The formula

Hanging mass at rest or moving at constant speed
T = mg
Hanging mass accelerating vertically
T = m(g + a)
Two masses over a pulley (Atwood machine)
a = (m₂ - m₁)g / (m₁ + m₂) and T = 2m₁m₂g / (m₁ + m₂)
General case, applied to each object separately
ΣF = ma

What the symbols mean

SymbolMeaningUnit
TTension in the ropeN
mMass hanging from the ropekg
gAcceleration due to gravity9.80 m/s²
aAcceleration of the mass, positive upward in the vertical casem/s²
m₁, m₂The two masses on either side of the pulley, with m₂ the heavier onekg
ΣFNet force on the object being analyzedN

When it applies

  • The rope is treated as massless and inextensible and the pulley as frictionless. A real rope with mass carries a tension that varies along its length.
  • Draw a free-body diagram for each object and apply ΣF = ma to each separately. The shared tension is what links the two equations.
  • Objects joined by a taut rope share the same magnitude of acceleration.

Worked example

Problem. A 15 kg crate hangs from a rope inside an elevator that is accelerating upward at 2.0 m/s². Find the tension, and compare it with the elevator at rest.

  1. Forces on the crate: tension T upward, weight mg downward. Take up as positive.
  2. Newton's second law along the vertical: T - mg = ma, so T = m(g + a).
  3. Substitute: T = (15 kg)(9.8 + 2.0) = (15)(11.8) = 177 N.
  4. At rest or at constant speed, a = 0 and T = mg = (15)(9.8) = 147 N, so accelerating upward adds 30 N.

Answer. 177 N while accelerating upward, against 147 N when the elevator is at rest.

Common mistakes

  • Assuming the tension always equals the weight. That holds only when the acceleration is zero.
  • Giving an ideal pulley a different tension on each side.
  • Forgetting that the rope pulls on both objects it connects, in opposite directions.
  • Adding the two weights on an Atwood machine to get the tension. Solve the two Newton's second law equations together instead.

Related formulas

Sources