Tension Force Formula: T = m(g + a) and Pulley Examples

The tension force formula for a mass hanging on a rope is $T = m(g + a)$ , where a is the upward acceleration. When the mass hangs still, a = 0 and the formula of tension reduces to $T = mg$ , so a 3.0 kg lamp pulls its cord tight with 29.4 N.

Tension is the pulling force a string, rope, chain or cable passes along its length. It always pulls, never pushes, and it acts along the string. The formula comes straight from Newton's second law, which places it among the laws of motion results. For the calculator below, g is fixed at 9.8 m/s² and a downward acceleration is entered as a negative number.

Quick facts

Hanging at rest
$T = mg$
Accelerating up
$T = m(g + a)$
Accelerating down
$T = m(g - a)$
SI unit
newton (N)
Dimensional formula
[M¹ L¹ T⁻²]
Direction
Along the string, pulling away from the body

Where the tension formula physics students use comes from

Draw the hanging mass on its own. Two forces act: tension T upward and weight mg downward. Take up as positive and apply $\sum F = ma$ :

$T - mg = ma \quad\Rightarrow\quad T = m(g + a)$

If the mass speeds up going upward, or slows down while going downward, a is positive and the rope pulls harder than the weight. If the mass speeds up going downward, a is negative and the rope pulls less. In free fall, a = −g and T = 0, so the rope goes slack.

Tension force formula: worked examples

Example 1: a lamp at rest

A 3.0 kg lamp hangs from a single cord. Find the tension.

$T = mg = 3.0 \times 9.8 = 29.4\ \text{N}$

Example 2: a lift cable

A 500 kg lift accelerates upward at 1.5 m/s². Find the cable tension. Then find it for a downward acceleration of 1.5 m/s².

Upward: $T = 500 \times (9.8 + 1.5) = 500 \times 11.3 = 5650\ \text{N}$ .

Downward: $T = 500 \times (9.8 - 1.5) = 500 \times 8.3 = 4150\ \text{N}$ .

The calculator gives 5650 N for m = 500, a = 1.5 and 4150 N for a = −1.5.

Example 3: two masses over a pulley

A 5.0 kg mass and a 3.0 kg mass hang on either side of a light, frictionless pulley. Find the acceleration and the tension.

$a = \frac{(m_1 - m_2)g}{m_1 + m_2} = \frac{(5.0 - 3.0) \times 9.8}{8.0} = 2.45\ \text{m/s}^2$

$T = \frac{2m_1 m_2 g}{m_1 + m_2} = \frac{2 \times 5.0 \times 3.0 \times 9.8}{8.0} = 36.75\ \text{N} \approx 36.8\ \text{N}$

Check with the lighter mass, which rises: T = 3.0 × (9.8 + 2.45) = 36.75 N. The two routes agree.

Tension in other set ups

Tension formulas by situation
Situation	Tension
Mass at rest on a vertical string	$T = mg$
Two blocks on a smooth floor, pulled by F on block 1, string to block 2	$T = \frac{m_2 F}{m_1 + m_2}$
Pendulum bob at the lowest point, speed v, length L	$T = mg + \frac{mv^2}{L}$
Mass hung from two strings, each at θ to the vertical	$T = \frac{mg}{2\cos\theta}$
Two masses over a pulley (Atwood machine)	$T = \frac{2m_1 m_2 g}{m_1 + m_2}$

The pendulum row shows why a swing's chains pull hardest at the bottom: they hold up the weight and also supply the centripetal force. The last but one row warns that a nearly horizontal washing line, with θ close to 90°, needs enormous tension to hold even a light shirt.

Ideal strings and what they assume

School problems assume a light, inextensible string over a light, frictionless pulley. Light means the string has no mass, so the tension is the same all along it. Inextensible means both ends move together with the same acceleration. A frictionless pulley changes the direction of the tension but not its size. If any of these fails, for example a heavy rope hanging under its own weight, the tension varies from point to point.

Tension shares its unit and dimensions with every other force; the dimensional formula of tension page shows the working.

Common tension slips

Assuming T = mg whenever something hangs. It is only true at zero acceleration.
Using the same sign of a for both masses on a pulley. One rises while the other falls.
Doubling the tension on a single string over a pulley. One string carries one tension.
Letting the tension come out negative. A string cannot push; a negative T means it has gone slack.

For every body in a pulley problem write its own ΣF = ma line. Two bodies, two equations, two unknowns: a and T.

More free body practice of this kind is gathered under the net force formula, with lifts and tug of war ropes.

Frequently asked questions

What is the formula of tension?

For a mass hanging still, T = mg. If the mass accelerates upward at a, T = m(g + a); downward, T = m(g − a).

Which unit is tension force measured in?

Newtons (N), the same as any other force.

Is tension the same throughout a string?

For a light string with no friction at any pulley, yes. A heavy rope or a rough pulley makes the tension change along the length.

Can tension be zero?

Yes. In free fall a string holding a mass carries no tension, and in a vertical circle the string can go slack at the top if the speed is too low.