Free-Body Diagram Builder

The block sits at the center of the diagram. Each nonzero push is drawn as a blue arrow from the edge of the block, with length proportional to its size (all applied forces and the net force use the same scale). The coral arrow from the center is the net force, and the dashed navy arrow is the acceleration, which always points the same way as the net force.

The readout gives F_net,x (right minus left), F_net,y (up minus down), the magnitude |F_net|, the direction as an angle measured from the +x axis, and |a| in m/s². Angles run from −180° to +180°, so a net force pointing down and to the left shows a negative angle. When everything cancels, the direction line reads “balanced (equilibrium)” and |a| is 0.

The simulator does not add gravity or a normal force for you; every force on the block is one you set. To model a 2 kg block resting on a level table (taking g ≈ 10 m/s²), set down = 20 N for its weight and up = 20 N for the normal force, then add horizontal pushes and watch only the x-component matter. Increase the mass and notice the dashed acceleration arrow shrink while the coral net-force arrow stays the same.

Newton's second law is applied one axis at a time: $$F_{net,x} = \sum F_x = m a_x \qquad F_{net,y} = \sum F_y = m a_y$$ With right and up taken as positive, $F_{net,x} = F_{right} - F_{left}$ and $F_{net,y} = F_{up} - F_{down}$.

The components combine like any vector: $$|F_{net}| = \sqrt{F_{net,x}^2 + F_{net,y}^2} \qquad \theta = \tan^{-1}\!\left(\frac{F_{net,y}}{F_{net,x}}\right)$$ and the acceleration has magnitude $a = |F_{net}|/m$ in the direction of the net force. Use the signs of the components to decide which quadrant the angle is in; a calculator's inverse tangent alone cannot tell up-left from down-right.

If $F_{net} = 0$, the block is in translational equilibrium: $a = 0$, so it is either at rest or moving at constant velocity (Newton's first law). Equilibrium does not require zero forces, only forces that cancel.

A 4 kg crate is pushed with 3 N to the right, 15 N to the left and 5 N downward, with no upward push. Find the net force and the acceleration.

Components. $F_{net,x} = 3 - 15 = -12$ N and $F_{net,y} = 0 - 5 = -5$ N.

Magnitude. $|F_{net}| = \sqrt{(-12)^2 + (-5)^2} = \sqrt{169} = 13$ N.

Direction. Both components are negative, so the net force points down and to the left. The angle below the −x axis is $\tan^{-1}(5/12) = 22.6°$. Measured from +x, that is $-157.4°$ (equivalently 202.6°).

Acceleration. $a = 13/4 = 3.25$ m/s², in the same direction as the net force.

Set right = 3, left = 15, up = 0, down = 5 and m = 4 in the simulator. The readout should show F_net,x = −12 N, F_net,y = −5 N, |F_net| = 13 N, a direction of −157.38° from +x and |a| = 3.25 m/s². Now change the mass to 8 kg: the net force is unchanged and the acceleration halves to about 1.63 m/s².

Free-body diagrams are the starting point for almost every dynamics problem in Unit 2 (Topic 2.2, Forces and Free-Body Diagrams), and they come back for circular motion, springs, torque and fluids. Free-response questions often ask you to draw the diagram first and then use it to write Newton's second law; the derivation is graded on whether your equation matches your diagram, so draw every force you will later use.