Net Force and Newton's Second Law Calculator

Give three forces with their directions and this adds them as vectors: the resultant’s magnitude, the angle it points in, and the acceleration it produces. F = ma is the last line of that, not the first — the marks are in the resolving, and in knowing that a resultant of zero means constant velocity and not necessarily rest.

Last updated 11 Nov 2026

Resolve three forces, then apply F = ma

Three forces at angles to one resultant

Force 1, magnitudeIn newtons. A magnitude, so keep it positive and put the direction in the angle field beside it. If the question gives a force in kgf or in grams-weight, multiply by g first: a kilogram-force is 9.806 65 N by the same 1901 convention that fixes standard gravity, and a mass is never a force.

…its direction, measured anticlockwise from the +x axisIn degrees. The convention here is the ordinary mathematical one: 0° points along +x (right), 90° along +y (up), 180° left, 270° down. Negative angles are accepted and mean the same as 360 minus themselves. The engine works in radians internally and converts for you, which is where this calculation most often goes wrong by hand — a calculator left in radian mode turns sin 30° into 0.5440 instead of 0.5 and the whole answer follows it.

Force 2, magnitudeIn newtons. Set it to zero to use the page with one force only.

…its directionIn degrees, same convention. For a force acting straight down — a weight — that is 270°, and for one acting straight up it is 90°.

Force 3, magnitudeIn newtons, and zero by default so the page starts as a two-force problem. The third slot is here because the commonest exam shape is not two forces but three: two applied and one weight, or two applied and a normal reaction.

…its directionIn degrees. Left at 270°, which is straight down, so that typing a weight into force 3 needs no change here.

Mass of the bodyIn kilograms, and only used for the acceleration rows. Grams are the usual slip: a 500 g trolley is 0.5 kg, and entering 500 makes the acceleration a thousand times too small.

45.8258NExample

50 N at 30°, 40 N at 150°, no third force, on a 5 kg body

Resolve, add the components, then recombine — and only then divide by the mass

F_x = ΣF cosθ · F_y = ΣF sinθ · |F| = √(F_x² + F_y²) · θ = atan2(F_y, F_x) · a = |F| / m

F, θ: each force as a magnitude and a direction. Forces add as vectors, which is why the resolving step exists at all: 50 N and 40 N can give anything from 10 N to 90 N depending on the angle between them
F_x, F_y: the components. These are signed, and the sign comes out of the cosine and sine on its own — a force at 150° has a negative x component without you deciding anything. Writing the signs in by hand on top of the trigonometry is how a resultant ends up on the wrong side
atan2: the two-argument arctangent, which is the reason this page gets the quadrant right. Plain atan(F_y/F_x) returns the same angle for (3, 4) and (−3, −4) and so puts half of all answers 180° out. atan2 looks at both signs
a = F/m: Newton’s second law, and the last line rather than the first. The F in it is the RESULTANT — the whole point of the three lines above. In its more general form the second law is F = dp/dt, force as the rate of change of momentum, which is the version that still works when the mass changes
N: the newton: kg m s⁻² in the SI Brochure’s Table 4. If an input arrives in kilogram-force, multiply by 9.806 65

Worked example

50 N at 30°, 40 N at 150°, no third force, on a 5 kg body
RESOLVE ALONG x. F_x = 50 cos 30° + 40 cos 150° = 43.3013 − 34.6410 = 8.6603 N. The second term is negative because 150° points up and to the LEFT; the cosine supplied that sign without being asked.
RESOLVE ALONG y. F_y = 50 sin 30° + 40 sin 150° = 25.0000 + 20.0000 = 45.0000 N. Both are positive: both forces point upwards, and the y components reinforce while the x components very nearly cancel.
RECOMBINE. |F| = √(8.6603² + 45.0000²) = √(75.00 + 2025.00) = 45.8258 N. Note what the arithmetic sum would have said: 50 + 40 = 90 N, which is almost exactly twice the truth. Adding magnitudes is the error this page exists to prevent.
DIRECTION. θ = atan2(45.0000, 8.6603) = 79.107° anticlockwise from +x — steeply up and slightly to the right, which is what you would sketch from two forces at 30° and 150° of unequal size. A single-argument arctangent would have given the same 79.107° here and would have been 180° wrong had both components been negative.
AND ONLY NOW THE SECOND LAW. a = 45.8258 / 5 = 9.1652 m/s², in the same direction as the resultant, 79.106°. Quote it to three significant figures as 9.17 m/s² unless the question asks otherwise: the inputs were given to two and three figures, and an answer carried to four claims a precision the question did not have.

