Free body diagrams — drawing all the forces — Physics

Before this

Newton's third law
Newton's second law
Sin, cos and tan · coming soon

This leads to

Pulleys and slopes
Resolving forces · coming soon
Friction — the model and its limits · coming soon

How many forces act on a box on a slope?

A box rests on a rough slope. It does not move. How many forces act on it?

Most students answer two: gravity pulling it down, and friction holding it up. Some add a third, "the force down the slope". Both answers are wrong, and in opposite ways. The first misses a force. The second invents one.

There are exactly three forces: the box's weight, the push of the slope at right angles to its surface, and friction along the surface. Forget the push of the slope, and every calculation fails. Add "the force down the slope" as an extra force, and you count part of the weight twice.

Mechanics problems rarely go wrong in the algebra. They go wrong earlier, when the list of forces is wrong. This lecture gives you a method for getting the list right every time: the free-body diagram.

One object, freed from everything else

The last lecture ended with a promise: a reliable method that shows only the forces on one object at a time. Here it is.

A free-body diagram is a drawing of one object, alone, with every force that acts on it drawn as an arrow. The word "free" means the object is freed from its surroundings. The table, the floor, the rope and the Earth are not drawn. Only the forces they exert on the object remain.

Why does this help? The third-law lecture showed that forces come in pairs, with each force of a pair acting on a different object. If you draw two objects together, you see both forces of each pair, and they look as if they cancel. They do not. A free-body diagram removes the confusion: it shows one object, so only one force of each pair can appear.

Look again at the book on the table from the first-law lecture. It has a mass of 1.5 kg, so its weight is $1.5 \times 9.8 = 14.7$ N.

The free-body diagram of a 1.5 kg book at rest on a table. Only the book is drawn. Two forces act on it: the table pushes it up, and the Earth pulls it down. Both arrows start on the book.

The book's push down on the table is not in the diagram. That force acts on the table, so it belongs in the table's free-body diagram. The book's pull on the Earth is not in the diagram either. It acts on the Earth. So the diagram shows exactly two forces, and they balance, because the book is not accelerating.

Where forces come from: a rule that catches mistakes

Here is the most useful question in all of mechanics: which object exerts this force? Every real force has an answer. A force is always one object acting on another. If you cannot name the other object, the force does not exist.

At this level, forces come from only two places:

Gravity, from the Earth. It acts on every object with mass, without touching. This is the weight, $W = m g$ .
Contact, from everything that touches the object. A surface, a rope, a hand, the air, water. Each thing in contact can push or pull.

That gives a method you can follow without guessing. Draw the weight first. Then trace round the edge of the object. At every place where something touches it, ask what force that contact makes.

Drawing a free-body diagram

1. Draw the chosen object alone, as a box or a dot.

2. Draw its weight, $W = m g$ , straight down.

3. Go round the object and add a force for each contact. A surface gives a normal force at right angles to it, and may give friction along it. A string gives tension along the string, away from the object. A hand, an engine or the air can give a push, a thrust or drag.

4. Start every arrow on the object. Make longer arrows for larger forces. Label each one.

5. Check each arrow: name the object that exerts it.

Some details on the contact forces. "Normal" means at right angles. A surface can only push straight out of itself, so the normal force is at right angles to the surface, which is not always vertical. Friction acts along the surface, against the sliding, or against the way the object would slide if nothing held it. Tension in a string always pulls, never pushes, because you cannot push with a string.

Now test the rule on a famous mistake. Make a prediction first.

Predict, then reveal — a ball thrown upwards

A ball has just left your hand. It is still rising. Ignore air resistance. Which forces act on the ball now?

The forces on a ball that has left the hand and is still rising, with air resistance ignored. Many people add an upward "force of the throw", but no object is exerting it.

Only one force acts: the ball's weight, straight down. The hand stopped pushing the moment the ball left it. The ball keeps rising because it is already moving upwards (the first law), and the weight reduces its upward speed by 9.8 m/s every second.

