\" y=\"200\" font-family=\"Georgia,serif\" font-size=\"15\" fill=\"#C8932A\" font-weight=\"bold\" font-style=\"italic\">v<\/text><line x1=\"598\" y1=\"195\" x2=\"626\" y2=\"195\" stroke=\"#C8932A\" stroke-width=\"2.5\"\/><polyline points=\"620,190 627,195 620,200\" fill=\"none\" stroke=\"#C8932A\" stroke-width=\"2.5\" stroke-linecap=\"round\" stroke-linejoin=\"round\"\/> <text x=\"525\" y=\"266\" font-family=\"Georgia,serif\" font-size=\"13\" fill=\"#1F2E47\" text-anchor=\"middle\" font-style=\"italic\">Velocity is not a force. It never gets an arrow.<\/text> <\/svg> <p style=\"text-align:center;font-size:13px;font-style:italic;color:#1F2E47;\">The phantom &#8220;force of motion&#8221; is the most common wrong arrow in introductory mechanics.<\/p> <h3>2. Centrifugal force<\/h3> <p>Swing a conker on a string and the string pulls the conker <em>inward<\/em>, toward your hand. There is no outward arrow on the conker. The outward feeling belongs to your hand, which the conker really does pull.<\/p> <h3>3. The third-law partner<\/h3> <p>Newton&#8217;s third law pairs always act on <em>different<\/em> bodies, so they can never appear on the same diagram. The book pushes down on the table; the table pushes up on the book. Only the second one belongs on the book&#8217;s diagram.<\/p> <h3>4. The net force itself<\/h3> <p>\u03a3F and ma are results, not inputs. Drawing an extra arrow labelled &#8220;ma&#8221; double-counts forces you have already drawn, and the equation collapses.<\/p> <h2>Common Forces and Which Way They Point<\/h2> <p>Most diagrams are built from the same short list of forces. This table is the one to memorise \u2014 direction first, magnitude second.<\/p> <div class=\"pf-table-scroll\" style=\"display:block;width:100%;max-width:100%;overflow-x:auto;-webkit-overflow-scrolling:touch;margin:1.5em 0;\"> <table style=\"width:100%;border-collapse:collapse;word-break:break-word;\"> <thead> <tr style=\"background:#0A1628;color:#FAF6EE;\"> <th style=\"padding:10px;border:1px solid #D9CFB8;text-align:left;\">Force<\/th> <th style=\"padding:10px;border:1px solid #D9CFB8;text-align:left;\">Symbol<\/th> <th style=\"padding:10px;border:1px solid #D9CFB8;text-align:left;\">Direction of the arrow<\/th> <th style=\"padding:10px;border:1px solid #D9CFB8;text-align:left;\">Magnitude<\/th> <\/tr> <\/thead> <tbody> <tr><td style=\"padding:10px;border:1px solid #D9CFB8;\">Weight<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">W<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">Always vertically down, whatever the surface is doing<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">mg, with g \u2248 9.81 m\/s\u00b2<\/td><\/tr> <tr style=\"background:#F5F2EA;\"><td style=\"padding:10px;border:1px solid #D9CFB8;\">Normal force<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">N<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">Perpendicular to the contact surface, pushing away from it<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">Whatever balances the perpendicular direction \u2014 <em>not<\/em> automatically mg<\/td><\/tr> <tr><td style=\"padding:10px;border:1px solid #D9CFB8;\">Friction<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">f<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">Along the surface, opposing sliding or the tendency to slide<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">Kinetic: \u03bc<sub>k<\/sub>N. Static: anything up to \u03bc<sub>s<\/sub>N<\/td><\/tr> <tr style=\"background:#F5F2EA;\"><td style=\"padding:10px;border:1px solid #D9CFB8;\">Tension<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">T<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">Along the rope, away from the body \u2014 ropes pull, never push<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">Usually an unknown you solve for<\/td><\/tr> <tr><td style=\"padding:10px;border:1px solid #D9CFB8;\">Applied