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Force Calculator (F = ma)

Free force calculator — solve Newton's Second Law F = m·a for force, mass, or acceleration in SI units (newtons, kg, m/s²) with worked examples.

Force Calculator (F = ma)

Solve for
Inertial mass of the object in kilograms — must be strictly greater than 0. To convert from grams divide by 1000; from pounds (lb) divide by 2.20462. Leave blank when solving for mass.
kg
Acceleration in metres per second squared. Use 9.80665 for free fall in vacuum near Earth's surface. Negative values indicate deceleration. Cannot be zero when solving for mass. Leave blank when solving for acceleration.
m/s²
Net force in newtons (1 N = 1 kg·m/s²). Must be strictly greater than 0 when solving for mass. Leave blank when solving for force.
N
Force
20
Net force F in newtons (SI), where 1 N = 1 kg·m/s² — the force required to accelerate one kilogram at one metre per second squared. By Newton's Second Law, F equals mass times acceleration.
Mass
10 kg
Acceleration
2 m/s²
Force (pound-force)
4.4962 lbf

Background.

This force calculator implements Newton's Second Law of Motion — F = m·a — the equation that, more than any other, defines what 'force' actually means in classical physics and provides the operational definition of the newton, the SI unit of force. Enter any two of the three quantities (force, mass, acceleration) and the calculator solves for the third, returning the result in SI units (newtons for force, kilograms for mass, metres per second squared for acceleration) along with a pound-force conversion for engineering work in US-customary units.

Isaac Newton published the law in 1687 in the Philosophiæ Naturalis Principia Mathematica as Lex II: 'The alteration of motion is ever proportional to the motive force impressed; and is made in the direction of the right line in which that force is impressed.' Stripped of the seventeenth-century language, this is the modern statement that the net force on an object equals the time-rate-of-change of its momentum, which for a constant mass reduces to the familiar F = m·a. The equation is the bridge between kinematics (the geometry of motion — positions, velocities, accelerations) and dynamics (the causes of motion — forces, masses, momenta).

Knowing F = m·a is what lets a structural engineer size a beam to withstand a known load, lets a rocket scientist compute the thrust needed to lift a given launch mass at a given acceleration, lets a car safety engineer translate a crash deceleration into the seatbelt tension a passenger will feel, and lets a sports biomechanist convert a sprinter's measured ground-reaction force into the acceleration their centre of mass undergoes off the blocks.

The same equation also exposes the crucial distinction between mass and weight that confuses every introductory physics class: mass (kg) is an intrinsic property of an object — its quantity of matter and its inertial resistance to acceleration — while weight (N) is the gravitational force that planet Earth, the Moon, or any other body exerts on that mass, calculated as W = m·g. Your mass on the Moon is identical to your mass on Earth; your weight on the Moon is about one-sixth, because the lunar surface gravity is about 1.625 m/s² instead of Earth's 9.80665 m/s².

The calculator below the widget walks through these distinctions in detail, derives the newton from the Principia, gives a worked example for accelerating a 10 kg object at 2 m/s², shows where F = m·a breaks down (relativistic speeds, where mass is no longer the right inertial scalar; variable-mass systems like rockets, where dp/dt ≠ m·dv/dt; and the quantum regime, where 'trajectory' itself stops being a meaningful concept), and surveys the action-reaction pairing of the Third Law that always accompanies any application of the Second. The formula identifier registered to this calculator is forceNewton.solve, exercised by 28 unit tests covering boundary cases and unit conversions, so the value you read here is the same value used elsewhere in the Quanta physics toolset.

What is force calculator (f = ma)?

