Newton laws and force superposition

Force is a vector quantity, so several forces applied to a body add by the rules of vector addition: superposition of forces means each force acts independently, and the total effect is set by the resultant $\vec{F}_{\text{net}} = \sum_i \vec{F}_i$. Newton's first law states that if the resultant is zero, the body keeps its state of rest or uniform straight-line motion in an inertial frame. Newton's second law links the resultant to acceleration: $\vec{F}_{\text{net}} = m\vec{a}$, and the acceleration points in the same direction as the resultant. Newton's third law says that the interaction forces between two bodies are equal in magnitude and opposite in direction and act on different bodies, so they never appear together in the net force of a single body. To add forces, choose axes, project each force using $\cos$ and $\sin$ of its angle, sum the projections along each axis, and then find the magnitude $F_{\text{net}} = \sqrt{F_x^2+F_y^2}$ and the direction $\tan\theta = F_y/F_x$. Weight $m\vec{g}$, elastic force, and friction follow their own laws ($F_{\text{el}} = kx$, $F_{\text{fr}} = \mu N$) and enter the sum as ordinary terms.

Two horizontal forces act on a body of mass $m = 4.0$ kg: $F_1 = 12$ N to the right and $F_2 = 5.0$ N to the left. Find the resultant and the acceleration of the body.

Two forces act on a body of mass $m = 2.0$ kg: $F_1 = 5.0$ N along the $+x$ axis and $F_2 = 5.0$ N at $60^\circ$ to the $+x$ axis. Find the magnitude and direction of the resultant, and the acceleration of the body.

A crate of mass $m = 10$ kg rests on a horizontal floor, friction coefficient $\mu = 0.30$. A force $F = 50$ N is applied at $30^\circ$ above the horizontal (upward and forward). Will the crate start to move? If so, find its acceleration. Take $g = 9.8$ m/s² and assume equal static and kinetic friction coefficients.

Self-study plan with a simulation (PhET “Forces and Motion: Basics”, Net Force tab); real loads are replaced by a safe model, so there is no need to measure heavy bodies or push them by hand.

1. Set the mass to $m = 2.0$ kg and apply two forces along one axis: first $F_1 = 8$ N and $F_2 = 3$ N in the same direction, then turn $F_2$ in the opposite direction. Record the resultant and the acceleration, and check the formula $a = F_{\text{net}}/m$.
2. Apply two mutually perpendicular forces $F_1 = 6$ N and $F_2 = 8$ N and verify that $F_{\text{net}} = \sqrt{6^2+8^2} = 10$ N, while the acceleration arrow points in the same direction as the resultant arrow.
3. Now add a third force that balances the first two: it must point opposite their resultant, i.e. have magnitude $F_3 = \sqrt{6^2+8^2} = 10$ N and direction $233.1^\circ$ from the $+x$ axis (about $127^\circ$ from the first force). Verify that $F_{\text{net}} = 0$; describe how the body moves when the resultant is zero and connect this with Newton's first law. Alternative: three equal forces of 10 N each, directed at $120^\circ$ to one another, also give a zero resultant.
4. Keep a table: $F_x$, $F_y$, $F_{\text{net}}$, direction, $a$. Compare your formula-based prediction with the simulation readout and note any discrepancies.

Show that forces add as vectors: the resultant is found by the polygon rule, its projections equal the algebraic sums of the individual projections, and the acceleration always points along the resultant.

A two-part diagram. On the left, a body at the centre with force arrows $\vec{F}_1$, $\vec{F}_2$, $\vec{F}_3$ drawn from a single point, with angles to the $x$ axis labelled. On the right, a tail-to-head vector polygon for the same forces, the dashed resultant arrow $\vec{F}_{\text{net}}$, and the projections $F_x$, $F_y$ shown on coordinate axes.

A football of mass $m=0.43\ \text{kg}$ rests on the ground. A player kicks it horizontally with force $F_{\text{kick}}=120\ \text{N}$. During the kick, gravity pulls down with $F_g = mg = 0.43 \times 9.8 \approx 4.2\ \text{N}$, the ground pushes up with normal force $N = 4.2\ \text{N}$, and friction opposes the sliding motion with $f=2\ \text{N}$. The vertical forces cancel: $N - F_g = 0$. The horizontal net force is $F_{\text{net}} = 120 - 2 = 118\ \text{N}$, so acceleration $a = F_{\text{net}}/m = 118/0.43 \approx 274\ \text{m/s}^2$.

This uses Newton's second law $F_{\text{net}} = ma$ and force superposition: weight, normal force, kick, and friction are added as vectors. Their vertical components sum to zero, and the horizontal resultant sets the acceleration.

A car of mass $m=1200\ \text{kg}$ moves at $v=20\ \text{m/s}$ (72 km/h) and brakes. On dry asphalt, kinetic friction is $f_k = 6000\ \text{N}$, air drag is $F_{\text{air}} = 300\ \text{N}$, and rolling resistance is $200\ \text{N}$. All three oppose motion, so the net horizontal force is $F_{\text{net}} = -(6000+300+200) = -6500\ \text{N}$, giving acceleration $a = -6500/1200 \approx -5.4\ \text{m/s}^2$. On wet asphalt friction drops to $2500\ \text{N}$; then $F_{\text{net}} = -(2500+300+200) = -3000\ \text{N}$ and $a \approx -2.5\ \text{m/s}^2$, roughly halving the deceleration.

Superposition of friction and drag as vectors along the direction of motion, followed by Newton's second law to find the deceleration.

A raindrop of mass $m=0.05\ \text{g} = 5\times 10^{-5}\ \text{kg}$ falls. Its weight is $F_g = mg = 5\times 10^{-5} \times 9.8 \approx 4.9\times 10^{-4}\ \text{N}$ downward. As speed grows, upward air drag $F_{\text{drag}}$ increases; when the drop reaches terminal velocity $v_t = 9\ \text{m/s}$, the drag equals the weight: $F_{\text{drag}} = 4.9\times 10^{-4}\ \text{N}$ upward. The net force is exactly zero, so acceleration is zero and the speed stays constant at $9\ \text{m/s}$.

This is force superposition producing zero net force — the condition for Newton's first law (equilibrium): $F_g + F_{\text{drag}} = 0$. It shows how the vector sum, not a single force, controls motion.