Tension & Ideal Connectors

Cut a taut string anywhere in your imagination and each piece pulls on the other; the size of that pull is the tension T at the cut. It is a contact force carried by the stretched fibres, so it acts along the string's own line and nowhere else. On a free-body diagram that settles the direction before any algebra: the string pulls the body toward the string, and the body pulls back on the string equally hard, a third-law pair. A 6.00 kg lamp hanging at rest from a light cord needs T = mg = 58.9 N at every cut, because whatever hangs below the cut is in equilibrium. Two things tension never does. It never pushes: T ≥ 0, and an equation that returns a negative T is reporting a slack string, not a compressed one. And it never acts sideways: a flexible string has no stiffness to resist bending, which is why a loaded string lies straight between the points it touches.

Take the string itself as the system. A rope of mass mₛ = 0.200 kg lies along a frictionless floor and drags a 4.00 kg trolley across it at 2.50 m s⁻², so the floor carries the rope's weight and only the pulls on its ends act along its length. The trolley needs T₁ = 4.00 × 2.50 = 10.0 N from the rope's far end. The hand has to accelerate rope and trolley together, so T₂ = 4.20 × 2.50 = 10.5 N. The second law for the rope alone says the same thing: T₂ − T₁ = mₛa = 0.50 N, a 5% disagreement between the ends. Swap the rope for 20 g of fishing line and the difference drops to 0.05 N, half a per cent. Let mₛ → 0 and T₁ = T₂ exactly, whatever the acceleration. That is the entire meaning of “light string” in a problem statement. It is not an equilibrium claim, since the string may be accelerating hard. It says that a body of zero mass can never carry a net force, so the pulls on its two ends must cancel and one symbol T serves for the whole string.

A pulley lets a string turn a corner. If the wheel has no mass and its bearing no friction, nothing is needed to spin it up, so the string must pull equally hard on both sides: a difference in tension would be a net turning effect on an object with no rotational inertia, the rotational twin of the massless-string argument. The ideal pulley therefore changes the direction of T and leaves its magnitude alone. It is not force-free, though. Each strand pulls on the wheel with T along its own line, and the axle must supply the opposite of their vector sum, of size 2T cos(φ/2) when the strands leave the wheel at angle φ to each other. A string that arrives horizontally and leaves vertically downward with T = 9.20 N pulls the wheel 9.20 N sideways and 9.20 N down, so the axle pushes back with √2 × 9.20 = 13.0 N, pointing up and away from the incoming strand at 45° to the horizontal. Two parallel strands hanging from a ceiling pulley load its axle with 2T. A real wheel with mass needs unequal tensions to change its spin, a job for the rotational dynamics later in this course.

Free-body diagrams for a cart on a track and a block hanging over the edge give two second-law equations with three unknowns: T and two accelerations. The missing equation is geometric. Let x be the cart's distance from the pulley and y the block's depth below it. The string's length is x + y plus the fixed arc on the wheel, and it cannot change, so dx/dt = −dy/dt and d²x/dt² = −d²y/dt². The cart closes on the pulley exactly as fast as the block descends, and the two share one acceleration magnitude a. The working rule is to let the positive direction follow the string, so a single symbol a serves both bodies. Rearranging the string changes the constraint, not the method. Hang a block from a movable pulley, fix one end of the string overhead and pull the other: both supporting strands must shorten when the block rises, so lifting the block a distance h takes up a length 2h of string, and the free end moves twice as far, twice as fast and with twice the acceleration of the block.

Put numbers in: a 2.50 kg cart on a frictionless track and a 1.50 kg block hanging, positive along the string. Cart: T = 2.50a. Block: 1.50g − T = 1.50a. Adding removes T, so a = 1.50 × 9.81/4.00 = 3.68 m s⁻², and then T = 2.50 × 3.68 = 9.20 N, well below the block's 14.7 N weight. The general results are a = m₂g/(m₁ + m₂) and T = m₂g × m₁/(m₁ + m₂), and their limits are worth more than the numbers. As m₁ → 0 there is nothing to pull: a → g, T → 0, and the block is in free fall under a limp string. As m₁ grows without limit, a → 0 and T → m₂g, the static value. Every real case sits between, with T the fraction m₁/(m₁ + m₂) of the hanging weight, here 0.625. The whole-system view agrees, since the only external force along the string's line is m₂g and it accelerates 4.00 kg, but that view hides T. To find a tension you must cut the system at the string and write the second law for one side.

Each idealisation fails in its own direction. Mass: a 3.00 m rope of mass 0.600 kg hoists a 9.40 kg bucket at 1.20 m s⁻². The rope hanging below a point y above the bucket has mass λy, with λ = 0.200 kg m⁻¹, so dT/dy = λ(g + a) and T(y) = (9.40 + 0.200y)(g + a): 103 N at the bucket and 110 N at the winch. The 6.4% rise is just the mass ratio 0.600/9.40, and whether it matters is your decision, made against the precision you need; for a mine cable heavier than its cage the massless model is useless. Pulley inertia and bearing friction: a wheel with mass needs a tension difference across it whenever its spin is changing, and a rubbing bearing needs one even at steady speed. Stretch: a bungee cord lets its ends move differently, so the constraint fails and the cord has become a spring. Slack: the constraint is one-sided and holds only while T stays positive. Strength: every cord has a maximum tension, and the second law turns it into a maximum acceleration. A 180 N cord lifting 15.0 kg allows a = 180/15.0 − 9.81 = 2.19 m s⁻² and no more.