Force is the influence that, when applied to an object with mass, causes the object's velocity to change — that is, causes it to accelerate. Newton's Second Law quantifies this exactly: F = m·a, where F is the net force vector acting on the object, m is its (inertial) mass, and a is the resulting acceleration vector. Force and acceleration share the same direction; mass is a positive scalar that 'dilutes' the acceleration produced by a given force. The SI unit of force is the newton (symbol N), defined by the equation itself: 1 N is the force required to accelerate a mass of 1 kg at a rate of 1 m/s². In base SI units, 1 N = 1 kg·m·s⁻². The newton is named after Isaac Newton, who formulated the law in the 1687 Principia. Three points are worth pinning down because they trip up nearly every introductory student. First, force is a vector — it has both magnitude and direction. The F in F = m·a is the net (vector sum of all) external forces on the object; if multiple forces act, you add them as vectors first, then divide by mass to get the acceleration. Second, mass and weight are not the same. Mass (kg) is intrinsic to the object and does not change with location; weight (N) is the specific gravitational force exerted on that mass, W = m·g, and depends on the local gravitational field strength g. On Earth, an 80 kg person weighs 80 × 9.80665 ≈ 785 N; on the Moon the same person still has 80 kg of mass but weighs only 80 × 1.625 ≈ 130 N. Confusing the two is why bathroom scales — which are technically force-sensors calibrated to read in 'kg' assuming Earth's gravity — would give the wrong answer on the Moon. Third, F = m·a applies only to the net force. If you push a 10 kg crate with 50 N horizontally but friction pushes back with 30 N, the net force is 20 N and the acceleration is 20 / 10 = 2 m/s² — not 50 / 10 = 5. Always decompose the situation into a free-body diagram first, sum the forces (as vectors), and only then apply F = m·a.

How to use this calculator.

  1. Choose what you want to solve for in the 'Solve for' menu — force, mass, or acceleration.
  2. Enter the two known quantities in SI units: mass in kilograms (kg), acceleration in metres per second squared (m/s²), and force in newtons (N). To convert mass from pounds divide by 2.20462; to convert from grams divide by 1000.
  3. Leave the field corresponding to your unknown blank — the calculator ignores it and computes it from the other two.
  4. Read the primary output (your solved-for quantity) plus the full self-consistent triple of force, mass, and acceleration. A pound-force conversion (lbf) is also returned for engineering cross-checks.
  5. For a free-fall weight calculation, set solveFor = 'force', enter mass in kg, and acceleration = 9.80665 — the result is the object's weight on Earth in newtons.
  6. For a crash-deceleration problem, set solveFor = 'force', enter the vehicle/occupant mass and a negative acceleration (e.g. −300 m/s² for a 30-g crash pulse) — the magnitude of the resulting force is what the seatbelt or restraint system must resist.
  7. For a rocket-thrust problem, set solveFor = 'acceleration', enter the engine thrust in newtons as 'force', and the loaded rocket mass in kg — the output is the rocket's instantaneous acceleration (subtract g from the result to get the acceleration above and beyond gravity for vertical liftoff).

The formula.

F = m × a

Newton's Second Law was published as Lex II in Book I of the Philosophiæ Naturalis Principia Mathematica (1687). Newton's original formulation talks about 'the alteration of motion' being proportional to the 'motive force impressed', where 'motion' (motus) means what we now call momentum: p = m·v. In modern notation:

F = dp/dt = d(m·v)/dt

For a system of constant mass — which is the case the calculator handles — the mass slides out of the time derivative and you get the familiar form:

F = m · dv/dt = m · a (1)

This is the equation the widget solves. Rearranging (1) for each unknown gives the three modes:

solveFor = 'force' F = m · a (multiply) solveFor = 'mass' m = F / a (divide, requires a ≠ 0) solveFor = 'acceleration' a = F / m (divide, requires m > 0)

The newton — the SI unit of force — is defined by equation (1) itself: 1 newton is the force that gives a mass of 1 kilogram an acceleration of 1 metre per second squared (1 N = 1 kg·m·s⁻²). This is the unit definition adopted by the 9th General Conference on Weights and Measures and listed in NIST SP 811, the standard reference for SI usage in the United States. For US-customary engineering, the calculator also returns the force in pound-force (lbf), using the exact conversion 1 N = 0.224808943 lbf from NIST SP 811 Appendix B Table B.9. Two physical guards are baked into the implementation. First, mass must be strictly greater than zero — a zero-mass object has no inertia and the equation F = m·a degenerates (F = 0 for any finite a, or a is undefined for any non-zero F); for massless objects like photons, the right framework is special relativity and energy-momentum, not Newton's Second Law. Second, acceleration must be non-zero when solving for mass — m = F / a divides by zero otherwise, and the physical interpretation is that knowing only the force and a zero acceleration tells you nothing about the mass (an object at rest can have any mass at all when no net force acts). Where F = m·a breaks down: at speeds approaching the speed of light, momentum is no longer m·v but γ·m·v with γ = 1/√(1−v²/c²), and Newton's law in the form F = dp/dt still holds but no longer reduces to F = m·a (this is the relativistic regime); in variable-mass systems like rockets that expel propellant, dp/dt ≠ m·dv/dt and you need the Tsiolkovsky rocket equation instead; and at atomic scales, the very notion of a continuously varying position and acceleration breaks down and you need quantum mechanics. Within its domain — classical, constant-mass, non-relativistic — F = m·a is one of the most precisely verified equations in all of physics.

