Newton-Law Application Studio

Before any algebra, draw a separate free-body diagram for every body whose motion you need, and name each force by the object that exerts it. A force between two bodies then appears twice, equal and opposite, which is how the third law enters. Take a 4.0 kg block on a table with μₖ = 0.30, tied over an edge pulley to a 2.0 kg hanging mass. The unknowns are the block's acceleration, the hanging mass's acceleration, the tension T, the normal force N and the friction f: five. The equations are the block's horizontal and vertical second law, the hanging mass's vertical second law, the rope constraint and the friction law f = μₖN: also five. Solving gives N = 39.2 N, f = 11.8 N, a = 1.31 m s⁻² and T = 17.0 N. If the count does not close, something is missing, usually a constraint. A pulley with mass adds one unknown, because the tension differs on its two sides, and one equation, its own rotation, so the books still balance.

A constraint is geometry that a connector enforces, and the safe way to write one is to express the connector's length in your coordinates and differentiate. For the table system, let x be the block's travel toward the pulley and y the depth of the hanging mass. The rope length (D − x) + y is fixed, so d²y/dt² = d²x/dt², with a sign you did not have to guess. The same two lines handle geometry where guessing fails. A cart is hauled by a rope that rises to a pulley H = 1.0 m above its tow hook and a horizontal distance x ahead of it, so ℓ = √(x² + H²) and dℓ/dt = (x/ℓ) dx/dt. At x = 2.4 m, ℓ = 2.6 m, and reeling in at a steady 0.50 m s⁻¹ moves the cart at 0.54 m s⁻¹. Differentiate again and a term in (dx/dt)² appears: even a steady winch speeds the cart up, here at 0.018 m s⁻². Every constraint also carries a condition, because it holds only while its connector works: a rope while T ≥ 0, a surface while N ≥ 0.

Every connector and contact has two states, and the state decides which equation you may write. A rope is taut, giving a constraint and an unknown T, or slack, with T = 0 and no constraint. The state is an assumption, so test it. Two 2.0 kg blocks slide down a 30° slope tied by a string, the lower one with μₖ = 0.40 and the upper one with μₖ = 0.20. Assume the string is taut: both share a = g sin 30° − g cos 30° × (0.40 + 0.20)/2 = 2.36 m s⁻², and the lower block's equation then asks the string for T = −1.70 N. A string cannot push, so that state is false. Re-solve with T = 0: the upper block accelerates at 3.21 m s⁻² and the lower at 1.51 m s⁻², so the string sags as the upper block closes the gap. Once it has caught up and the two move together, the connector is a contact, and a contact can push: it presses on the lower block with 1.70 N. The same sign that condemned the string is exactly right for a contact force.

A stage ends at an event: a mass lands, a rope goes slack, a block leaves a surface, sliding stops. At the event the force list changes, so the acceleration jumps, but position and velocity carry over. Return to the tethered 4.0 kg block, μₖ = 0.30, with the 2.0 kg mass released 0.50 m above the floor. Stage 1, rope taut: a₁ = 1.31 m s⁻², and the constraint says the block has moved exactly 0.50 m when the mass lands, after √(2h/a₁) = 0.87 s, moving at √(2a₁h) = 1.14 m s⁻¹. Stage 2 starts from that position and speed with a new force list: the rope is slack, friction alone acts, a₂ = −μₖg = −2.94 m s⁻², and the block slides 1.14²/(2 × 2.94) = 0.22 m further, stopping 0.72 m from its start, 0.39 s after the landing. A third state follows: at rest with nothing pulling, static friction is zero, not μₖN. Since d(v²)/dx = 2v dv/dx = 2a, a graph of v² against x is a straight line in each stage, of slope 2a₁ = 2.62 m s⁻² and then −2μₖg = −5.89 m s⁻².

When a stage runs along a curve, switch at the event to radial and tangential axes. The radial equation, ΣFᵣ = mv²/r toward the centre, now decides the normal force, and through it the friction that the tangential equation sees. A 60 kg sledder on snow with μₖ = 0.10 crosses the crest of a hump of radius 8.0 m at 6.0 m s⁻¹. Radially, mg − N = mv²/r, so N = 60 × (9.81 − 4.50) = 319 N instead of the 589 N on level snow. That is the reading a scale would give in a lift accelerating downward at 4.50 m s⁻²: the radial equation is an apparent-weight equation. Along the path at the crest, friction μₖN is the only force, so dv/dt = −0.10 × 319/60 = −0.53 m s⁻², barely half the −0.98 m s⁻² on the flat. In a dip of the same radius N rises to 859 N and the sled slows at 1.43 m s⁻². Then test the state: N ≥ 0 at the crest requires v ≤ √(gr) = 8.86 m s⁻¹. Any faster and the sled leaves the snow, N and friction vanish, and the next stage is projectile motion.

A finished answer is tested before it is trusted. Limits: a = (m₂ − μₖm₁)g/(m₁ + m₂) becomes the frictionless m₂g/(m₁ + m₂) at μₖ = 0, approaches g as m₁ → 0, and falls to zero at m₂ = μₖm₁. Bounds: T = 17.0 N lies between zero and m₂g = 19.6 N, as it must for a mass speeding up downward. Idealizations: judge each by the size of what it drops, through δa/a = δFₙₑₜ/Fₙₑₜ − δM/M. Here Fₙₑₜ = 7.85 N and M = 6.0 kg. A 0.40 kg uniform-disc pulley acts as an extra 0.20 kg on the rope, a result borrowed from the rotation unit, and lowers a by 3.2%. Drag on a block of frontal area 0.010 m² at 1.14 m s⁻¹ is about ½ × 1.2 × 1.0 × 0.010 × 1.14² = 0.008 N, a thousandth of Fₙₑₜ. But μₖ is rarely known better than ±0.02, which shifts Fₙₑₜ by 0.78 N, or 10%. Lower the hanging mass to 1.4 kg and Fₙₑₜ is only 1.96 N, so the same ±0.02 moves a by 40% while the pulley still costs under 4%. The closer the drive is to the friction it fights, the more the prediction belongs to μ.