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Dimensional Formula of Force

Dimensional formulas of physical quantitiesClass 11Units and measurementUpdated 1 Nov 2026

[F]=[M1L1T−2][F] = [M^{1} L^{1} T^{-2}]
SI unit
newton (N) = kg m/s²
Dimensions
[M¹ L¹ T⁻²]
where
  • M dimension of mass kg
  • L dimension of length m
  • T dimension of time s

The dimensional formula of force is [M¹ L¹ T⁻²], usually written [MLT⁻²]. It means force depends on mass to the power 1, length to the power 1 and time to the power minus 2, with no dependence on current, temperature or amount of substance.

You get it in one line from Newton's second law, F = ma, because mass has dimension [M] and acceleration has dimension [LT⁻²]. The same result explains the SI unit: one newton is one kilogram metre per second squared. Every formula on the force formula page must reduce to these dimensions.

Quick facts

  • Dimensional formula

    [M¹ L¹ T⁻²], short form [MLT⁻²]

  • Defining formula

    F=maF = ma

  • SI unit

    newton (N), equal to kg m s⁻²

  • CGS unit

    dyne, where 1 N = 10⁵ dyne

  • Vector or scalar

    Force is a vector, but its dimensions carry no direction

  • Same dimensions as

    Weight, tension, thrust, friction and every other kind of force

Derivation of the dimensional formula of force

Start from the definition and replace each quantity with its dimensions.

  1. Newton's second law: F=m×aF = m \times a.
  2. Mass is a base quantity: [m]=[M][m] = [M].
  3. Acceleration is velocity per unit time, and velocity is length per unit time, so [a]=[LT−1][T]=[LT−2][a] = \frac{[L T^{-1}]}{[T]} = [L T^{-2}].
  4. Multiply: [F]=[M]×[LT−2]=[M1L1T−2][F] = [M] \times [L T^{-2}] = [M^{1} L^{1} T^{-2}].

Any other correct force formula gives the same answer. Weight is W=mgW = mg, and g is an acceleration. Force as rate of change of momentum, F=ΔpΔtF = \frac{\Delta p}{\Delta t}, gives [MLT−1][T]\frac{[MLT^{-1}]}{[T]}, again [MLT⁻²]. The dimensional formula of acceleration page shows the middle step in more detail.

How the dimensions explain the newton

Each letter in the bracket maps to an SI base unit: M to the kilogram, L to the metre and T to the second. Reading [M¹ L¹ T⁻²] as units gives kg m s⁻², which is exactly one newton. That is how dimensions turn a long unit into a named one.

The same mapping converts between unit systems. In CGS the base units are the gram, centimetre and second, so the unit of force is g cm s⁻², called the dyne. Using the dimensions:

1 N=(1000 g)(100 cm)(1 s)−2=105 dyne1\ \text{N} = (1000\ \text{g})(100\ \text{cm})(1\ \text{s})^{-2} = 10^{5}\ \text{dyne}

The exponent on each base unit in the conversion is the exponent from the dimensional formula. That is the whole method.

Quantities with the same dimensions as force

Quantities sharing [M¹ L¹ T⁻²]
QuantityTypical formulaWhy it matches
WeightW=mgW = mgMass times an acceleration
TensionT=mgT = mg for a hanging loadA pulling force in a string
Frictionf=μNf = \mu Nμ has no dimensions, N is a force
ThrustRate of change of momentum of ejected gasMomentum per time
Rate of change of momentumΔpΔt\frac{\Delta p}{\Delta t}[MLT⁻¹] divided by [T]

Quantities that only look similar do not match. Pressure is force per area, [M¹ L⁻¹ T⁻²]. Work is force times distance, [M¹ L² T⁻²]. Momentum is [M¹ L¹ T⁻¹], one power of time short. Mixing these up is the most common error in a dimensional formula of force question.

Checking an equation with dimensions

Example: is F = mv²/r dimensionally correct?

Right side: [M][LT−1]2[L]=[ML2T−2][L]=[MLT−2]\frac{[M][LT^{-1}]^2}{[L]} = \frac{[M L^2 T^{-2}]}{[L]} = [M L T^{-2}].

Left side: [MLT⁻²]. Both sides match, so the centripetal force formula passes the check.

Example: is F = mv/t² correct?

Right side: [M][LT−1][T2]=[MLT−3]\frac{[M][LT^{-1}]}{[T^{2}]} = [M L T^{-3}], which is not [MLT⁻²]. The equation is wrong, and dimensions find the mistake without any numbers.

A dimensional check proves an equation is wrong when the sides differ, but a match does not prove it is right. It cannot catch a missing number such as the 1/2 in kinetic energy.

Exam slips with force dimensions

  • Writing [MLT²]. Time is in the denominator of acceleration, so its power is minus 2.
  • Using g as a dimension. g is an acceleration with dimensions [LT⁻²], not a base quantity.
  • Confusing force and pressure. Pressure divides by area, which adds L⁻².
  • Leaving out powers of 1 when the question asks for the full form. Some boards expect [M¹ L¹ T⁻²]; the short form [MLT⁻²] means the same thing.

For a full list of quantities and their brackets, use the dimensional formulas page and its dimension finder.

Frequently asked questions

What is the dimensional formula of force?

It is [M¹ L¹ T⁻²], often written [MLT⁻²]. It comes from F = ma, mass [M] times acceleration [LT⁻²].

What is the SI unit of force and how does it follow from the dimensions?

The SI unit is the newton. Replacing M, L and T with kilogram, metre and second gives kg m s⁻², and one newton is defined as exactly that.

Do force and weight have the same dimensional formula?

Yes. Weight is the force of gravity on a mass, W = mg, so it also has dimensions [MLT⁻²] and is measured in newtons.

What is the dimensional formula of force constant?

The spring force constant is force per extension, k = F/x, so its dimensions are [MLT⁻²] divided by [L], which is [M¹ L⁰ T⁻²].

Is the dimensional formula of force the same as impulse?

No. Impulse is force multiplied by time, so it has dimensions [M¹ L¹ T⁻¹], the same as momentum.