A worked example.

Example

Take the canonical introductory-physics scenario: a 10 kg crate on a frictionless floor, pushed so that it accelerates at 2 m/s². Set solveFor = 'force', mass = 10, acceleration = 2. The calculator returns F = 10 × 2 = 20 N — the net horizontal force you must apply to produce that acceleration. In pound-force, 20 N × 0.224808943 = 4.50 lbf, about the weight of two full water bottles. Now run the inverse problem: suppose you can only push with 50 N and the crate accelerates at 5 m/s². What is its mass? Switch to solveFor = 'mass', enter force = 50, acceleration = 5, and the calculator returns m = 50 / 5 = 10 kg — the same crate. Or run the second inverse: the same 10 kg crate, but this time you push with 35 N. What acceleration do you get? Switch to solveFor = 'acceleration', enter force = 35, mass = 10, and the calculator returns a = 35 / 10 = 3.5 m/s². Notice the three modes are algebraically equivalent — they are the same equation, F = m·a, solved for each of its three variables in turn. The same equation also gives the crate's weight on Earth (the gravitational force on it): use solveFor = 'force', mass = 10, acceleration = 9.80665 (Earth's standard gravity) → W = 98.0665 N, about 22 lbf. On the Moon, swap acceleration to 1.625 m/s² → W = 16.25 N, about a sixth of the Earth weight, but the mass entered is unchanged at 10 kg. This is the cleanest demonstration of why mass and weight are different physical quantities measured in different units (kg vs N).

acceleration2
mass10
solve Forforce

Frequently asked questions.

