2.2 Forces and Free-Body Diagrams

Picture stories

Press Next (or Play) to walk through each story one small step at a time. The numbers come last.

1. A book on a desk

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A heavy textbook lies on your desk. It is not moving. Does that mean no forces act on it? If no force acted, why does it not fall through the desk?

Later you help a friend move. You push a box of books up the ramp of a moving truck. Earth pulls the box down, the ramp pushes on it, and you push on it. Some of those pushes help you and some fight you.

The question this topic answers: what forces act on an object, and how can one simple picture show all of them so you can add them up?

2. A box pushed across a gym floor

3. A box on a ramp

Quick check. Diego pushes a box across the gym floor at a steady speed in a straight line. How does his push compare with friction on the box?

4. A bag held by its strap

5. A ball in flight

6. Pressing down on a box

Quick check. Kenji presses down on a box of books resting on his desk. How big is the desk's normal force on the box now?

7. Check yourself

Think of your answer first, then tap to see it.

a) A lamp hangs at rest from a cord. Which forces act on the lamp?

Show answer

Gravity pulls it down and the cord's tension pulls it up. They are equal, so the lamp stays still.

b) You throw a ball. While it flies through the air (ignore air), which forces act on it?

Show answer

Only gravity. Your hand stopped touching it when it left, so there is no "force of the throw" in the free-body diagram.

c) A 5.0 kg box rests on a level floor. How big is the normal force?

Show answer

It balances gravity: N = m g = 5.0 kg × 9.8 m/s² = 49 N, pointing up.

Already know this?

Three quick questions. Get all three right on the first try and you can skip ahead to the simulation.

1. A book rests on a table. Which forces act on the book?

2. You toss a ball straight up. Ignoring air, which forces act on it at the very top of its path?

3. A 3.0 kg box sits on a level floor. You pull straight up on it with 10 N, but it does not leave the floor. How big is the normal force?

The idea in plain words

A force is a push or a pull that one object exerts on another. Every force has a size (in newtons, N) and a direction, so it is a vector. Every force needs two objects: the one pushing and the one being pushed. If you cannot name the object that exerts a force, that force does not exist.

The forces you will see most in AP Physics 1:

A free body diagram (FBD) shows one object as a dot or a box, with one arrow for each force on that object, pointing in the force's direction, starting at the object. Label each arrow with what exerts it. Do not draw velocity, acceleration or "ma" on it: those are results of the forces, not forces.

Trap: drawing a force the object exerts (the box pushing on the floor) on the object's own diagram. Instead: draw only forces on the object; read each arrow as "something on the box".

Net force is the vector sum of all the forces. Add the parts along each axis. On a ramp, it is easier to tilt your axes: one axis along the ramp, one perpendicular to it. Then weight splits into two parts:

along the ramp: mg sinθ (down the slope)    perpendicular: mg cosθ (into the ramp)

The components replace the weight arrow; they are not extra forces.

Trap: drawing the weight arrow perpendicular to the ramp. Instead: weight always points straight down; only the normal force is perpendicular to the ramp.

Worked numbers: pushing a box up a ramp

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A 5.0 kg box sits on a smooth (frictionless) ramp tilted at 30°. You push it with 30 N, parallel to the ramp, up the slope. Take "up the ramp" as positive.

  1. Draw the FBD: weight (Earth on box) down, normal force (ramp on box) perpendicular to the ramp, your push up along the ramp. Why: three objects touch or pull on the box: Earth, the ramp and you. So three forces.
  2. Weight: Fg = mg = 5.0 kg × 9.8 m/s² = 49 N. Why: weight is always mg, straight down.
  3. Split the weight. Along the ramp: 49 N × sin 30° = 24.5 N, down the slope. Into the ramp: 49 N × cos 30° = 42.4 N. Why: our axes are tilted, so weight is the only force not lined up with an axis.
  4. Perpendicular to the ramp the box does not move, so the forces balance: FN = 42.4 N. Why: the normal force is only as big as needed. It is less than 49 N on a ramp.
  5. Along the ramp: net force = 30 N − 24.5 N = +5.5 N (up the ramp). Why: your push is up the slope, the weight part is down the slope.
  6. So the box speeds up up the ramp: a = 5.5 N / 5.0 kg = 1.1 m/s² (that is Newton's second law, topic 2.5). Check in the lab below: set m = 5, angle 30, push 30, push angle 0.