force<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">F<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">Whichever way the push or pull acts<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">Given in the problem<\/td><\/tr> <tr style=\"background:#F5F2EA;\"><td style=\"padding:10px;border:1px solid #D9CFB8;\">Drag<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">F<sub>d<\/sub><\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">Directly opposite the velocity through the fluid<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">Grows with speed<\/td><\/tr> <tr><td style=\"padding:10px;border:1px solid #D9CFB8;\">Spring force<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">F<sub>s<\/sub><\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">Opposite the stretch or compression, back toward natural length<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">kx<\/td><\/tr> <\/tbody> <\/table> <\/div> <p>Notice how many entries describe direction by a <em>rule<\/em> rather than a picture. That is deliberate \u2014 the rules survive when the geometry gets strange, and a good grasp of the <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/types-of-forces\/\">different types of forces in physics<\/a> is what lets you populate a diagram quickly.<\/p> <h2>Free Body Diagrams on an Inclined Plane<\/h2> <p>On an inclined plane, tilt your axes so x runs along the slope and y runs perpendicular to it, then resolve the weight into mg sin \u03b8 down the slope and mg cos \u03b8 into the surface. This one choice removes almost all the algebra.<\/p> <p>Why does it help so much? Because on a slope, N and f already lie along the tilted axes. Leave the axes horizontal and you must split three forces instead of one.<\/p> <svg viewBox=\"0 0 700 340\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" role=\"img\" aria-label=\"Free body diagram of a block on an inclined plane with axes tilted along the slope, showing weight resolved into mg sine theta down the slope and mg cosine theta into the surface\" style=\"width:100%;height:auto;max-width:700px;\"> <defs> <marker id=\"incArrInk\" markerWidth=\"10\" markerHeight=\"10\" refX=\"9\" refY=\"3.5\" orient=\"auto\"><polygon points=\"0 0, 10 3.5, 0 7\" fill=\"#0A1628\"\/><\/marker> <marker id=\"incArrWine\" markerWidth=\"10\" markerHeight=\"10\" refX=\"9\" refY=\"3.5\" orient=\"auto\"><polygon points=\"0 0, 10 3.5, 0 7\" fill=\"#7A1F2B\"\/><\/marker> <marker id=\"incArrGold\" markerWidth=\"10\" markerHeight=\"10\" refX=\"9\" refY=\"3.5\" orient=\"auto\"><polygon points=\"0 0, 10 3.5, 0 7\" fill=\"#C8932A\"\/><\/marker> <\/defs> <rect x=\"0\" y=\"0\" width=\"700\" height=\"340\" fill=\"#F5F2EA\"\/> <rect x=\"0\" y=\"0\" width=\"700\" height=\"340\" fill=\"none\" stroke=\"#D9CFB8\" stroke-width=\"2\"\/> <text x=\"350\" y=\"30\" font-family=\"Georgia,serif\" font-size=\"16\" fill=\"#7A1F2B\" text-anchor=\"middle\" font-weight=\"bold\">Tilt the axes, then split the weight<\/text> <path d=\"M 60 80 L 60 270 L 430 270 Z\" fill=\"#142139\" opacity=\"0.09\"\/> <path d=\"M 60 80 L 60 270 L 430 270 Z\" fill=\"none\" stroke=\"#0A1628\" stroke-width=\"2.5\"\/> <path d=\"M 372 270 A 58 58 0 0 1 378 244\" fill=\"none\" stroke=\"#0A1628\" stroke-width=\"1.8\"\/> <text x=\"384\" y=\"264\" font-family=\"Georgia,serif\" font-size=\"16\" fill=\"#0A1628\">\u03b8<\/text> <g transform=\"translate(215,160) rotate(27.2)\"> <rect x=\"-32\" y=\"-28\" width=\"64\" height=\"28\" fill=\"#142139\" opacity=\"0.20\"\/> <rect x=\"-32\" y=\"-28\" width=\"64\" height=\"28\" fill=\"none\" stroke=\"#0A1628\" stroke-width=\"1.6\"\/> <\/g> <line x1=\"222\" y1=\"147\" x2=\"338\" y2=\"206\" stroke=\"#C5D0DC\" stroke-width=\"1.6\" stroke-dasharray=\"5,4\"\/> <text x=\"344\" y=\"212\" font-family=\"Georgia,serif\" font-size=\"14\" fill=\"#1F2E47\">x<\/text> <line x1=\"222\" y1=\"147\" x2=\"222\" y2=\"257\" stroke=\"#7A1F2B\" stroke-width=\"4.5\" marker-end=\"url(#incArrWine)\"\/> <text