Connected Objects

Whether a force is internal or external is not a property of the force; it depends on where you draw the boundary. A kayaker tows two loaded rafts in line: her 150 N tow rope pulls a 40 kg raft, which drags a 60 kg raft on a short painter, and the water resists them with 30 N and 45 N. Draw the boundary round both rafts. The painter lies wholly inside it, pulling the rear raft forward and the front raft back with equal force, so the pair sums to zero. What crosses the boundary horizontally is 150 N forward and 75 N back: a = 75 N ÷ 100 kg = 0.75 m s⁻². The painter's tension never appeared. Now draw the boundary round the rear raft alone. The painter crosses it and becomes external: T − 45 N = (60 kg)(0.75 m s⁻²), so T = 90 N. The front raft was never used, which makes it a free check: 150 − 30 − 90 = 30 N, and (40 kg)(0.75 m s⁻²) = 30 N.

Two bodies on a line give two second-law equations but three unknowns: a₁, a₂ and the force in the link. The missing equation is not one of Newton's laws. Put the bodies at x₁ and x₂ on one axis. A rigid bar, a taut string that does not stretch, or a contact that stays closed holds x₁ − x₂ = ℓ, a constant, and differentiating twice with respect to time gives v₁ = v₂ and then a₁ = a₂. Each link enforces this one-sidedly. A string stops the separation growing but not shrinking, so it delivers a₁ = a₂ only while its tension is positive. A contact stops the bodies overlapping but not parting, so only while the normal force between them is positive. Only a bar works both ways. Treat the shared acceleration as a hypothesis: solve with it, then check the sign of the link force before believing the answer. The next topic runs the same differentiation on strings that bend round pulleys, where the accelerations match in size but not in direction.

Give every body the same positive direction and write each link force twice: a light string pulls the rear body with +T and the front body with −T. The sizes match because the string is light: each pull has a third-law partner acting on the string, and a string of negligible mass needs no net force. For a pull F on the front body, F − f₁ − T = m₁a and T − f₂ = m₂a. Add them and T cancels, leaving the composite equation, which is therefore not a new law but the sum of the individual ones; a stray 2T in your sum means a sign is wrong. The same discipline handles a vertical chain. A winch lifts a 9.0 kg crate with a 6.0 kg crate slung beneath it, and the cable pulls up with 180 N. With up positive for both, 180 − (15.0)(9.81) = 15.0a gives a = +2.19 m s⁻², and the sling obeys T − (6.0)(9.81) = (6.0)(2.19), so T = 72.0 N. Cut the cable force to 120 N and a = −1.81 m s⁻²: the crates accelerate downward, yet T = (6.0)(9.81 − 1.81) = 48.0 N is still positive and the sling stays taut. A negative a is only a direction. It is a negative T that would break the model.

To find one particular link force, draw a boundary that cuts that link and no other, then keep the side with fewer forces, usually the side away from the applied pull. There the link is the only thing driving the motion, so it equals the mass beyond it times a, plus any resistance that mass meets. A tractor hauls three trailers of 1500 kg, 1000 kg and 500 kg, in that order, with a drawbar force of 3600 N and negligible rolling resistance. The whole train gives a = 3600 N ÷ 3000 kg = 1.20 m s⁻². The coupling behind the first trailer drags 1500 kg, so it carries 1800 N; the coupling behind the second drags 500 kg and carries 600 N. Keeping the near side gives the same answer with more terms: 3600 − (1500)(1.20) = 1800 N. Read along the train and the force steps down 3600, 1800, 600, 0 N, each step one trailer's mass times a. With N identical cars the steps are equal and the coupling behind car k carries F(N − k)/N: in a ten-car train the coupling behind the first car takes 90% of the pull and the one behind the ninth takes 10%.

Give each body its own kinetic friction, μ₁m₁g under the front body that receives F and μ₂m₂g under the towed one. The composite system gives a = [F − (μ₁m₁ + μ₂m₂)g]/M, and the towed body alone then gives T = [m₂F + (μ₂ − μ₁)m₁m₂g]/M. Take the figure's defaults: F = 40 N on a 5.0 kg front block with μ₁ = 0.40, towing 3.0 kg. With μ₂ = 0.40 as well, a = 1.08 m s⁻² and T = 15.0 N, exactly the 40 × 3.0/8.0 share a frictionless floor would give: friction and inertia are both proportional to mass, so the string supplies both in one ratio. Make the towed block slicker, μ₂ = 0.10, and a rises to 2.18 m s⁻² while T falls to 9.48 N, because that block now needs less help. T reaches zero at F = (μ₁ − μ₂)m₁g = 14.7 N. Pull more gently than that while both still slide and the formula returns a negative tension: the slick rear block would slow less sharply than the front one, so it closes the gap and the string goes slack. The honest solution then treats each block alone, with T = 0.

Shrink the cars and multiply them and a train becomes a rope. For a rope of mass per length λ, pulled with force F at x = 0 and accelerating at a across a frictionless surface, the slice between x and x + dx has mass λ dx, so T(x) − T(x + dx) = λa dx and dT/dx = −λa: the tension falls linearly. A 6.0 m rope of mass 3.0 kg hauling a 27 kg sled over smooth ice with 60 N has a = 2.0 m s⁻² and λ = 0.50 kg m⁻¹, so its tension drops 1.0 N per metre, from 60 N at the hand to 54 N at the sled. The fractional drop is mᵣₒₚₑ/M, 10% here and 0.55% for a 150 g cord, and that ratio is what a light string has to mean. Rigidity is an approximation too. A push travels through steel at about 5000 m s⁻¹, so a 0.40 m block settles to one acceleration in roughly 10⁻⁴ s. But a train with 0.10 m of free play in each of 60 couplings has moved its locomotive 6.0 m before the last car stirs, and until then there is no single a to speak of.