Physics

How to draw free body diagrams

A free body diagram is a simple sketch of one object with an arrow for every force acting on it. Each arrow starts on the object, points the way the force pushes or pulls, and gets a label such as weight, normal force, friction, tension or applied force. You then add the forces in each direction and use Newton's second law, ΣF=ma\Sigma F = ma, to find what you need.

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Key ideas

The forces you will draw most
ForceSymbolDirectionSize
Weight (gravity)FgF_g or mgmgStraight downmgmg
Normal forceFNF_NAway from the surface, at a right angle to itWhatever stops the object sinking into the surface
FrictionffAlong the surface, against slidingUp to μFN\mu F_N, see friction
TensionTTAlong the rope or string, pulling away from the objectSame along a light rope
Applied forceFAF_AThe way the push or pull goesGiven in the problem
Air resistance (drag)FDF_DOpposite to the velocityGrows with speed
  1. Draw the object as a dot or a box. Draw only that one object.
  2. Draw weight first, straight down from the center.
  3. Find everything touching the object. Each contact can give a normal force, friction, tension or a push.
  4. Draw each force as an arrow starting on the object, longer arrows for bigger forces, and label it.
  5. Choose axes. For a ramp, tilt them: one axis along the slope, one perpendicular to it.
  6. Split any angled force into components, then write ΣFx=max\Sigma F_x = ma_x and ΣFy=may\Sigma F_y = ma_y.
ΣFx=maxΣFy=may\Sigma F_x = m a_x \qquad \Sigma F_y = m a_y

Here ΣFx\Sigma F_x and ΣFy\Sigma F_y are the sums of the force components in each direction (N), mm is the mass (kg) and axa_x, aya_y are the accelerations (m/s²). On a ramp at angle θ\theta, weight splits into mgsin⁡θmg\sin\theta down the slope and mgcos⁡θmg\cos\theta into the slope.

Worked examples

Example 1: a book at rest

Problem A 1.5 kg book rests on a table. Draw its free body diagram and find each force.

  1. Forces: weight down, and the table's normal force up. Nothing else touches it.
  2. Weight.
    Fg=mg=(1.5 kg)(9.8 m/s2)=14.7 NF_g = mg = (1.5\ \text{kg})(9.8\ \text{m/s}^2) = 14.7\ \text{N}
  3. The book does not accelerate, so the vertical forces balance.
    FN−Fg=0⇒FN=14.7 NF_N - F_g = 0 \quad\Rightarrow\quad F_N = 14.7\ \text{N}

Answer Weight 15 N down and normal force 15 N up

Example 2: a pull at an angle

Problem A 10.0 kg box on a frictionless floor is pulled by a rope with 50.0 N at 30.0° above the horizontal. Find the normal force and the acceleration.

  1. Forces: weight down, normal force up, tension at 30.0° up and to the right.
  2. Split the tension.
    Tx=50.0cos⁡30.0∘=43.30 NTy=50.0sin⁡30.0∘=25.0 NT_x = 50.0\cos 30.0^\circ = 43.30\ \text{N} \qquad T_y = 50.0\sin 30.0^\circ = 25.0\ \text{N}
  3. Vertical: no vertical acceleration, so the up forces equal the down force.
    FN+25.0−(10.0)(9.8)=0⇒FN=73.0 NF_N + 25.0 - (10.0)(9.8) = 0 \quad\Rightarrow\quad F_N = 73.0\ \text{N}
  4. Horizontal: only TxT_x acts.
    a=43.30 N10.0 kg=4.33 m/s2a = \frac{43.30\ \text{N}}{10.0\ \text{kg}} = 4.33\ \text{m/s}^2
  5. The normal force is less than the weight (98 N) because the rope lifts a little.

Answer Normal force 73.0 N, acceleration 4.33 m/s²

Example 3: a frictionless ramp

Problem A 4.0 kg block slides down a frictionless ramp tilted 25.0°. Find the normal force and the acceleration.