The "force of the throw" is the most common invented force in mechanics. Others are "the force of motion", "the force of momentum" and "the force down the slope". None of them has an object that exerts it. Moving does not need a force. Changing how you move needs one. That is the first law, and the free-body diagram makes you obey it.

From diagram to answer

A free-body diagram is not decoration. It is the first half of the solution. The second half is Newton's second law, applied in each direction separately:

Choose a positive direction for each line, and say so: "take right as positive", "take up as positive".
Add the forces along each line to find the resultant force in that direction.
If the object does not accelerate in a direction, the forces along it balance. If it does, the resultant equals $ma$ .

Worked Example 1 — A pushed block

A 5 kg block on a horizontal floor is pushed with a horizontal force of 20 N. Friction on the block is 8 N. Draw the free-body diagram and find the acceleration.

Diagram. Four arrows. Vertically: weight $W = 5 \times 9.8 = 49$ N down (from the Earth) and the normal force $N$ up (from the floor). Horizontally: the 20 N push to the right (from the hand) and friction 8 N to the left (from the floor).

Vertical, up positive. The block does not rise or sink, so the forces balance: $N - 49 = 0$ , and $N = 49$ N.

Horizontal, right positive. Resultant $= 20 - 8 = 12$ N. So $a = 12 \div 5 = 2.4$ m/s² to the right.

Notice that the floor exerts two forces: the normal force and friction. They come from the same contact but act in different directions, so they are drawn as two arrows.

Worked Example 2 — Standing in a lift

A 60 kg person stands on scales in a lift. The lift accelerates upwards at 1.5 m/s². What do the scales read?

Diagram of the person. Two arrows: weight $60 \times 9.8 = 588$ N down; the normal force $N$ from the scales, up. The scales show the size of $N$ .

Vertical, up positive. The person accelerates upwards, so the forces do not balance: $N - 588 = 60 \times 1.5 = 90$ . So $N = 678$ N.

The scales read 678 N, more than the weight. You feel heavier as a lift starts upwards because the floor must push harder than your weight to speed you up. If the lift accelerated downwards at 1.5 m/s², the same method gives $588 - N = 90$ , so $N = 498$ N.

Stop and think

Is the normal force on an object always equal to its weight?

Show the answer

No. The normal force is as large as it needs to be to stop the object moving into the surface. It equals the weight only when the surface is horizontal, nothing else pushes or pulls vertically, and there is no vertical acceleration. In the lift, it was 678 N for a weight of 588 N. Pull a box upwards with a rope at an angle, and the normal force becomes smaller than the weight, because the rope takes part of the load. On a slope, it is smaller again, as the next section shows. Never write $N = m g$ without checking the other vertical forces first.

On a slope: tilt the directions with the surface

Now return to the box on the slope. Its free-body diagram has three arrows: weight straight down, the normal force at right angles to the slope, and friction along the slope. Two of these lie along the slope's own directions. Only the weight does not.

So choose your two directions to match the slope: one along the slope and one at right angles to it. Then split the weight into two parts, one in each direction. This is called resolving the force. You did the same with velocity in the projectile lectures, when you split a launch speed into horizontal and vertical parts.

A block on a slope at angle θ. The weight mg (solid arrow) is replaced by its two parts (dashed arrows): mg sin θ down the slope and mg cos θ into the slope. The normal force N is at right angles to the slope. The angle between the weight and the line into the slope is also θ.

This is the hard part of the lecture. Spend time here. Which part gets sine and which gets cosine depends on one angle: the angle between the weight and the line into the slope. That angle is θ, the same as the angle of the slope. Here is why. When the slope is flat, the line into the slope points straight down, exactly along the weight, and the angle between them is zero. Tilt the slope by θ, and the line at right angles to it tilts by θ too. So it ends up θ away from the weight.

Now use the right-angled triangle made by the weight and its two parts. The weight is the hypotenuse. The part into the slope is next to the angle θ, so it is $m g cos θ$ . The part down the slope is opposite θ, so it is $m g sin θ$ .

down the slope: into the slope: m g sin θ m g cos θ

Check with the two extreme cases, because a check is better than a memory. On flat ground, $θ = 0$ : $sin 0 = 0$ , so nothing pulls the box along, and $cos 0 = 1$ , so the whole weight presses into the ground. That is right. On a vertical wall, $θ = 9 0^{\circ}$ : $sin 9 0^{\circ} = 1$ , so the whole weight pulls along the "slope", straight down, and nothing presses into it. That is right too. If you ever swap sine and cosine, these two checks catch it at once.