Lab: the box, the ramp and your push

Set the ramp angle to 0 for a flat table. Change the push and its angle (measured from the surface: positive tilts the push away from the surface, negative pushes into it). Positive x and v are up the ramp. Watch how the normal force changes. The ramp is frictionless.

Examples

Example 1: the book on the desk basic

A 1.5 kg book rests on a level desk. Draw its free body diagram in words and find each force.

Show solution
Read the steps as text
  1. Forces: weight (Earth on book) down, normal force (desk on book) up. Nothing else touches it.
  2. Fg = 1.5 × 9.8 = 14.7 N down.
  3. The book is at rest and stays at rest, so the net force is zero: FN = 14.7 N up. Why: the desk pushes up just hard enough to stop the book sinking in.

Example 2: pulling a sled with a slanted rope medium

You pull a 10 kg sled across smooth ice with a rope at 30° above the horizontal. The tension is 50 N. Find the normal force and the sled's acceleration.

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Read the steps as text
  1. Split the tension: horizontal 50 cos 30° = 43.3 N, vertical 50 sin 30° = 25 N (up).
  2. Vertical: the sled does not leave the ice, so FN + 25 N = mg = 98 N → FN = 73 N. Why: the rope lifts a little, so the ice needs to push less. FN is less than mg.
  3. Horizontal: net force 43.3 N, so a = 43.3 / 10 = 4.3 m/s² forward.

Example 3: pushing down on a box medium

You push a 4.0 kg box on a smooth floor with 40 N directed 20° below the horizontal. Find the normal force and the acceleration.

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Read the steps as text
  1. Components of the push: horizontal 40 cos 20° = 37.6 N, vertical 40 sin 20° = 13.7 N (down).
  2. Vertical balance: FN = mg + 13.7 = 39.2 + 13.7 = 52.9 N. Why: you press the box into the floor, so the floor pushes back harder than the weight.
  3. a = 37.6 N / 4.0 kg = 9.4 m/s². Try it in the lab: angle 0, push 40, push angle −20, m = 4.
Trap: writing FN = mg = 39.2 N here. Instead: find FN from the vertical balance: your push presses the box down, so FN = 39.2 + 13.7 = 52.9 N.

Example 4: a block held on a ramp, then the rope is cut AP

An 8.0 kg block rests on a frictionless 25° ramp, held by a rope parallel to the ramp. (a) Find the tension and the normal force. (b) The rope is cut. Find the block's acceleration. (c) Does the normal force change when the rope is cut?

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Read the steps as text
  1. Fg = 8.0 × 9.8 = 78.4 N. Along the ramp: 78.4 sin 25° = 33.1 N down the slope. Perpendicular: 78.4 cos 25° = 71.1 N.
  2. (a) At rest, along the ramp: FT = 33.1 N. Perpendicular: FN = 71.1 N.
  3. (b) Without the rope the only force along the ramp is 33.1 N down the slope: a = 33.1 / 8.0 = 4.1 m/s² down the ramp (that is g sin 25°).
  4. (c) No. The rope was parallel to the ramp, so it had no perpendicular part. FN stays 71.1 N. Why: only forces with a perpendicular part change the normal force.
Trap: writing FN = mg = 78.4 N on the ramp. Instead: only the perpendicular part of the weight presses into the ramp: FN = mg cos 25° = 71.1 N.

Practice (AP style)

1. A book rests on a level table. Which list gives all the forces exerted on the book?

Show answer

Two objects interact with the book: Earth (weight) and the table (normal force). The book's push on the table acts on the table, not on the book, so it does not belong on the book's diagram.

2. A 2.0 kg book rests on a table. You press straight down on it with 5.0 N. What is the normal force on the book?

Show answer

Weight = 2.0 × 9.8 = 19.6 N down, your push 5.0 N down. The book stays at rest, so FN = 19.6 + 5.0 = 24.6 N up.