  1. Forces: weight straight down and normal force perpendicular to the ramp. Tilt the axes along the ramp.
  2. Perpendicular to the ramp, the forces balance.
    FN=mgcos⁡θ=(4.0)(9.8)cos⁡25.0∘=35.5 NF_N = mg\cos\theta = (4.0)(9.8)\cos 25.0^\circ = 35.5\ \text{N}
  3. Along the ramp, only mgsin⁡θmg\sin\theta acts.
    a=mgsin⁡θm=gsin⁡θ=9.8sin⁡25.0∘=4.14 m/s2a = \frac{mg\sin\theta}{m} = g\sin\theta = 9.8\sin 25.0^\circ = 4.14\ \text{m/s}^2

Answer Normal force 36 N, acceleration 4.1 m/s² down the ramp

Common mistakes and how to fix them

  • Drawing forces the object exerts. The diagram shows forces on the object. The book pushing down on the table belongs on the table's diagram.
  • Adding a force in the direction of motion. A sliding box keeps moving because of inertia, not a force. Fix: only contact forces and gravity.
  • Assuming the normal force always equals mg. It changes on ramps, with angled pulls and in elevators. Fix: find it from ΣFy=may\Sigma F_y = ma_y.
  • Using sin and cos the wrong way on a ramp. Fix: check with θ=0\theta = 0. On flat ground, the force along the surface must be 0 (sin⁡0=0\sin 0 = 0) and the normal force must be mgmg (cos⁡0=1\cos 0 = 1).

Practice problems

  1. A skydiver falls at a constant (terminal) speed. Which forces belong on her free body diagram?

    1. Weight down only
    2. Weight down and air resistance up, equal in size
    3. Weight down and a larger air resistance up
    4. Weight down, air resistance up and a downward force of motion
    Show answer

    Answer: Weight down and air resistance up, equal in size

    Constant velocity means zero net force, so air resistance exactly balances weight. A "force of motion" does not exist.

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

    Show answer

    Answer: 29 N

    Two forces: tension up and weight down. They balance: T=mg=3.0×9.8=29.4T = mg = 3.0 \times 9.8 = 29.4 N, which is 29 N.

  3. On a ramp at angle θ\theta, which part of the weight pulls a block down the slope?

    1. mgcos⁡θmg\cos\theta
    2. mgsin⁡θmg\sin\theta
    3. mgtan⁡θmg\tan\theta
    4. mgmg
    Show answer

    Answer: mgsin⁡θmg\sin\theta

    Check with a flat surface: at θ=0\theta = 0, nothing should pull along the surface, and sin⁡0=0\sin 0 = 0.

  4. A 5.0 kg box sits on the floor while you push straight down on its top with 20. N. What is the normal force from the floor?

    Show answer

    Answer: 69 N

    Down: weight 5.0×9.8=495.0 \times 9.8 = 49 N plus your 20. N push. The box is at rest, so the floor pushes up with 49+20.=6949 + 20. = 69 N.

  5. A box slides to the right across a rough floor and is slowing down. Which force should NOT be on its diagram?

    1. Weight
    2. Normal force
    3. Friction to the left
    4. A force to the right that keeps it moving
    Show answer

    Answer: A force to the right that keeps it moving

    Nothing is pushing it to the right anymore. It slows down because friction to the left is the only horizontal force.

Frequently asked questions

Do I draw the arrows from the center of the object?

For most high school and AP Physics 1 problems, yes: draw the object as a dot or box and start every arrow on it. Some teachers want each force drawn where it acts, such as friction at the bottom surface. Follow your class rules, but always start the arrow on the object.

How do I know how long to make each arrow?

Make bigger forces longer arrows. If forces balance, like weight and normal force on a resting book, draw them the same length. If the object speeds up in a direction, the arrows in that direction should be longer than the ones against it.

Why tilt the axes on a ramp?

The block moves along the ramp, so its acceleration is along the slope. Tilting the axes puts the acceleration on one axis and leaves only weight to split into parts. Otherwise you would have to split the normal force, friction and the acceleration too.

Sources

  1. OpenStax College Physics 2e, 4.5 Normal, Tension, and Other Examples of Forces, accessed October 1, 2026
  2. OpenStax College Physics 2e, 4.6 Problem-Solving Strategies, accessed October 1, 2026

Try asking Ducky

  • “Check my free body diagram for the ramp problem. Did I miss a force?”
  • “I drew a forward force on a sliding box and got it wrong. Why?”
  • “Walk me through splitting the tension into components on number 4, but let me do the trig.”

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