Free-body diagrams, Newton's laws, friction and inclines: the one method for every force problem

A free-body diagram is one object drawn as a dot with every force on it drawn as an arrow from the dot, and nothing else: no other objects, no velocities, no "centrifugal" force. Draw it, choose axes along and across the motion, split any tilted force into components, and write Newton's second law once per axis. That is the whole method, and it is the same for a hanging mass, a pushed box and a block on a ramp. Below, the five steps and one incline worked to the end with the lines students get wrong.

What a free-body diagram is, and what never goes on it

A free body diagram (the hyphen comes and goes; it is one thing) is a drawing of one object, reduced to a dot, with every force acting on that object drawn as an arrow starting at the dot. Each arrow is labelled by the thing that exerts it, because that label is how you know you have not invented a force.

The arrows you will draw most often:

• Gravity, exerted by Earth, straight down, size mg.
• The normal force, exerted by a surface, perpendicular to that surface and pushing away from it.
• Friction, exerted by a surface, along the surface, opposing the sliding or the tendency to slide.
• Tension, exerted by a rope or string, along the rope and pulling toward it.
• An applied force, exerted by a hand, a motor, a push.

What never goes on the diagram:

• The other object. The table, the rope, the person are not drawn. Only their force on your object is.
• A velocity arrow. Velocity is not a force. If you want to note the direction of motion, write it in words beside the dot.
• "ma" as a force. The product ma is what the forces add up to. It is the right-hand side of the equation, never an arrow on the left.
• A force the object exerts on something else. The box pushes down on the ramp, but that force acts on the ramp. Your diagram is the box's.

One line to carry: the normal force is not always mg. It is whatever the perpendicular axis needs it to be, and you find it, not assume it.

Newton's three laws in the words you will use

First law. If the net force on an object is zero, its velocity does not change. A moving object with balanced forces keeps moving at the same speed in the same direction. It does not slow down, and it does not need a force to keep going.

Second law. The net force equals mass times acceleration, ΣF = ma, and you write it once per axis: ΣFₓ = maₓ and ΣFᵧ = maᵧ. On an axis with no motion, a = 0 and the forces balance.

Third law. Forces come in pairs that act on two different objects. The box pushes on the ramp, the ramp pushes on the box, equal in size and opposite in direction. Because the two act on different objects, they never appear on the same free-body diagram and they never cancel each other on one.

The first law is the one students resist. Clement (1982) asked introductory students about a coin tossed upward and found most of them drew a force in the direction of motion while the coin was rising, a force that does not exist; Hestenes, Wells and Swackhamer (1992) found the same belief, that motion needs a force to sustain it, across thousands of students with the Force Concept Inventory. On the diagram it shows up as an extra arrow pointing the way the object is going. If you have drawn one, ask what exerts it. If nothing does, erase it.

The five steps, every problem

1. The dot and the arrows. One object, every force on it, each labelled by what exerts it.
2. Axes. One axis along the motion (or along the surface, for an incline), one perpendicular to it. Tilting the axes to match the surface is what keeps the algebra short.
3. Components. Any force not already along an axis is split into a part along each axis. On an incline, that force is gravity.
4. ΣF = ma on each axis. On the axis with no motion, a = 0. On the axis with motion, a is the unknown or the given.
5. Solve, then check. Units in newtons and m/s². Sense: a negative a means the acceleration points the other way along your axis, not that you made a mistake.

Friction: static, kinetic, and the ≤ that students drop

Friction depends on how hard the two surfaces are pressed together, which is the normal force. That is why N appears in both friction rules, and why friction on an incline, where N is less than mg, is smaller than friction on the flat.

Kinetic friction acts while the surfaces slide past each other: fₖ = μₖN, where μₖ is the coefficient of friction for sliding. It has one value for a given pair of surfaces.

Static friction acts while they are not sliding: fₛ ≤ μₛN. The maximum is a ceiling, not the value. A box sitting on a ramp has exactly as much static friction as it needs to stay put, and no more; only when the needed amount would exceed μₛN does it start to slide. Writing fₛ = μₛN for a box that is simply resting is the most common friction error, and the ≤ is the whole point of the rule.

Friction opposes sliding. That is not always "opposite the push". A box carried on a truck bed that accelerates forward has friction pointing forward, because without friction the box would slide backward relative to the bed.

Worked example: a 5.0 kg box slides down a 30.0° incline, μₖ = 0.20

A 5.0 kg box slides down a rough incline at 30.0° to the horizontal. The coefficient of kinetic friction is 0.20. Find the acceleration.

Givens: m = 5.0 kg, θ = 30.0°, μₖ = 0.20, g = 9.8 m/s². The data carry two significant figures (5.0, 0.20) and three (30.0°); the answer is given to two, and the intermediate lines keep three so that the rounding happens once, at the end.

Step 1, the diagram. A dot. Three arrows: gravity straight down (Earth); the normal force perpendicular to the incline, pointing away from it (the surface); kinetic friction along the incline, pointing up the slope, because the box slides down (the surface).

Step 2, the axes. The x-axis along the incline, down the slope positive. The y-axis perpendicular to the incline, away from it positive.

Step 3, the components. Gravity is the only tilted force.