Why the resolving is the whole exercise, and F = ma is one line at the end

F = ma is not the hard part and it is not where the marks are. A typical question gives you two or three forces pointing in different directions and asks for the acceleration. Every step that matters happens before the second law appears: choose axes, resolve each force onto them, add the components with their signs, recombine into a magnitude, and report a direction. Divide by the mass last. The reason this page takes three forces and an angle each, rather than one force and one mass, is that the one-force version is arithmetic you can do in your head and a search engine will do for you on its own results page. The three-force version is the exam question.

Signs come from the trigonometry, not from you. Measure every angle anticlockwise from the +x axis and the cosine and sine hand you the correct sign automatically: at 150° the cosine is negative and the sine is positive, so the force pulls left and pushes up, which is what 150° means. Students who decide the signs separately and then also take the trigonometry at face value end up subtracting twice. The second habit worth building is to use the two-argument arctangent for the direction. atan(F_y/F_x) cannot tell (3, 4) from (−3, −4), so it is right in two quadrants out of four; atan2 reads both signs and is right in all four.

A resultant of zero does not mean the body is at rest. This is the sentence that costs marks. Newton’s first law says a body with no resultant force on it keeps its velocity — which may be zero, and may equally be 30 m/s in a straight line. A lift descending at a steady speed has zero resultant force on it, and so does a lift parked at the ground floor. The CBSE Class XI list names “equilibrium of concurrent forces” as its own item for exactly this reason. When the page tells you the forces balance, the right next sentence is “so the velocity is unchanged”, not “so it does not move”.

What this page does not do. It adds the forces you give it and nothing else. It does not add the weight — type it in as a force at 270° if the body has one, which on a horizontal-plane problem it usually does, balanced by a normal reaction at 90°. It does not know about friction, which depends on that normal reaction and on a coefficient; the friction pages in this category take that on. It works in two dimensions, because school questions are two-dimensional. And it treats the mass as constant, which is the assumption hiding inside writing the second law as F = ma rather than as F = dp/dt. For a rocket losing fuel or a trolley being loaded as it moves, the momentum form is the correct one and the force is the rate of change of momentum.

Units, and the one conversion worth memorising. A newton is kg m s⁻²: the SI Brochure’s Table 4 says so, and it means a force expressed in kilograms is not a force. Papers and older textbooks sometimes give a load in kilogram-force or in grams-weight, and the conversion is 1 kgf = 9.806 65 N exactly — the same 1901 convention, by the same committee, that both fixed g_n and formally separated mass from weight.

Frequently asked questions

Can I just add the forces up?

Only if they act along the same line and you keep track of direction. 50 N and 40 N give 90 N when they point the same way, 10 N when they oppose, and 45.8 N at the angles in the worked example above — which is roughly half the arithmetic sum. The page prints that sum as a secondary row precisely so you can see how far from the answer it is. Forces are vectors; the sum of the magnitudes is only ever an upper bound.

Which way do I measure the angles?

Anticlockwise from the +x axis, which here means 0° right, 90° up, 180° left and 270° down. This is the mathematical convention and the one nearly every physics board uses. It is not universal: a bearing is measured clockwise from north, so a bearing of 060° is 30° in this page’s convention. If a question gives angles from the vertical, or as “20° above the horizontal, pointing left”, convert before typing — that one is 160°.

