Newton's Second Law

The Σ in ΣF = m a is a vector sum over the forces that objects outside your chosen body exert on it, and nothing else. Forces that one part of the body exerts on another come in equal and opposite pairs with both members inside the system, so they add to zero in the sum, and the product m a is not in the list at all: it is the response, not one of the interactions, so it never appears as an arrow on the diagram. Take a cyclist and bicycle as one 75.0 kg body on a level road. The road pushes forward on the rear tyre with 110 N, air drag pulls back with 25 N and rolling resistance with 10 N. With forward as +x, ΣFₓ = 110 N − 25 N − 10 N = 75 N, so aₓ = 75 N ÷ 75.0 kg = 1.0 m s⁻². The 110 N drive does not set the acceleration; the 75 N that survives the subtraction does. The rider's push on the pedals and pull on the handlebars appear nowhere, because both ends of those forces lie inside the system. Once drag has grown until the three forces sum to zero, the same equation gives a = 0 and the rider holds a steady speed.

The m in the law is defined by the law itself. Apply the same net force to two bodies and compare: m₂/m₁ = a₁/a₂. The force needs no calibration, only to be repeatable, such as a fan unit running at a fixed setting. Suppose the fan gives a 0.500 kg air-track glider an acceleration of 0.720 m s⁻², and the same glider carrying an unknown block 0.450 m s⁻². The loaded glider has mass 0.500 kg × 0.720 ÷ 0.450 = 0.800 kg, so the block is 0.300 kg, and the fan's thrust was 0.500 kg × 0.720 m s⁻² = 0.360 N. No balance and no gravity entered, which is why the same method works in orbit. Mass found this way is a positive scalar, it adds when bodies are joined, and at everyday speeds it depends neither on where the body is nor on how fast it is moving. That the same number also fixes the pull of gravity on the body is a separate experimental fact, taken up with weight.

A vector equation is true component by component, so choose axes, resolve every force, and write ΣFₓ = m aₓ and ΣF_y = m a_y. The two equations share nothing but the mass and the clock. A 25.0 kg sledge on level, frictionless ice is pulled by a rope with 60.0 N at 40.0° above the horizontal. Along x only the rope acts: 60.0 N × cos 40.0° = 46.0 N, so aₓ = 46.0 N ÷ 25.0 kg = 1.84 m s⁻². Along y the sledge stays on the ice, so a_y = 0 is known before any force is, and n + 60.0 N × sin 40.0° − (25.0 kg)(9.81 m s⁻²) = 0 returns the unknown support force: n = 245.3 N − 38.6 N = 207 N, less than the 245 N weight because the rope carries part of it. That division of labour is typical: one axis has a known acceleration and delivers a force, the other has known forces and delivers the acceleration. It works only if every force, the tilted ones included, is resolved onto the same pair of axes before anything is added.

Because a = dv/dt, the net force tells you how the velocity is changing and nothing about what it currently is. A 0.145 kg ball thrown straight up feels, once it leaves the hand and ignoring air, one force: its weight, 1.42 N downward. Its acceleration is therefore 9.81 m s⁻² downward on the way up, on the way down, and at the top where v = 0 for an instant; if the net force vanished there, a ball at rest would stay at rest and hang in mid-air. In two dimensions the same point looks like this. A 0.50 kg puck gliding east at 3.0 m s⁻¹ receives a steady net force of 2.0 N due north. Its acceleration is 4.0 m s⁻² due north, exactly along the force. After 0.50 s its velocity is (3.0, 2.0) m s⁻¹, pointing 34° north of east, a direction that matches neither the force nor the original motion. What lines up with ΣF is Δv, the change, here 2.0 m s⁻¹ due north. Speeding up, slowing down and turning are all accelerations, and each one needs a net force.

Written as m d²x/dt² = ΣFₓ, the law is an equation for the unknown function x(t), completed by the starting position and velocity. A constant force gives back the constant-acceleration formulas; anything else must be integrated. A 2.00 kg cart starts from rest under a push that grows as F(t) = (6.00 N s⁻¹) t. Then a(t) = (3.00 m s⁻³) t, v(t) = ∫₀ᵗ a dt′ = (1.50 m s⁻³) t², and x(t) = (0.500 m s⁻³) t³, so after 3.00 s the cart moves at 13.5 m s⁻¹ and has covered 13.5 m. Using v = at with the final acceleration of 9.00 m s⁻² would have claimed 27.0 m s⁻¹, double the truth. The law also runs backward. If video tracking of a 1.5 kg body fits x(t) = (2.0 m s⁻²) t² − (0.25 m s⁻³) t³, two derivatives give a(t) = 4.0 m s⁻² − (1.5 m s⁻³) t, so the net force was 6.0 N − (2.25 N s⁻¹) t. It passes through zero at t = 2.7 s, which is the instant the velocity peaks, at 5.3 m s⁻¹, not the instant the body stops.

A second-law result should survive three quick tests. Units: a force over a mass is N kg⁻¹ = (kg m s⁻²) kg⁻¹ = m s⁻², so an acceleration written as F × m, or as F/g, fails at once. Limits: in the sledge result aₓ = F cos θ/m and n = mg − F sin θ, setting θ = 0 gives aₓ = F/m and n = mg, a level pull on a fully supported body, while θ = 90° gives aₓ = 0 and n = mg − F, a straight lift. Let m grow without bound and a falls to zero: a very massive body barely responds to the same force. Range: n = mg − F sin θ turns negative once F sin θ exceeds mg, which for the 25.0 kg sledge at 40.0° means a pull above 382 N. Ice cannot pull downward, so a negative n announces that the assumption a_y = 0 has failed and the sledge has left the surface. The law did not break; the constraint fed into it did, and with that constraint removed the same two equations describe the airborne sledge.