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Mechanics · Forces & Newton's laws

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Newton's laws and F = ma

Newton's three laws and the one equation that runs all of dynamics — the RESULTANT force equals mass times acceleration — applied along a line and to forces given as 2-D vectors, why weight is mg and the normal reaction is whatever balances the rest, and how a lift problem is just F = ma done vertically.

Mechanics · Forces & Newton's laws

Newton's laws and F = ma

Newton's three laws and the one equation that runs all of dynamics — the RESULTANT force equals mass times acceleration — applied along a line and to forces given as 2-D vectors, why weight is mg and the normal reaction is whatever balances the rest, and how a lift problem is just F = ma done vertically.

Why it works

Three laws, then one master equation.
  • Newton's first law. A body stays at rest or moves with constant velocity
unless a resultant force acts. So constant velocity needs no net force — the forces are balanced ([[forces.force-diagrams]]). A car cruising at a steady 30 m s−130\ \text{m s}^{-1} has its driving force exactly cancelled by resistance.
  • Newton's second law. A resultant force makes a body accelerate, in the same
direction, with Fnet=ma.\mathbf{F}_{\text{net}} = m\mathbf{a}. This is the workhorse. The force in it is the resultant of all forces, never just one of them.
  • Newton's third law. If A pushes B, then B pushes A with an equal and opposite
force. The two forces of the pair act on different bodies, so they never appear on the same free-body diagram and never "cancel each other out."

Weight is a force; mass is not. Mass mm (in kg) measures how much matter is there; weight W=mgW = mg (in newtons) is the force gravity exerts on that mass. Putting weight into F=maF = ma as if it were mass — or vice versa — is the most common slip in the topic. Keep the units straight: kilograms in mm, newtons in FF.

Working in a straight line. Pick a positive direction (usually the direction of motion), add up the forces with signs to get the resultant, and divide by the mass: a=resultant forcem.a = \frac{\text{resultant force}}{m}. A 4 kg4\,\text{kg} box pulled by 20 N20\,\text{N} against a 8 N8\,\text{N} resistance has resultant 20−8=12 N20 - 8 = 12\,\text{N}, so a=12/4=3 m s−2a = 12/4 = 3\ \text{m s}^{-2}. A negative resultant just means a deceleration.

The normal reaction works itself out. Stand a body on the floor and push or pull it vertically: RR adjusts so the vertical equation of motion holds. If the body isn't accelerating vertically, the vertical forces balance and RR is whatever makes that true — found from F=maF = ma with a=0a = 0 vertically, not assumed.

The lift problem is F=maF = ma done vertically. A person of mass mm stands in a lift; the floor pushes up with RR, gravity pulls down with mgmg, and the lift (and person) accelerate together. Taking up as positive, R−mg=ma  ⇒  R=m(g+a).R - mg = ma \;\Rightarrow\; R = m(g + a).RW = mga*Free-body diagram of the person in a lift accelerating upward (not accurately drawn). RR is the push of the floor, W=mgW = mg the weight; aa marks the acceleration, not a force.* Accelerating up, R>mgR > mg (you feel heavier); accelerating down, R=m(g−a)<mgR = m(g - a) < mg (you feel lighter); in free fall a=ga = g and R=0R = 0 — weightlessness. The reading on the scales is RR, the apparent weight, not the true weight mgmg.

Motion in two dimensions. When forces are given as i,j\mathbf{i}, \mathbf{j} vectors, F=maF = ma holds component-by-component. Find the resultant by adding components, then a=1mFnet\mathbf{a} = \frac{1}{m}\mathbf{F}_{\text{net}}. The acceleration points along the resultant, and its magnitude is ∣a∣=∣Fnet∣m|\mathbf{a}| = \frac{|\mathbf{F}_{\text{net}}|}{m}.