That is it. That is the whole idea. The slope's parts of the weight are $m g sin θ$ along it and $m g cos θ$ into it, and the normal force balances the part into it. Explore it below.

Explore — the weight on a slope

Change the angle of the slope. The block has a mass of 3 kg and there is no friction. Watch the two dashed parts of the weight and the normal force. Try 0°, then a steep slope.

Angle of slope: 25°

The two parts of the weight of a 3 kg block (29.4 N) on a smooth slope, and its acceleration down the slope:

Angle	Down the slope, mg sin θ	Into the slope, mg cos θ = N	a = g sin θ
0°	0 N	29.4 N	0 m/s²
10°	≈ 5.1 N	≈ 29.0 N	≈ 1.70 m/s²
25°	≈ 12.4 N	≈ 26.6 N	≈ 4.14 m/s²
45°	≈ 20.8 N	≈ 20.8 N	≈ 6.93 m/s²
60°	≈ 25.5 N	≈ 14.7 N	≈ 8.49 m/s²

As the slope steepens, more of the weight pulls along it and less presses into it.

As the slope gets steeper, the part down the slope grows and the part into the slope shrinks. At 45° they are equal. The normal force follows the part into the slope, so a steep slope presses on the block less than a gentle one.

Worked Example 3 — Sliding down a smooth slope

A 3 kg block slides down a frictionless slope at 25° to the horizontal. Find the normal force and the acceleration.

Diagram. Two forces: weight $3 \times 9.8 = 29.4$ N down, and the normal force at right angles to the slope. No friction.

At right angles to the slope. The block does not lift off or sink in, so $N = m g cos 2 5^{\circ} = 29.4 \times cos 2 5^{\circ} \approx 26.6$ N.

Along the slope, down positive. Resultant $= m g sin 2 5^{\circ} = 29.4 \times sin 2 5^{\circ} \approx 12.4$ N. So $a \approx 12.4 \div 3 \approx 4.14$ m/s².

Keep unrounded values until the end. Then $a = \frac{m g sin θ}{m} = g sin θ$ : the mass cancels. A 3 kg block and a 30 kg block slide down a smooth slope together, just as a heavy and a light stone fall together.

Worked Example 4 — A rope at an angle

A child pulls a 10 kg sledge across flat snow with a rope at 30° above the horizontal. The tension is 50 N, and friction is 20 N. Find the acceleration and the normal force.

Diagram. Four forces: weight 98 N down; normal force $N$ up; tension 50 N along the rope, at 30° above the horizontal; friction 20 N backwards.

Resolve the tension. Horizontal part: $50 cos 3 0^{\circ} \approx 43.3$ N. Vertical part: $50 sin 3 0^{\circ} = 25$ N.

Horizontal, forwards positive. Resultant $\approx 43.3 - 20 = 23.3$ N. So $a \approx 23.3 \div 10 \approx 2.33$ m/s².

Vertical, up positive. No vertical acceleration: $N + 25 - 98 = 0$ , so $N = 73$ N. The rope lifts part of the sledge's weight, so the snow pushes up less than 98 N.

Three classic mistakes

Trap 1 — Adding "the force down the slope"

$m g sin θ$ is part of the weight, not an extra force. Draw either the weight, or its two parts, but never both in the same sum. Ask the usual question: which object exerts "the force down the slope"? Only the Earth, through the weight.

Trap 2 — Writing N = mg everywhere

The normal force is at right angles to the surface, and its size depends on the other forces. On a slope it is $m g cos θ$ . In an accelerating lift, or with a rope pulling upwards at an angle, it changes again. Work it out from the forces at right angles to the surface each time.

