Newton’s Laws · Chapter 4
Forces at Work: Free-Body Diagrams
Almost nothing feels just one force. A book on a shelf is pulled down by gravity and pushed up by the shelf. A sled feels gravity, the snow, the rope and friction all at once. Newton’s second law only cares about their sum, and the free-body diagram is how you find it: draw the object as a dot, draw every force on it as an arrow, then add the arrows. Here you’ll tilt a ramp until a block slips, ride an elevator standing on a bathroom scale, race two weights over a pulley, and draw diagrams of your own.
Ramp
Live numbers
Missions
What you’re seeing
The ramp. A block sits on a board that tilts on a hinge. Drag the gold knob at the top, or use the slider, to change the angle θ. The surfaces menu sets what rubs on what, with the friction numbers measured for real pairs of materials. The ramp is 6 m long; if the block reaches either end, it goes back to the middle.
The arrows. Each color is one force, and each arrow’s length shows its size. Pink is gravity, the weight mg. Its two dashed parts are the pull along the ramp, mg sin θ, and the press into the ramp, mg cos θ. Blue is the normal force N, the ramp pushing back. Red is friction. Green is your push F. Gold is the net force: the sum of all the others.
The free-body diagram. The box in the corner shows the same forces with the block shrunk to a dot, on axes tilted to match the ramp. That is how physicists draw it: one object, every force on it, nothing else.
The friction meter. Static friction is lazy: it pushes only as hard as it must to stop sliding, up to a limit, μsN. Watch the blue bar fill as you tilt. When it reaches the gold line, the block slips, and the bar drops to the smaller kinetic friction, μkN. The graph next to it shows the block’s speed: a flat line means constant speed, a sloping line means acceleration.
The forces you’ll meet most
A force is a push or a pull, measured in newtons (N). One newton is about the weight of a small apple. Almost every force in everyday life is one of these:
Earth pulls every mass straight down. On Earth g = 9.81 m/s², so a 10 kg block weighs 98.1 N. It acts without touching.
A surface pushes back on whatever presses on it, straight out of the surface (“normal” means at right angles). It is only as big as it has to be to stop you sinking in.
Acts along a surface, against sliding. Static friction: fs ≤ μsN, as much as needed up to a limit. Kinetic friction while sliding: fk = μkN.
A rope, cable or chain pulls along its own length. It can pull but never push: you can’t push a rope.
A push or pull from a hand, a motor or anything else, in whatever direction it acts.
Air or water pushes against motion through it. It grows quickly with speed, which is why a falling skydiver reaches a top speed.
Stretch or squeeze a spring by x and it pushes or pulls back in proportion (Hooke’s law). k is the spring’s stiffness in N/m.
Drawing a free-body diagram, and solving with it
- Pick one object. Draw it as a dot.
- Add every force on it as an arrow from the dot: gravity, plus one force for each thing that touches it (a surface gives a normal force and maybe friction, a rope gives tension). Leave out forces the object puts on other things, and don’t draw the net force as an extra arrow.
- Choose axes. Point one along the way the object moves or would move. On a ramp, that means one axis along the ramp and one across it.
- Split any force that isn’t along an axis into two parts (components), using sine and cosine.
- Apply the second law along each axis, then solve. Check that the units and the size of the answer make sense.
For the block on the ramp, with x pointing up the ramp:
along: F − mg sin θ + f = m a
With no push, the block starts to slip when mg sin θ is bigger than μsmg cos θ, which means
The mass cancels out. A heavy crate and a light one of the same materials slip at the same angle.
A 20 kg wooden crate rests on a wooden ramp tilted 15°. For wood on wood, μs = 0.5 and μk = 0.3.
- Weight: mg = 20 kg × 9.81 m/s² = 196.1 N.
- Along the ramp: mg sin 15° = 50.8 N. Into the ramp: mg cos 15° = 189.4 N, so N = 189.4 N.
- Most static friction can give: μsN = 0.5 × 189.4 = 94.7 N. That is more than the 50.8 N trying to slide it, so the crate stays put, and static friction is exactly 50.8 N, pointing up the ramp.
- To start it moving up the ramp, a push must beat gravity’s 50.8 N and the full static friction: 50.8 + 94.7 = 145.5 N.
- Once it moves, friction drops to μkN = 56.8 N, so 50.8 + 56.8 = 107.6 N keeps it going at a steady speed. Starting is harder than keeping going. Try it: the 📦 setup is this crate.
Waxed wood on wet snow has μs = 0.14 and μk = 0.10. A sled at rest starts to slide by itself on any hill steeper than tan⁻¹(0.14) = 8.0°. On a 10° hill, once it is going:
After 5 seconds it is moving at 3.7 m/s (about 8 mph) and has gone 9.2 m. On a 5° hill, a sled at rest stays put; give it a shove and it slows down and stops, because tan 5° = 0.087 is less than μk.