3. You throw a ball straight up. Air resistance is negligible. While the ball is rising, after it leaves your hand, which forces act on it?

Show answer

Once the ball leaves your hand, your hand no longer touches it, so it exerts no force. Only Earth acts on it. The ball still moves up because it already has an upward velocity, not because a force carries it.

4. A 2.0 kg block sits on a frictionless ramp at 30°. What is the normal force on the block?

Show answer

Perpendicular to the ramp the forces balance: FN = mg cos 30° = 19.6 × 0.866 = 17.0 N. The 9.8 N choice is mg sin 30°, the part along the ramp.

5. A child pulls a wagon across level ground with a handle tilted up at an angle. Compared with the weight of the wagon, the normal force from the ground is

Show answer

The handle force has an upward component that holds up part of the wagon, so the ground pushes up less: FN = mg − F sinθ.

6. A 3.0 kg lamp hangs at rest from a single vertical cord. What is the tension in the cord?

Show answer

Two forces act: weight 3.0 × 9.8 = 29.4 N down and tension up. At rest the net force is zero, so FT = 29.4 N.

7. A block slides down a frictionless ramp of angle θ. Which expression gives the size of the net force on it?

Show answer

Perpendicular to the ramp, FN = mg cosθ cancels that part of the weight. The only part left is the weight along the ramp, mg sinθ. Check the limits: θ = 0 gives zero (flat table), θ = 90° gives mg (free fall).

8. (Short free response: translation between representations) A free body diagram for a 5.0 kg crate shows three forces: 49 N down labelled "Earth on crate", 49 N up labelled "floor on crate", and 20 N to the right labelled "rope on crate". (a) Describe a real situation this could show. (b) Find the net force. (c) Can you tell which way the crate is moving? Explain.

Show answer

(a) A crate on a smooth level floor being pulled to the right by a horizontal rope.

(b) Vertical forces cancel (49 N up, 49 N down). Net force = 20 N to the right.

(c) No. The diagram tells you the acceleration is to the right (4.0 m/s²), not the velocity. The crate could be moving right and speeding up, or moving left and slowing down, or momentarily at rest. A free body diagram shows forces, not motion.

9. (Short free response: qualitative reasoning) A block rests on a ramp. A student claims "the normal force on the block equals its weight, because the normal force always balances gravity". Explain whether the student is right, using the directions of the forces. Then say how the normal force changes as the ramp is made steeper.

Show answer

The student is wrong. The normal force is perpendicular to the ramp surface, while weight is straight down, so they cannot cancel each other on their own. Only the part of the weight perpendicular to the ramp, mg cosθ, is balanced by the normal force, so FN = mg cosθ, which is less than mg. As the ramp gets steeper, cosθ gets smaller, so the normal force decreases (toward zero for a vertical wall).

10. (Short free response: mathematical routines) You pull a 6.0 kg box across a level frictionless floor with a rope at 37° above the horizontal. The tension is 40 N. (a) Calculate the normal force on the box. (b) Calculate the box's acceleration. (c) Calculate the largest tension, at the same angle, that keeps the box on the floor.

Show answer
  1. Weight: mg = 6.0 × 9.8 = 58.8 N. Tension parts: vertical 40 sin 37° = 24.1 N up, horizontal 40 cos 37° = 31.9 N.
  2. (a) Vertical balance: FN + 24.1 = 58.8, so FN = 34.7 N (less than mg because the rope lifts).
  3. (b) a = 31.9 N / 6.0 kg = 5.3 m/s² forward.
  4. (c) The box leaves the floor when FN = 0: T sin 37° = 58.8 N, so T = 58.8 / 0.602 = 98 N.

Point guide (4 points):

  • 1 point: splits the tension into components correctly.
  • 1 point: FN = 34.7 N from the vertical balance (not mg).
  • 1 point: a = 5.3 m/s² using only the horizontal part.
  • 1 point: sets FN = 0 and gets T ≈ 98 N.