x=\"231\" y=\"262\" font-family=\"Georgia,serif\" font-size=\"17\" fill=\"#7A1F2B\" font-weight=\"bold\">W = mg<\/text> <line x1=\"267\" y1=\"170\" x2=\"222\" y2=\"257\" stroke=\"#C5D0DC\" stroke-width=\"1.4\" stroke-dasharray=\"4,3\"\/> <line x1=\"177\" y1=\"234\" x2=\"222\" y2=\"257\" stroke=\"#C5D0DC\" stroke-width=\"1.4\" stroke-dasharray=\"4,3\"\/> <line x1=\"222\" y1=\"147\" x2=\"267\" y2=\"170\" stroke=\"#C8932A\" stroke-width=\"4\" marker-end=\"url(#incArrGold)\"\/> <text x=\"280\" y=\"166\" font-family=\"Georgia,serif\" font-size=\"16\" fill=\"#C8932A\" font-weight=\"bold\">mg sin \u03b8<\/text> <line x1=\"222\" y1=\"147\" x2=\"177\" y2=\"234\" stroke=\"#C8932A\" stroke-width=\"4\" marker-end=\"url(#incArrGold)\"\/> <text x=\"88\" y=\"252\" font-family=\"Georgia,serif\" font-size=\"16\" fill=\"#C8932A\" font-weight=\"bold\">mg cos \u03b8<\/text> <line x1=\"222\" y1=\"147\" x2=\"262\" y2=\"69\" stroke=\"#0A1628\" stroke-width=\"4\" marker-end=\"url(#incArrInk)\"\/> <text x=\"269\" y=\"64\" font-family=\"Georgia,serif\" font-size=\"17\" fill=\"#0A1628\" font-weight=\"bold\">N<\/text> <text x=\"288\" y=\"80\" font-family=\"Georgia,serif\" font-size=\"13\" fill=\"#1F2E47\">(y)<\/text> <line x1=\"222\" y1=\"147\" x2=\"158\" y2=\"114\" stroke=\"#0A1628\" stroke-width=\"4\" marker-end=\"url(#incArrInk)\"\/> <text x=\"136\" y=\"106\" font-family=\"Georgia,serif\" font-size=\"17\" fill=\"#0A1628\" font-weight=\"bold\">f<\/text> <circle cx=\"222\" cy=\"147\" r=\"7\" fill=\"#0A1628\"\/> <rect x=\"470\" y=\"72\" width=\"218\" height=\"184\" fill=\"#FAF6EE\" stroke=\"#D9CFB8\" stroke-width=\"2\" rx=\"4\"\/> <text x=\"486\" y=\"100\" font-family=\"Georgia,serif\" font-size=\"15\" fill=\"#7A1F2B\" font-weight=\"bold\">Along the slope (x):<\/text> <text x=\"486\" y=\"126\" font-family=\"Georgia,serif\" font-size=\"15\" fill=\"#0A1628\">mg sin \u03b8 &#8211; f = ma<\/text> <text x=\"486\" y=\"158\" font-family=\"Georgia,serif\" font-size=\"15\" fill=\"#7A1F2B\" font-weight=\"bold\">Into the slope (y):<\/text> <text x=\"486\" y=\"184\" font-family=\"Georgia,serif\" font-size=\"15\" fill=\"#0A1628\">N = mg cos \u03b8<\/text> <line x1=\"486\" y1=\"202\" x2=\"672\" y2=\"202\" stroke=\"#D9CFB8\" stroke-width=\"1.5\"\/> <text x=\"486\" y=\"224\" font-family=\"Georgia,serif\" font-size=\"13\" fill=\"#1F2E47\">N is smaller than mg \u2014 the<\/text> <text x=\"486\" y=\"242\" font-family=\"Georgia,serif\" font-size=\"13\" fill=\"#1F2E47\">slope carries only part of it.<\/text> <text x=\"350\" y=\"320\" font-family=\"Georgia,serif\" font-size=\"13\" fill=\"#1F2E47\" text-anchor=\"middle\" font-style=\"italic\">With x along the slope, only the weight needs splitting \u2014 N and f already lie on an axis.<\/text> <\/svg> <p style=\"text-align:center;font-size:13px;font-style:italic;color:#1F2E47;\">Tilting the axes on an incline turns a three-force resolution into a one-force resolution.<\/p> <p>The perpendicular equation is where the insight hides. Since the block does not accelerate into the ramp, N = mg cos \u03b8 \u2014 smaller than the weight, and smaller still as the slope steepens.<\/p> <p>That is why friction fades on a steep ramp: friction depends on N, and N is shrinking exactly when gravity&#8217;s pull along the slope is growing. Our full guide to <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/inclined-plane-physics\/\">inclined plane physics<\/a> works through the sliding condition in detail.<\/p> <h2>Real-World Examples of Free Body Diagrams<\/h2> <p>These diagrams are not a classroom ritual. Every one of the situations below is solved this way in professional practice.<\/p> <h3>A lift accelerating upward<\/h3> <p>Standing in a lift, you feel heavier as it starts to rise. Your diagram has just two arrows: N up from the floor, W down. Because you accelerate upward, N must exceed W \u2014 and N is exactly what a bathroom scale reads.