The normal force is not a fixed amount. It is whatever push the surface needs to stop the object from sinking into it, and it equals mg only for an object resting on level ground with nothing else pushing up or down. On a ramp, the block presses in with only part of its weight, so N = mg cos θ: at 60°, that’s half the weight. Push down on a book and the table pushes up harder. Pull a sled with a rope angled upward and the snow pushes up less. In an elevator that speeds up going up, the floor pushes up with more than your weight, which is the next experiment. Always find N from the forces across the surface, never by habit.
The elevator scale: what a scale really measures
A bathroom scale doesn’t measure your weight. It measures how hard it has to push up on your feet, and shows that as kilograms by dividing by g. Standing still, that push equals your weight. In an elevator, Newton’s second law for you says N − mg = ma, so
Ride
Speeding up going up (or slowing down going down): the acceleration points up, so the floor must push harder than your weight. You feel heavy. In this elevator, a = 1.2 m/s², and a 50 kg rider’s scale reads 50 × (9.81 + 1.2) = 550 N, which it shows as 56.1 kg.
Cruising at a steady speed, up or down: a = 0, and the scale reads your weight, 490 N for 50 kg. Moving and accelerating are different things. Only acceleration changes the reading.
Speeding up going down (or slowing down going up): the floor pushes less than your weight. The same rider reads 430 N, shown as 43.9 kg.
The drop tower. On a drop-tower ride, the seats are let go and fall freely, with a = −g. Then N = m(g − g) = 0: the scale reads nothing, and riders feel weightless. Gravity hasn’t switched off. You and the seat fall together, so the seat has nothing to push on. In this model, brakes then stop the car at 2g, and the scale shows three times the rider’s weight. Astronauts in orbit feel weightless for the same reason: they are falling too, as Chapter 6 explains.
A pulley race: the Atwood machine
Hang two masses on a rope over a pulley. The heavier one sinks, the lighter one rises, and both accelerate at the same rate, much more gently than free fall. The English mathematician George Atwood described this machine in 1784 as a way to slow falling down enough to time it.
Masses
Draw one free-body diagram for each mass. Both feel the same tension T up, because a light rope over a frictionless pulley pulls equally at both ends. Taking the direction the heavier side moves as positive:
Add the two equations and T cancels:
With 1.20 kg and 0.80 kg: a = (0.40 / 2.00) × 9.81 = 1.96 m/s², one fifth of g. The tension is 2 × 1.20 × 0.80 × 9.81 / 2.00 = 9.41 N, between the two weights (11.77 N and 7.85 N), as it must be: it holds the heavy one back and lifts the light one. Each mass moves 1 m in √(2 × 1 / 1.96) = 1.01 s. Press Release and compare.
Build your own free-body diagrams
Pick a scene. Drag forces from the tray onto the picture, or tap a force and then tap where it should point. Drag an arrow’s tip to aim it. Tap a force again to remove it. Then press Check. The arrows here all have the same length: this is about which forces act and which way they point.
Newton’s rule for adding forces
Right after his three laws, Newton gave a rule for what happens when two forces act at once. Draw each force as an arrow, complete the parallelogram, and the object goes along its diagonal, exactly as if the two had been replaced by one. Today we call that adding vectors, and every free-body diagram relies on it.
Corpus viribus conjunctis diagonalem parallelogrammi eodem tempore describere, quo latera separatis.
“A body by two forces conjoined will describe the diagonal of a parallelogram, in the same time that it would describe the sides, by those forces apart.”
Isaac Newton, Philosophiæ Naturalis Principia Mathematica (1687), Laws of Motion, Corollary I. English: Andrew Motte’s translation, 1729.Friction was measured long before Newton. Leonardo da Vinci sketched friction experiments around 1500, and Guillaume Amontons published the rules in 1699: friction is proportional to how hard the surfaces press together and doesn’t depend on how big the contact area is. Charles-Augustin de Coulomb tested these rules with great care in 1781, including the difference between starting a slide and keeping it going. The μsN and μkN you used above are their rules, still good approximations for dry surfaces.
Where you see it
Tires and brakes. Rubber on dry concrete has μs ≈ 1.0, but on wet concrete it can drop to about half that. Antilock brakes keep the tires rolling instead of skidding, because a rolling tire grips with static friction, which is stronger than the kinetic friction of a skid. See how brakes work and wheels and tires in the cars course.
Climbing and walking. Rock-climbing shoes use soft, sticky rubber to raise μs. Walking on ice is hard because shoes on ice have μs ≈ 0.1: you can only push back on the ground with a tenth of your weight before your foot slips.
Ropes and pulleys. Cranes, elevators and the counterweights in old windows are Atwood machines at heart. An elevator’s counterweight hangs on the other side of the pulley so the motor only has to lift the difference. The Trebuchet Workshop uses a falling counterweight too.