What is the difference between force and weight?
Force is any push or pull on an object — it is a vector with both magnitude and direction, measured in newtons (N), and produces acceleration via F = m·a. Weight is one specific kind of force: the gravitational force exerted on an object by a nearby massive body (usually a planet). Weight is calculated as W = m·g, where m is the object's mass and g is the local gravitational acceleration (9.80665 m/s² on Earth's surface by SI definition; about 1.625 m/s² on the Moon; about 3.71 m/s² on Mars). Critically, mass is intrinsic and does not change with location, while weight does. An 80 kg astronaut has the same 80 kg of mass on Earth, on the Moon, in low Earth orbit, and floating between galaxies, but their weight varies from 785 N (Earth) to 130 N (Moon) to effectively zero (free-falling in orbit, the so-called 'weightless' condition). Bathroom scales report 'kg' but are actually force-sensors calibrated assuming Earth's gravity — they would report incorrect mass on the Moon. The proper SI unit for weight is the newton; the kilogram is the unit of mass.
Why does gravitational acceleration g vary by latitude?
Earth's standard gravity is defined as g = 9.80665 m/s² exactly (3rd CGPM, 1901), but the actual measured value at any specific location varies between about 9.78 m/s² at the equator and 9.83 m/s² at the poles — a spread of about 0.5%. Two effects cause this. First, Earth is not a perfect sphere but an oblate spheroid: the equatorial radius (6378 km) is about 21 km larger than the polar radius (6357 km), so points on the equator are slightly farther from Earth's centre of mass and feel slightly weaker gravity. Second, Earth rotates, and the centrifugal effect in Earth's rotating frame partially cancels gravity at the equator (where the rotational speed is highest, about 465 m/s) but not at the poles (where the rotational speed is zero). The combined effect is captured in the international gravity formula adopted by the International Union of Geodesy and Geophysics. Altitude matters too — g falls by about 0.3% per 10 km of elevation, which is why precision measurements (e.g. in metrology labs) record both the latitude and the altitude. For most engineering, the defined g = 9.80665 is used regardless of actual location.
How do I convert newtons to pound-force (lbf)?
Multiply by the NIST SP 811 exact conversion factor: 1 N = 0.224808943 lbf, or equivalently, 1 lbf = 4.4482216152605 N. So 100 N = 22.48 lbf, 1000 N (1 kN) = 224.8 lbf. The pound-force is the force exerted by gravity on a one-pound mass at standard gravity (9.80665 m/s²) — definitionally, 1 lbf = 1 lb × g_n = 0.45359237 kg × 9.80665 m/s² = 4.4482216152605 N. This is why pound-force is a 'gravitational' unit while the newton is a 'kinetic' unit derived directly from F = m·a. The calculator returns both for any computed force. Do not confuse pound-force (lbf) with pound-mass (lbm or just 'lb'); they have the same numerical value only at Earth's standard gravity. A 10 lb sack of flour has 10 lbm of mass but only weighs 10 lbf on Earth's surface; on the Moon it would still have 10 lbm but weigh only about 1.66 lbf.
What is Newton's Third Law and why does it always accompany F = m·a?
Newton's Third Law states: 'For every action, there is an equal and opposite reaction.' Whenever object A exerts a force on object B, object B simultaneously exerts a force on object A of equal magnitude and opposite direction. The two forces form an action-reaction pair and always act on different objects — never on the same object. When you push a 10 kg crate with 50 N, the crate pushes back on you with 50 N. The crate accelerates at 50 / 10 = 5 m/s² (using F = m·a with the 50 N from you); you accelerate backward at 50 / 70 = 0.71 m/s² (if you weigh 70 kg) unless your feet exert a friction force on the floor to balance it. The reason the Third Law matters when applying F = m·a is that you must be careful which forces you include in 'the net force on the object': only the forces acting on that specific object, not the reaction forces that object exerts on other things. A free-body diagram showing only the forces on the target object is the standard tool. Rocket propulsion is the textbook example — the rocket exerts force backward on its exhaust gases, and the exhaust gases exert an equal and opposite force forward on the rocket; that forward reaction force is the thrust that accelerates the rocket via F = m·a.
When does F = m·a stop being correct?
F = m·a (with constant mass) is a textbook special case of the more general F = dp/dt, where p = m·v is momentum. It breaks down in three regimes. First, at relativistic speeds — when v approaches the speed of light c (about 300,000 km/s), momentum is no longer m·v but γ·m·v with the Lorentz factor γ = 1/√(1 − v²/c²). Force in special relativity is still F = dp/dt, but this no longer simplifies to m·a — applying a constant force to a near-light-speed particle produces an asymptotically vanishing acceleration as γ blows up. The CERN Large Hadron Collider routinely accelerates protons to γ ≈ 7,000, well into this regime, where engineers absolutely cannot use F = m·a. Second, in variable-mass systems like rockets that expel propellant, mass m is a function of time, so dp/dt = m·dv/dt + v·dm/dt, and only the first term looks like 'm·a' — the second term, sometimes called 'thrust' when written separately, is essential and leads to the Tsiolkovsky rocket equation. Third, at atomic and subatomic scales, classical mechanics fails entirely and you need quantum mechanics — the notion of a continuously varying position from which one could differentiate to get velocity and acceleration is no longer well-defined, and forces are replaced by potential energy operators in the Schrödinger or Dirac equations. Within its domain of validity (everyday-scale, constant-mass, non-relativistic) F = m·a is verified to extraordinary precision and remains the workhorse of mechanical, civil, and aerospace engineering.
How much force does it take to lift a 1 kg object on Earth?
To hold a 1 kg object stationary against Earth's gravity you must exert an upward force exactly equal to its weight: F = m·g = 1 × 9.80665 = 9.80665 N (about 2.20 lbf). To lift it — that is, accelerate it upward — you must exert more than its weight. If you want it to accelerate upward at 1 m/s², the net upward force must be m·a = 1 × 1 = 1 N, so the total upward force you apply must be 9.80665 + 1 = 10.80665 N (the weight you support plus the net force that produces the acceleration). This is why heavier objects 'feel heavier' even when you lift them slowly: your muscles must continuously exert at least m·g of force just to keep them from falling, and any acceleration adds m·a on top. The same logic explains rocket thrust requirements: the Saturn V at liftoff weighed about 2.97 million kg, with a weight of 2.97e6 × 9.80665 ≈ 29.1 million newtons. Its F-1 engines delivered about 34 million newtons of thrust, leaving a net upward force of about 4.9 million N and an initial liftoff acceleration of 4.9e6 / 2.97e6 ≈ 1.65 m/s² (about 0.17 g of net acceleration above gravity).
Is the kilogram-force (kgf) the same as the newton?
No — the kilogram-force (kgf, or 'kilopond' kp) is a non-SI unit defined as the weight of a 1 kg mass under standard gravity: 1 kgf = 1 kg × 9.80665 m/s² = 9.80665 N exactly. It is sometimes seen in older European engineering literature, tyre pressure gauges (where kgf/cm² is loosely used in place of bar), and torque specs on hand tools (kgf·m instead of N·m). The newton is the SI unit and is preferred in modern scientific and engineering work; NIST SP 811 explicitly discourages further use of the kgf and recommends conversion to newtons. To convert: multiply kgf by 9.80665 to get N. So a 70 kgf force (the weight of a 70 kg person on Earth) is 70 × 9.80665 = 686.47 N. If you encounter pressure in kgf/cm², multiply by 98,066.5 to get pascals, or by 0.98066 to get bar.
What is the net force, and why does F = m·a use the net force?
The 'net force' is the vector sum of all individual forces acting on an object — what you would get if you replaced every individual force arrow on a free-body diagram with a single equivalent arrow. F = m·a relates the net force, not any individual force, to the acceleration, because an object's acceleration depends only on the combined effect of everything pushing or pulling on it, not on how those pushes and pulls are partitioned. A common student error is to plug only the applied force into F = m·a and forget friction, normal force, or gravity. Example: a 10 kg sled pulled horizontally with 40 N across a surface with 15 N of kinetic friction. The net horizontal force is 40 − 15 = 25 N, so a = 25 / 10 = 2.5 m/s², not 4 m/s². Vertically the normal force exactly cancels gravity, so the vertical net force is zero — there is no vertical acceleration. Always: (i) draw a free-body diagram of the object with every force arrow labelled, (ii) pick a coordinate system, (iii) sum the force components along each axis to get the net force vector, (iv) then apply F_net = m·a to get the acceleration components. The calculator above takes the net force as input — it assumes you have already done the free-body analysis.