<\/p> <h3>A climber on a rope<\/h3> <p>A hanging climber has weight down and tension up, and while they hang still the two are equal. The moment they are lowered with acceleration, tension drops below weight \u2014 which is why a controlled lower feels gentler than a sudden stop.<\/p> <h3>A plane in level flight<\/h3> <p>Four arrows: lift up, weight down, thrust forward, drag back. Cruising at constant speed and height means both pairs cancel exactly, and the whole of aerodynamics starts from that balance.<\/p> <h3>A crate on a lorry that brakes<\/h3> <p>Here friction is the only horizontal force on the crate, and it must point <em>backward<\/em> to slow the crate with the lorry. If the required friction exceeds \u03bc<sub>s<\/sub>N, the crate slides forward \u2014 the calculation that sets load-securing rules.<\/p> <h3>A parked car on a hill<\/h3> <p>Weight down, normal force out of the slope, friction up the slope, and nothing accelerating. Engineers size handbrakes by asking whether friction alone can supply mg sin \u03b8.<\/p> <h2>Common Misconceptions About Free Body Diagrams<\/h2> <h3>&#8220;The normal force always equals the weight&#8221;<\/h3> <p>It equals the weight only in the narrow case of a flat surface with nothing else acting vertically. Tilt the surface, add a rope pulled at an angle, or accelerate vertically, and N changes immediately.<\/p> <p>This is the single most expensive mistake in force problems, because friction depends on N. Get N wrong and every friction value downstream is wrong.<\/p> <h3>&#8220;Static friction equals \u03bc<sub>s<\/sub>N&#8221;<\/h3> <p>That product is the <em>maximum<\/em> static friction, not its actual value. Static friction is whatever it needs to be to prevent sliding, right up until it cannot manage any more.<\/p> <p>A book resting on a gentle slope needs only a few newtons of friction, even if the surface could supply thirty. Read more in our guide to <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/what-is-friction\/\">what friction is and how it works<\/a>.<\/p> <h3>&#8220;If it&#8217;s moving, something must be pushing it&#8221;<\/h3> <p>Constant velocity means zero acceleration, which means the forces balance exactly. A car at a steady 70 mph has thrust and drag in perfect opposition \u2014 it is a case of <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/equilibrium-physics\/\">equilibrium<\/a>, not of a winning force.<\/p> <h3>&#8220;Two cables each carry half the weight&#8221;<\/h3> <p>Only if both cables hang vertically. Angle them and each carries more than half, because only the vertical component of each tension does the lifting \u2014 a point Worked Problem 4 below makes concrete.<\/p> <h2>How Free Body Diagrams Fit With Newton&#8217;s Laws<\/h2> <p>Each of <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/newtons-laws-of-motion\/\">Newton&#8217;s three laws of motion<\/a> maps onto a different part of the diagram, which is why the technique feels natural once it clicks.<\/p> <ul> <li><strong>First law<\/strong> \u2014 tells you what a balanced diagram means: at rest or at constant velocity, every arrow cancels.<\/li> <li><strong>Second law<\/strong> \u2014 turns an unbalanced diagram into numbers through \u03a3F = ma.<\/li> <li><strong>Third law<\/strong> \u2014 tells you which arrows are forbidden, since a reaction force lives on the other body&#8217;s diagram.<\/li> <\/ul> <p>The diagram is also where vector skills earn their keep. Forces are vectors, so direction carries as much information as size, and confidence with <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/kinematics\/scalar-vector-quantities\/\">vector and scalar quantities<\/a> is what makes resolving components feel routine.