Every other force problem. Engineers checking a bridge, a ladder against a wall or a crane’s cable all begin with a free-body diagram. So does the Book Cart balancing game, and the Rocket Builder, where thrust, weight and drag fight it out.
How the simulation works
The ramp model. The block is a single particle on a straight ramp. Gravity is split into parts along and across the ramp. Your push acts along the ramp, so it doesn’t change the normal force, and N = mg cos θ. When the block is at rest, the program finds the friction needed to keep it still. If that is no more than μsN, friction supplies exactly that much. If not, the block slips, and kinetic friction μkN opposes the sliding.
No guessing in the steps. While the block slides one way, every force is constant, so the motion is exactly uniform acceleration: x = x₀ + v₀t + ½at². The only special moment is when a sliding block comes to rest. The program finds that moment exactly, stops the block there and asks static friction again. So the simulation has no step-size error at all.
What’s simplified. The friction coefficients come from Table 6.1 of OpenStax University Physics. They are approximate, given to one or two digits. Real friction changes a little with speed, temperature and how clean the surfaces are. For rubber on wet concrete, the table gives a range (0.5–0.7 and 0.3–0.5), and the simulation uses the middle values. The ramp doesn’t bend, and the block doesn’t tip or roll.
How it was checked (node tests, all passed):
- Rubber on dry concrete holds at 45° minus a millionth of a radian and slips at 45° plus one. Slip angles: 45° (μs 1.0), 26.565° (0.5), 5.711° (0.1).
- Wood on wood at 30°, from rest: after 1 s the speed is 2.355483 m/s, matching g(sin 30° − 0.3 cos 30°) to 4 × 10⁻¹⁵.
- A block shoved up a 10° ramp at 2 m/s stops after 0.434763 s and 0.434763 m, as the formulas give, and stays put.
- At tan θ = μk, a sliding block keeps exactly 1 m/s for 3 s. A push of mg(sin θ + μk cos θ) = 68.108 N (10 kg, 25°) keeps 0.5 m/s uphill, but can’t start the block from rest (that takes 85.88 N).
- Energy: over 200 random runs, kinetic energy gained equals the work of the push, gravity and friction to a relative error below 2 × 10⁻¹⁴.
- Elevator: each ride ends at rest; free fall reads 0 N. Atwood: the formulas satisfy both masses’ equations, and a simulated stiff rope gives the same speed after 1 s (1.96133 m/s).
Check yourself
A 10 kg box rests on a ramp tilted 20°. How big is the static friction on it?
Show the answer
Static friction is only as big as it must be. Gravity pulls the box down the ramp with mg sin 20° = 33.5 N, so friction pushes up the ramp with exactly 33.5 N. It reaches μsN only at the moment of slipping.
A box sits on a level floor and you push straight down on its top with 20 N. The normal force from the floor is…
Show the answer
The box isn’t accelerating, so the forces on it balance: the floor pushes up with the box’s weight plus your 20 N. The normal force is whatever the surface needs to push.
You stand on a scale in an elevator that is going up but slowing down to stop at a floor. The scale reads…
Show the answer
Slowing down while going up means the acceleration points down, so a is negative and N = m(g + a) is less than mg. The direction of motion doesn’t matter; the direction of acceleration does.
Which of these belongs on the free-body diagram of a book lying on a table?
Show the answer
A free-body diagram shows only forces on the object: gravity, and the table’s normal force up on the book. The book’s push on the table acts on the table. The net force is the sum of the arrows, not another one. Nothing tries to slide the book, so there is no friction.
An Atwood machine holds 3 kg on one side and 1 kg on the other. What is the acceleration?
Show the answer
a = (3 − 1)g / (3 + 1) = g/2. The 2 kg difference in weight has to accelerate all 4 kg.
Why does it take a bigger push to start a heavy crate sliding than to keep it sliding?
Show the answer
For most pairs of surfaces μs is bigger than μk: wood on wood is 0.5 against 0.3. Once the crate slides, friction drops, so a smaller push balances it.
Sources
- OpenStax, University Physics Volume 1, 6.2 Friction: Table 6.1, the friction coefficients used here; fs ≤ μsN and fk = μkN.
- OpenStax, 5.6 Common Forces: weight, the normal force on an incline (N = mg cos θ), tension and Hooke’s law.
- OpenStax, 5.7 Drawing Free-Body Diagrams: the steps and conventions for free-body diagrams.
- OpenStax, 6.1 Solving Problems with Newton’s Laws: the elevator and scale, the Atwood machine, the problem-solving strategy.
- NASA, Earth Fact Sheet: standard gravity g = 9.80665 m/s² (exact by definition), used in all the calculations.
- I. Newton, Philosophiæ Naturalis Principia Mathematica (London, 1687), scan at the Internet Archive: Corollary I. A. Motte, The Mathematical Principles of Natural Philosophy (London, 1729), scan at the Internet Archive.