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Free Body Diagram Maker

Add forces with magnitude and direction, see them drawn on a free body diagram and get the net force, its direction and the equilibrium steps. Free FBD tool for AP Physics and GCSE.

Presets:

Angles are measured counter-clockwise from the +x axis (0° = right, 90° = up, 180° = left, 270° = down).

The diagram and steps update as you type.

Net force

ΣF = 12.393 N at 53.794°
ΣFx = 7.321 N, ΣFy = 10 N · Balancing force: 12.393 N at 233.794°

Free body diagram (arrow length ∝ magnitude; dashed = resultant)

+x+yWeight (50 N)Normal (50 N)Pull (20 N)Friction (10 N)ΣF = 12.393 N
  1. 1

    Resolve every force into x and y components

    Each force F at angle θ (counter-clockwise from +x) has components Fx = F cos θ and Fy = F sin θ. All values in newtons.

    ForceFθFcos⁡θFsin⁡θWeight50270∘0−50Normal5090∘050Pull2030∘17.32110Friction10180∘−100\begin{array}{l|r|r|r|r} \text{Force} & F & \theta & F\cos\theta & F\sin\theta \\ \hline \text{Weight} & 50 & 270^\circ & 0 & -50 \\ \text{Normal} & 50 & 90^\circ & 0 & 50 \\ \text{Pull} & 20 & 30^\circ & 17.321 & 10 \\ \text{Friction} & 10 & 180^\circ & -10 & 0 \end{array}
  2. 2

    Weight: 50 N at 270°

    Fx=50cos⁡270∘=0 N,Fy=50sin⁡270∘=−50 NF_x = 50\cos 270^\circ = 0\ \text{N},\qquad F_y = 50\sin 270^\circ = -50\ \text{N}
  3. 3

    Normal: 50 N at 90°

    Fx=50cos⁡90∘=0 N,Fy=50sin⁡90∘=50 NF_x = 50\cos 90^\circ = 0\ \text{N},\qquad F_y = 50\sin 90^\circ = 50\ \text{N}
  4. 4

    Pull: 20 N at 30°

    Fx=20cos⁡30∘=17.321 N,Fy=20sin⁡30∘=10 NF_x = 20\cos 30^\circ = 17.321\ \text{N},\qquad F_y = 20\sin 30^\circ = 10\ \text{N}
  5. 5

    Friction: 10 N at 180°

    Fx=10cos⁡180∘=−10 N,Fy=10sin⁡180∘=0 NF_x = 10\cos 180^\circ = -10\ \text{N},\qquad F_y = 10\sin 180^\circ = 0\ \text{N}
  6. 6

    Add the components

    The net force components are the sums of all x-components and all y-components.

    ∑Fx=0+0+17.321+(−10)=7.321 N,∑Fy=(−50)+50+10+0=10 N\sum F_x = 0 + 0 + 17.321 + (-10) = 7.321\ \text{N},\qquad \sum F_y = (-50) + 50 + 10 + 0 = 10\ \text{N}
  7. 7

    Resultant magnitude

    Combine the perpendicular components with Pythagoras.

    ∣Fnet∣=(∑Fx)2+(∑Fy)2=7.3212+102=12.393 N|F_{\text{net}}| = \sqrt{(\sum F_x)^2 + (\sum F_y)^2} = \sqrt{7.321^2 + 10^2} = 12.393\ \text{N}
  8. 8

    Resultant direction

    Use atan2(ΣFy, ΣFx) so the angle lands in the correct quadrant, measured counter-clockwise from +x.

    θnet=atan2⁡(10, 7.321)=53.794∘\theta_{\text{net}} = \operatorname{atan2}(10,\ 7.321) = 53.794^\circ
  9. 9

    Equilibrium check

    The net force is not zero, so the object is NOT in equilibrium — it accelerates in the direction of the resultant.

    ∣Fnet∣=12.393 N≠0|F_{\text{net}}| = 12.393\ \text{N} \neq 0
  10. 10

    Balancing force (equilibrant)

    To reach equilibrium, add a force equal in size and opposite in direction to the resultant.

    Fbal=12.393 N at 53.794∘+180∘=233.794∘F_{\text{bal}} = 12.393\ \text{N at } 53.794^\circ + 180^\circ = 233.794^\circ

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FAQ

How is the net force found?

Each force is split into x and y components; the components are added and the resultant is √(Fx² + Fy²) at angle atan2(Fy, Fx).

What is equilibrium?

When the net force is zero — the body stays at rest or moves with constant velocity.

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