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.
2 · Watch
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8 questions wait behind this lesson, each with its answer explained. Every answer feeds the sky: stars light as they are learned, and dim when it is time to come back.