11. (Short free response: experimental design and analysis) A student claims that the normal force on a block resting on a ramp equals mg cos θ. Available: a 2.0 kg block, a board that can be tilted, a protractor, and a flat digital force plate (it reads the push on its top face) that can be fixed to the board. (a) Describe a procedure to test the claim. (b) The student gets these readings: 0°: 19.6 N; 15°: 18.9 N; 30°: 17.0 N; 45°: 13.9 N; 60°: 9.8 N. What should be graphed to get a straight line, and what should its slope be? (c) The block stays still on the plate because of friction. Does friction spoil the reading? Explain.

Show answer
  1. (a) Fix the force plate flat on the board. Put the block on it. Set the board at several angles (0° to 60°), measure each angle with the protractor and record the plate reading once the block is still. Repeat each angle and average.
  2. (b) Plot the reading against cos θ. If the claim is right, the points lie on a straight line through the origin with slope mg = 2.0 × 9.8 = 19.6 N. Check: cos 60° = 0.50 and 19.6 × 0.50 = 9.8 N; cos 30° = 0.866 and 19.6 × 0.866 = 17.0 N.
  3. (c) No. Friction acts along the surface of the plate, so it has no part perpendicular to the plate. The plate reads only the perpendicular push, the normal force.

Point guide (4 points):

  • 1 point: varies the angle and measures the angle and the plate reading at each.
  • 1 point: repeats or uses several angles (enough data for a graph).
  • 1 point: graph of reading vs cos θ, slope mg = 19.6 N.
  • 1 point: friction is parallel to the surface, so it does not change the perpendicular reading.

12. (Short free response: qualitative/quantitative translation) A sled of mass m is pulled across smooth level ice by a rope at angle θ above the horizontal. The tension T stays the same while the student changes θ. (a) Without equations, explain how the normal force and the sled's acceleration change as θ grows. (b) Derive expressions for FN and a in terms of m, g, T and θ. (c) Show that your expressions agree with (a), and say what happens at θ = 90°. (d) For m = 10 kg and T = 50 N, calculate FN and a at 30°.

Show answer
  1. (a) A steeper rope lifts more, so the ice needs to push up less: FN gets smaller. Less of the rope pulls forward, so a gets smaller.
  2. (b) Vertical: FN + T sin θ = mg, so FN = mg − T sin θ. Horizontal: a = T cos θ / m.
  3. (c) As θ grows, sin θ grows (FN falls) and cos θ falls (a falls), matching (a). At 90°: a = 0 and FN = mg − T, the rope only lifts.
  4. (d) FN = 98 − 50 × 0.50 = 73 N; a = 50 × 0.866 / 10 = 4.3 m/s².

Point guide (4 points):

  • 1 point: correct directions of change for both FN and a with a reason.
  • 1 point: FN = mg − T sin θ.
  • 1 point: a = T cos θ / m and links the trends back to (a).
  • 1 point: FN = 73 N and a = 4.3 m/s².

Common mistakes

The mistake: drawing a "force of motion" or "force of the throw" on a moving object.

Why it is wrong: every force needs an object that exerts it right now. Moving does not need a force.

How to spot it: for each arrow, name "what on what". If you cannot name the "what", erase the arrow.

The mistake: setting the normal force equal to mg every time.

Why it is wrong: FN is whatever the surface needs. It changes on ramps, with angled pushes and pulls, and in elevators.

How to spot it: write the perpendicular balance every time and solve for FN.

The mistake: putting forces the object exerts (the book's push on the table) on its own diagram.

Why it is wrong: a free body diagram shows only forces acting on the chosen object.

How to spot it: every label must end in "... on (my object)".

The mistake: counting weight twice: the full weight arrow plus its two components.

Why it is wrong: components are a replacement for the force, not extra forces.

How to spot it: draw components dashed, and use either the force or its components in your sums, never both.

The mistake: drawing weight perpendicular to the ramp, or the normal force straight up on a ramp.

Why it is wrong: weight is always straight down (toward Earth's center); the normal force is always perpendicular to the surface.

How to spot it: on a ramp, the weight arrow and the normal arrow should not be on the same line.

More explanations: free resources.