<\/p> <p>One more connection worth holding on to: ropes and pulleys are just tension arrows with a geometric constraint attached. Our guide to <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/tension-force\/\">tension force<\/a> covers the ideal-rope assumptions that let you carry one symbol T across two diagrams.<\/p> <h2>Worked Problems<\/h2>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 1<\/div><div class=\"pf-problem-question\">A 2.0 kg book rests on a level table. Draw the free body diagram and find the normal force. Take g = 9.81 m\/s\u00b2.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<strong>Solution:<\/strong>\nStep 1: Two arrows only \u2014 W down, N up. The book is at rest, so a = 0 and \u03a3Fy = 0.\nStep 2: \u03a3Fy = N \u2212 W = 0, so N = W = mg.\nStep 3: N = 2.0 kg \u00d7 9.81 m\/s\u00b2 = 19.62 N.\n<strong>Answer: N = 19.6 N (3 s.f.), directed vertically upward.<\/strong>\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 2<\/div><div class=\"pf-problem-question\">A 65 kg person stands in a lift accelerating upward at 1.2 m\/s\u00b2. Find the normal force from the floor.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<strong>Solution:<\/strong>\nStep 1: Arrows are N up and W down. Take up as positive; the acceleration is upward, so ay = +1.2 m\/s\u00b2.\nStep 2: \u03a3Fy = N \u2212 mg = ma, so N = m(g + a).\nStep 3: N = 65 kg \u00d7 (9.81 + 1.2) m\/s\u00b2 = 65 \u00d7 11.01 = 715.65 N.\nSanity check: at rest N would be 637.65 N, so the person genuinely feels heavier.\n<strong>Answer: N = 716 N (3 s.f.).<\/strong>\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 3<\/div><div class=\"pf-problem-question\">A 12 kg crate on a level floor is pushed horizontally with 45 N. The coefficient of kinetic friction is 0.25. Find the acceleration.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<strong>Solution:<\/strong>\nStep 1: Vertical direction first \u2014 no vertical acceleration, so N = mg = 12 \u00d7 9.81 = 117.72 N.\nStep 2: Kinetic friction f = \u03bc<sub>k<\/sub>N = 0.25 \u00d7 117.72 = 29.43 N, opposing the push.\nStep 3: \u03a3Fx = 45 \u2212 29.43 = 15.57 N, so a = 15.57 \/ 12 = 1.2975 m\/s\u00b2.\n<strong>Answer: a = 1.30 m\/s\u00b2 (3 s.f.), in the direction of the push.<\/strong>\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 4<\/div><div class=\"pf-problem-question\">An 8.0 kg sign hangs from two cables, each making 30\u00b0 with the horizontal. Find the tension in each cable.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<strong>Solution:<\/strong>\nStep 1: Three arrows \u2014 W down, and two tensions T pulling up and outward at 30\u00b0 above horizontal. By symmetry both tensions are equal.\nStep 2: Horizontal components cancel. Vertically: 2T sin 30\u00b0 \u2212 W = 0.\nStep 3: W = 8.0 \u00d7 9.81 = 78.48 N, so T = 78.48 \/ (2 \u00d7 0.500) = 78.48 N.\nNote the result: each cable carries the <strong>full<\/strong> weight, not half \u2014 because at 30\u00b0 only half of each tension acts vertically.\n<strong>Answer: T = 78.5 N (3 s.f.) in each cable.<\/strong>\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 5<\/div><div class=\"pf-problem-question\">A 5.0 kg block slides down a frictionless 30\u00b0 incline. Find its acceleration and the normal force.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<strong>Solution:<\/strong>\nStep 1: Tilt the axes \u2014 x down the slope, y perpendicular. Resolve W into mg sin \u03b8 along x and mg cos \u03b8 along \u2212y.\nStep 2: Along x: \u03a3Fx = mg sin \u03b8 = ma, so a = g sin \u03b8 = 9.81 \u00d7 0.500 = 4.905 m\/s\u00b2.\nStep 3: Along y: N \u2212 mg cos \u03b8 = 0, so N = 5.0 \u00d7 9.81 \u00d7 cos 30\u00b0 = 49.05 \u00d7 0.8660 = 42.48 N.\nNote that a does not depend on mass \u2014 every object slides at the same rate on a frictionless slope.\n<strong>Answer: a = 4.91 m\/s\u00b2 down the slope; N = 42.5 N (3 s.f.).<\/strong>\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 6<\/div><div class=\"pf-problem-question\">A 3.0 kg block sits on a 20\u00b0 incline with \u03bcs = 0.45. Does it slide? If not, what is the friction force acting on it?