Connected Bodies: Tension, Pulleys and Lifts
Two bodies joined by a rope share one acceleration, so each gets its own equation and the tension appears in both. Adding the pair of equations removes the tension, and substituting back returns it.
What a learner can do afterwards
- Draws a separate free-body diagram for each body and writes F = ma for each
- Solves the pair of equations for the shared acceleration and then for the tension
- Explains the changing reading of a bathroom scale in a lift that starts and stops
1 · Read
A bathroom scale in a lift reads differently when it starts moving, though your weight stays the same. Draw one free body diagram per object, showing every force on it alone, and never mix two bodies in one sketch. Tension is the pull along the string, and it is the same throughout one light string. Both bodies move as one unit, so they share the same size of acceleration along the string.
Write F is m a for each body separately, then solve the pair together. Masses of 3 kg and 1 kg hanging over a pulley with g as 10 give 3 times 10 minus T is 3 a and T minus 1 times 10 is 1 a. Adding removes T: 20 is 4 a, so a is 5 metres per second squared. Back-substituting gives T as 15 N.
A bathroom scale reads apparent weight, which changes with acceleration. A 60 kg person standing still reads 60 times 9.8, which is 588 N. In a lift accelerating up at 2 metres per second squared the floor must push 60 times 11.8, which is 708 N. In free fall person and lift fall together, so the scale reads zero.
Add the pair of equations to remove the tension, then substitute back for it. If your tension comes out different from the two bodies, one equation has a sign error.
One diagram and one equation per body, add to kill tension, then read the scale with F is m a.
The rewritten readings were written for this dataset.
More to read
Choose the system before writing equations
Apply Newton's second law by first choosing the object or group of objects to analyse. Identify external forces acting on that system, then write net F = ma for each direction. Internal force pairs cancel in the system's total force. For connected objects, analyse each separately and solve the resulting equations together. When their connection constrains them to equal acceleration magnitudes, this lets you find both acceleration and tension.
Recognising contact forces
A string exerts a pull. A surface exerts a normal force perpendicular to itself, while friction acts along contact and opposes relative sliding or its tendency. An ideal massless string has uniform tension when no external tangential force acts along it. This tension connects the bodies' equations. Normal force prevents interpenetration and may change as a lift accelerates. Identify these forces to draw only the relevant arrows.
Applying force equations to familiar setups
Pulley systems, slopes and lifts require careful force accounting. An ideal frictionless pulley redirects a massless string's tension without altering its magnitude. A taut, inextensible string over a fixed pulley gives the two hanging bodies equal acceleration magnitudes. For a slope, choose axes parallel and perpendicular to it. For a mass m resting on a scale in a lift, the reading is m(g + a) for upward acceleration of magnitude a, or m(g - a) for downward acceleration of magnitude a no greater than g.
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2 · Watch
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