The resultant came out as zero. Does that mean nothing happens?

It means nothing CHANGES. Zero resultant force means zero acceleration, and zero acceleration means constant velocity — which is rest only if the body was already at rest. A parachutist at terminal velocity has no resultant force and is falling at 50 m/s. Write “the body continues at constant velocity” and you will never lose the mark; write “the body does not move” and you will lose it whenever the body was moving.

Do I need to include the weight?

If the question has gravity in it, yes, and you type it in yourself as a force of mg at 270°. This page adds the forces it is given and does not invent any. On a horizontal surface the weight is usually balanced by a normal reaction of the same size at 90°, so the two cancel and the net force is horizontal — but that cancellation is a result, not an assumption, and it fails the moment the surface tilts or something pushes down on the body as well.

Why does the page say F = dp/dt is the more general law?

Because F = ma assumes the mass is constant, and the momentum form does not. Newton’s second law as a statement about momentum — force is the rate of change of momentum — still holds for a rocket, a leaking tank or a conveyor being loaded. Cambridge 9702 asks for both: topic 3.1 wants you to “recall F = ma and solve problems using it” and, separately, to “define and use force as rate of change of momentum”. The impulse page in this category is the one that uses the second form.

Related calculators

References

Fundamental Physical Constants — Complete Listing, NIST Reference on Constants, Units and Uncertainty, 2022 CODATA adjustment (read 11 November 2026). The line this batch depends on is “standard acceleration of gravity 9.806 65 (exact) m s⁻²”. The word exact is the point: g_n is a conventional value, not a measured one, so it carries no uncertainty and it is not the g where you are standing. The same listing gives the Newtonian constant of gravitation as 6.674 30 × 10⁻¹¹ m³ kg⁻¹ s⁻² with an uncertainty of 0.000 15 × 10⁻¹¹, which is 22 parts per million — the worst-known of the constants a school paper quotes, and a useful reminder that “constant” and “exact” are different claims.
The International System of Units (SI), 9th edition, 2019, Bureau International des Poids et Mesures (read 11 November 2026). Table 4 gives the newton as kg m s⁻², which is the only definition of force this site uses: a newton is not a unit of mass and a kilogram is not a unit of force. Appendix 1 lists the 3rd CGPM of 1901, whose declaration both separated mass from weight and fixed the conventional value of g_n — the two things a force question most often conflates. Cited, not reproduced.
PHYSICS (042) Curriculum, Class XI–XII, Central Board of Secondary Education, session 2025–26 (read 11 November 2026). Unit III, Laws of Motion, reads in part: “Intuitive concept of force, Inertia, Newton’s first law of motion; momentum and Newton’s second law of motion; impulse”, “Equilibrium of concurrent forces, Static and kinetic friction, laws of friction, rolling friction, lubrication” and “Dynamics of uniform circular motion: Centripetal force, examples of circular motion (vehicle on a level circular road, vehicle on a banked road)”. Unit V, Motion of System of Particles and Rigid Body, reads in part: “Moment of a force, torque, angular momentum, law of conservation of angular momentum and its applications” and “Moment of inertia, radius of gyration, values of moments of inertia for simple geometrical objects (no derivation)”. Quoted to establish what is in scope, not to claim any endorsement by the board.
Cambridge International AS & A Level Physics 9702, syllabus for examination in 2025, 2026 and 2027, Cambridge Assessment International Education (read 11 November 2026). Two exclusions on this syllabus are worth more than most inclusions, because they explain why a question from one board looks unanswerable to a student taught on another. Topic 3.2 asks only for “a qualitative understanding of frictional forces and viscous/drag forces including air resistance” and states that “no treatment of the coefficients of friction and viscosity is required”. Topic 3.3 states that “knowledge of the concept of coefficient of restitution is not required”, while still requiring that for an elastic collision “the relative speed of approach is equal to the relative speed of separation”. Its data sheet gives the acceleration of free fall as 9.81 m s⁻².