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<strong>Solution:<\/strong>\nStep 1: Find the pull along the slope: mg sin \u03b8 = 3.0 \u00d7 9.81 \u00d7 sin 20\u00b0 = 29.43 \u00d7 0.3420 = 10.066 N.\nStep 2: Find the maximum static friction available: \u03bc<sub>s<\/sub>N = \u03bc<sub>s<\/sub>mg cos \u03b8 = 0.45 \u00d7 29.43 \u00d7 0.9397 = 12.445 N.\nStep 3: Since 10.07 N < 12.44 N, the block stays put. Static friction supplies only what is needed to hold it, not its maximum.\nEquivalent check: the block slides only when tan \u03b8 > \u03bc<sub>s<\/sub>, and tan 20\u00b0 = 0.364 < 0.45.\n<strong>Answer: It does not slide. Friction = 10.1 N (3 s.f.) up the slope \u2014 not 12.4 N.<\/strong>\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 7<\/div><div class=\"pf-problem-question\">A 4.0 kg block on a frictionless table is joined by a light cord over a frictionless pulley to a 2.0 kg block hanging freely. Find the acceleration and the cord tension.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<strong>Solution:<\/strong>\nStep 1: Draw two diagrams. Block 1 (on the table): T forward, N up, W down. Block 2 (hanging): T up, m<sub>2<\/sub>g down. The cord fixes both accelerations to the same magnitude a.\nStep 2: Block 1 horizontally: T = m<sub>1<\/sub>a. Block 2 vertically: m<sub>2<\/sub>g \u2212 T = m<sub>2<\/sub>a.\nStep 3: Add the two equations to eliminate T: m<sub>2<\/sub>g = (m<sub>1<\/sub> + m<sub>2<\/sub>)a, so a = (2.0 \u00d7 9.81) \/ 6.0 = 19.62 \/ 6.0 = 3.27 m\/s\u00b2.\nStep 4: Back-substitute: T = m<sub>1<\/sub>a = 4.0 \u00d7 3.27 = 13.08 N.\nCheck with block 2: m<sub>2<\/sub>g \u2212 T = 19.62 \u2212 13.08 = 6.54 N, and m<sub>2<\/sub>a = 2.0 \u00d7 3.27 = 6.54 N. Consistent.\n<strong>Answer: a = 3.27 m\/s\u00b2; T = 13.1 N (3 s.f.). Note T is less than the hanging weight of 19.6 N.<\/strong>\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 8<\/div><div class=\"pf-problem-question\">A 20 kg sled is pulled by a 60 N force at 30\u00b0 above the horizontal across ground with \u03bck = 0.15. Find the normal force and the acceleration.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<strong>Solution:<\/strong>\nStep 1: Resolve the pull. Fx = 60 cos 30\u00b0 = 51.96 N; Fy = 60 sin 30\u00b0 = 30.00 N upward.\nStep 2: Vertically there is no acceleration, so N + Fy \u2212 mg = 0, giving N = mg \u2212 Fy = 196.2 \u2212 30.0 = 166.2 N.\nStep 3: Friction f = \u03bc<sub>k<\/sub>N = 0.15 \u00d7 166.2 = 24.93 N. Then \u03a3Fx = 51.96 \u2212 24.93 = 27.03 N and a = 27.03 \/ 20 = 1.3516 m\/s\u00b2.\nThe lesson: the upward component of the pull lightens the sled, so N is 166 N rather than 196 N \u2014 pulling at an angle beats pushing down at one.\n<strong>Answer: N = 166 N; a = 1.35 m\/s\u00b2 (3 s.f.).<\/strong>\n<\/div><\/details><\/div>\n\n<h2>Frequently Asked Questions<\/h2>\n\n<details class=\"pf-faq-item\"><summary>What is a free body diagram in simple terms?<\/summary><div class=\"pf-faq-item-answer\">\nA free body diagram is a sketch of one object on its own, with an arrow drawn for every force pushing or pulling on it. You remove everything around the object \u2014 floors, ropes, other blocks \u2014 and replace each removed contact with the force it was applying. The arrows then add up as vectors to give the net force.\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>What forces should be included in a free body diagram?<\/summary><div class=\"pf-faq-item-answer\">\nInclude only external forces acting on the chosen body: weight, plus one contact force for each thing physically touching it. A rough surface contributes two, a normal force and friction. Exclude any force the body exerts on something else, any net force or ma term, and any invented force such as a forward &#8220;force of motion&#8221; or centrifugal force.