• mg = 5.0 kg × 9.8 m/s² = 49 N (the weight)
• Along the slope: mg sin 30.0° = 49 N × 0.500 = 24.5 N, pointing down the slope (the component of weight along the incline is the sine of the angle)
• Into the slope: mg cos 30.0° = 49 N × 0.866 = 42.4 N, pointing into the surface (the component perpendicular to the incline is the cosine)

Step 4, Newton's second law per axis.

• y-axis, no motion, a = 0: N − 42.4 N = 0, so N = 42.4 N (the normal force balances the perpendicular part of the weight, and only that part)
• Friction: fₖ = μₖN = 0.20 × 42.4 N = 8.49 N, up the slope (kinetic friction from the normal force just found)
• x-axis, down the slope positive: 24.5 N − 8.49 N = 5.0 kg × a (the gravity component pulls down the slope, friction holds back)
• 16.0 N = 5.0 kg × a (subtracting the forces)

Step 5, solve and check.

• a = 16.0 N ÷ 5.0 kg = 3.20 m/s², which is 3.2 m/s² down the slope to two significant figures (N ÷ kg = m/s²)

The check by sense: with no friction the acceleration would be a = g sin θ = 9.8 × 0.500 = 4.9 m/s², so friction has cost 1.7 m/s². The box still speeds up, which is right, because the gravity component (24.5 N) is larger than the friction (8.49 N). If the question then asks how fast the box is going after sliding 3.0 m, that is a job for the kinematics equations with this a as the given.

The four wrong lines a marker sees

Each has a symptom you can check your own page for.

1. N = mg on an incline. The normal force is mg cos θ, not mg. The symptom is a friction force that is too large: here it would be 0.20 × 49 = 9.8 N instead of 8.49 N, and the acceleration comes out 2.9 m/s².
2. Sin and cos swapped. The component along the slope is mg sin θ, the one into the slope is mg cos θ. The check that settles it: let θ go to 0°. A box on a flat floor does not slide, so the along-slope component must vanish, and sin 0° = 0 does. Swapped, the box on a flat floor would feel its whole weight pulling it sideways.
3. Friction drawn down the slope "because the box moves down". Friction opposes sliding, so for a box sliding down it points up the slope. The symptom is an acceleration larger than the frictionless 4.9 m/s², which friction can never produce.
4. Friction added to the gravity component instead of subtracted. The same symptom as line 3 with the arrow drawn correctly and the sign wrong in the equation: 24.5 + 8.49 = 33.0 N gives a = 6.6 m/s², and nothing on a ramp accelerates faster than g sin θ without a motor.

A fifth, for the static case: a box at rest on the ramp with a push smaller than it would take to move it does not have fₛ = μₛN. It has fₛ equal to whatever balances the other forces along the slope. The ≤ in the rule is a reminder that you solve for static friction; you do not look it up.

A sibling to try cold

Cover the example. An 8.0 kg box slides down a 25.0° incline with μₖ = 0.15. Find the acceleration. Draw the diagram first, three arrows with their labels, and only then write ΣF for each axis. Practice problems in the order that teaches says why the second problem should be the same shape as the first with different numbers: you are practising the method, and the method is what the test marks.

Check

• Weight: mg = 8.0 kg × 9.8 m/s² = 78.4 N
• Into the slope: 78.4 N × cos 25.0° = 78.4 × 0.906 = 71.1 N, so N = 71.1 N
• Friction: fₖ = 0.15 × 71.1 N = 10.7 N, up the slope
• Along the slope: 78.4 N × sin 25.0° = 78.4 × 0.423 = 33.1 N, down the slope
• Net force down the slope: 33.1 N − 10.7 N = 22.5 N
• a = 22.5 N ÷ 8.0 kg = 2.8 m/s² down the slope (two significant figures)

The frictionless value would have been 9.8 × 0.423 = 4.1 m/s². If your answer is above that, one of the four wrong lines is on your page.

Where this sits in the course

This page covers ideas 2 and 3 of 25 in SPH4U: Newton's laws; free-body diagrams; net force and Friction and motion on inclined planes, both in strand B. Before them is idea 1, Two-dimensional kinematics and projectile motion, which built the same two-axis habit without any forces: pick axes, treat each separately, share what they share.

After them is idea 4, Uniform circular motion and centripetal force, where the diagram is drawn exactly as here and the one new thing is that the net force must point to the centre of the circle. The same diagram with the net force pointing to the centre is the next page. The SPH4U course map shows all 25 ideas in teaching order, and every idea of the three science courses has a page like this one.

Free-body diagrams on one screen

• The diagram. One dot, one arrow per force, each labelled by what exerts it. Never the other object, never a velocity, never ma, never a force the object exerts on something else.
• The five steps. Dot and arrows; axes along and across; components; ΣF = ma per axis with a = 0 across; solve and check.
• Friction. fₛ ≤ μₛN (a ceiling); fₖ = μₖN; opposes sliding; depends on N, which on an incline is mg cos θ.
• The four wrong lines. N = mg on a slope; sin and cos swapped; friction drawn with the motion; friction added instead of subtracted.
• The habit. Draw before you write, every time; how to study physics is built on deriving the equation from the diagram rather than memorising one per problem type.