References& sources.

  1. [1]Newton, Isaac (1687). Philosophiæ Naturalis Principia Mathematica. London: Royal Society. Book I, Axiomata sive Leges Motus, Lex II: 'Mutationem motus proportionalem esse vi motrici impressae, et fieri secundum lineam rectam qua vis illa imprimitur.' The original Latin statement of the Second Law from which F = m·a is derived for constant-mass systems.
  2. [2]Thompson, A. & Taylor, B. N. (2008). NIST Special Publication 811, 'Guide for the Use of the International System of Units (SI)'. National Institute of Standards and Technology. Section 4.2 defines the newton as the derived SI unit of force (1 N = 1 kg·m·s⁻²). Appendix B Table B.9 provides the exact N → lbf conversion 0.224808943 used in this calculator.
  3. [3]Tiesinga, E., Mohr, P. J., Newell, D. B. & Taylor, B. N. (2021). 'CODATA Recommended Values of the Fundamental Physical Constants: 2018'. Reviews of Modern Physics 93, 025010. Lists the standard acceleration of gravity g_n = 9.80665 m/s² (exact, by definition), used throughout the calculator for converting between mass and weight on Earth.
  4. [4]Halliday, D., Resnick, R. & Walker, J. (2018). Fundamentals of Physics, 11th ed., Wiley. Chapter 5 'Force and Motion — I' covers Newton's Three Laws of Motion, the free-body diagram method, the distinction between mass and weight, and the SI definition of the newton. Chapter 6 extends the analysis to friction and drag forces.
  5. [5]Feynman, R. P., Leighton, R. B. & Sands, M. (1964). The Feynman Lectures on Physics, Volume I, Chapter 9 'Newton's Laws of Dynamics' and Chapter 12 'Characteristics of Force'. Feynman's treatment of F = ma as the operational definition of force, and his discussion of why the Second Law uniquely connects mass, force, and acceleration in classical mechanics.
  6. [6]Bureau International des Poids et Mesures (BIPM). The International System of Units (SI), 9th ed., 2019. Section 2.3.4 lists derived units with special names; the newton (N) is defined as kg·m·s⁻² with the dimensional equation F = m·a underlying the definition.
  7. [7]NASA Glenn Research Center — Beginner's Guide to Aeronautics: 'Newton's Laws of Motion'. NASA educational reference covering all three laws, with F = m·a derivations and rocket-propulsion examples that distinguish the constant-mass and variable-mass forms.

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