\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>Do you include acceleration in a free body diagram?<\/summary><div class=\"pf-faq-item-answer\">\nNo \u2014 acceleration is not a force and does not get an arrow among the forces. It is the result the diagram helps you calculate through \u03a3F = ma. Many textbooks allow you to note the acceleration direction beside the diagram, clearly separated and often in a different colour, so it is never mistaken for a force.\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>Why is the normal force not always equal to the weight?<\/summary><div class=\"pf-faq-item-answer\">\nThe normal force equals the weight only when the surface is horizontal, nothing else acts vertically, and there is no vertical acceleration. On a slope it becomes mg cos \u03b8. In an accelerating lift it becomes m(g \u00b1 a). If a rope pulls partly upward, the normal force drops by that upward component.\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>How do you draw a free body diagram for two connected objects?<\/summary><div class=\"pf-faq-item-answer\">\nDraw a separate diagram for each object, never one combined picture. Use the same symbol T for the tension in a light cord, since it is the same throughout an ideal rope over a frictionless pulley. Each diagram gives one equation, and the cord supplies a constraint linking the accelerations, so two equations solve for two unknowns.\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>What is the difference between a free body diagram and a force diagram?<\/summary><div class=\"pf-faq-item-answer\">\nThe terms are used almost interchangeably, but a free body diagram is stricter. It isolates exactly one body, usually shrinks it to a point, and shows only external forces acting on that body. A general force diagram may keep the original picture and show forces on several objects at once, which makes it easier to mix up which forces belong where.\n<\/div><\/details>\n\n<p>Free body diagrams reward practice more than memorisation. Draw one for the next five problems you meet \u2014 even the easy ones \u2014 and the harder ones stop looking hard. For a rigorous parallel treatment, the <a href=\"https:\/\/openstax.org\/books\/university-physics-volume-1\/pages\/5-7-drawing-free-body-diagrams\" target=\"_blank\" rel=\"noopener\">OpenStax University Physics section on drawing free-body diagrams<\/a> works through coupled blocks in detail, and <a href=\"https:\/\/ocw.mit.edu\/courses\/8-01sc-classical-mechanics-fall-2016\/pages\/week-2-newtons-laws\/\" target=\"_blank\" rel=\"noopener\">MIT OpenCourseWare&#8217;s 8.01 Week 2 materials<\/a> include filmed worked examples on stacked blocks and pulley systems.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>A free body diagram shows one object on its own with an arrow for every external force acting on it. This guide covers the five steps to draw one, the four forces that never belong, and eight worked problems solved in full.<\/p>\n","protected":false},"author":1,"featured_media":1809,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[2],"tags":[],"class_list":["post-1808","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-mechanics"],"_links":{"self":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts\/1808","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/comments?post=1808"}],"version-history":[{"count":2,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts\/1808\/revisions"}],"predecessor-version":[{"id":1830,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts\/1808\/revisions\/1830"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/media\/1809"}],"wp:attachment":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/media?parent=1808"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/categories?post=1808"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/tags?post=1808"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}