Trap 3 — Drawing forces the object exerts

A free-body diagram shows forces on the object. The book's push on the table, or your foot's push on the ground, act on other objects. Putting them in adds a third-law partner to the diagram, and the two will seem to cancel.

Try it yourself

Take $g = 9.8$ N/kg. Draw the free-body diagram first for each one.

A 2.5 kg lamp hangs at rest from a ceiling by a single cord. Draw its free-body diagram and find the tension in the cord.

Hint 1
Two objects act on the lamp: the Earth and the cord.

Hint 2
The lamp is at rest, so the forces balance.

Full answer
Weight $2.5 \times 9.8 = 24.5$ N down; tension $T$ up. At rest, so $T - 24.5 = 0$ , and $T = 24.5$ N. Check: the lamp's pull on the cord is the third-law partner of $T$ ; it acts on the cord, so it is not in this diagram.
A 1200 kg car on a level road has a driving force of 3000 N and total resistive forces of 600 N. Find its acceleration and the normal force from the road.

Hint 1
Four forces: weight, normal force, driving force, resistance.

Hint 2
Treat the vertical and horizontal directions separately.

Full answer
Vertical: no acceleration, so $N = 1200 \times 9.8 = 11760$ N. Horizontal, forwards positive: resultant $= 3000 - 600 = 2400$ N, so $a = 2400 \div 1200 = 2.0$ m/s².
A 2 kg block rests, without moving, on a rough slope at 30°. Find the friction force and the normal force.

Hint 1
Three forces: weight, normal force, friction. The block is at rest, so the forces balance in both directions.

Hint 2
Friction balances $m g sin 3 0^{\circ}$ . The normal force balances $m g cos 3 0^{\circ}$ .

Full answer
Weight $= 19.6$ N. Along the slope: friction $= 19.6 sin 3 0^{\circ} = 9.8$ N, up the slope. At right angles: $N = 19.6 cos 3 0^{\circ} \approx 17.0$ N. Check: the three forces form a closed triangle, because their resultant is zero. Friction points up the slope because the block would slide down without it.
A 4 kg picture hangs from a nail by a wire. The two halves of the wire each make 30° with the horizontal, and the tension is the same in both. Find the tension. Then find it again when each half makes only 10° with the horizontal. Why is a tight, nearly flat wire risky?

Hint 1
Draw the picture alone: its weight, and two tensions along the wire. Resolve each tension into vertical and horizontal parts.

Hint 2
The horizontal parts cancel. The two vertical parts, each $T sin θ$ , hold up the weight: $2 T sin θ = m g$ .

Full answer
Weight $= 4 \times 9.8 = 39.2$ N. At 30°: $2 T sin 3 0^{\circ} = 39.2$ , so $T = 39.2$ N, equal to the whole weight. At 10°: $2 T sin 1 0^{\circ} = 39.2$ , so $T \approx 113$ N, nearly three times the weight. A flat wire has only a small vertical part to hold the picture up, so the tension must be huge. That is why a wire pulled almost straight can snap, and why no rope with a weight hanging from it can ever be pulled perfectly straight.

The bigger idea

A free-body diagram is an argument made of arrows. Each arrow is a claim: "this object exerts this force on that one". The diagram is right when every claim can be defended. Engineers use the same habit on a much larger scale. To check one beam of a bridge, they draw that beam alone, with every load and every support that acts on it. Then they apply the same balance of forces in each direction.

The method also scales up. The next lecture connects objects with strings over pulleys and places them on slopes. Each object gets its own diagram. The string's tension appears in both, and Newton's laws link them. The diagrams do most of the work before any equation is written.

Key concepts in this lesson

A free-body diagram shows one object alone, with every force that acts on it, each arrow starting on the object and labelled.
Every force is exerted by another object: the Earth through gravity, or something in contact through a normal force, friction, tension, a push or drag.
The normal force is at right angles to the surface, and its size depends on the other forces, so it is not always equal to the weight.
On a slope at angle θ, the weight splits into mg sin θ down the slope and mg cos θ into the slope.
After drawing the diagram, apply Newton's second law separately in two directions at right angles, with a stated positive direction for each.