The parallelogram law of forces states that if two forces acting at a point are represented in magnitude and direction by the two adjacent sides of a parallelogram drawn from that point, their resultant is represented in magnitude and direction by the diagonal of the parallelogram drawn from the same point. For two forces P and Q with an angle θ between them, the resultant is R = √(P² + Q² + 2PQ cos θ), and it acts at an angle α to P given by tan α = Q sin θ / (P + Q cos θ).
Throughout this page, θ is the angle between the two forces, measured at the point where they both act. That single convention decides every sign in the working, and getting it wrong is the most common source of a wrong answer.
Statement of the parallelogram law of forces
Two forces acting at the same point can always be replaced by one force that has the same effect. Finding that single force is called composition of forces, and the parallelogram law is the rule for doing it.
Draw the two forces P and Q to a chosen scale as the sides OA and OB of a parallelogram, both starting from the common point O and pointing along their own lines of action. Complete the parallelogram OACB. The diagonal OC, measured to the same scale, is the resultant R. Its length gives the magnitude and its direction gives the line along which the single equivalent force acts.
The law applies to any two concurrent forces, meaning forces whose lines of action meet at a point. It is not restricted to forces: the same construction adds velocities, accelerations, momenta and any other vector quantity, which is why it is also taught as the parallelogram law of vector addition.
Derivation of the resultant formula
Let P act along OA and Q act along OB, with θ as the angle AOB between them. Complete the parallelogram OACB, so that AC is parallel to OB and equal in length to Q. Extend OA and drop a perpendicular from C onto that extension, meeting it at D.
Because AC is parallel to OB, the angle CAD equals θ. From the right-angled triangle ACD:
- CD = AC sin θ = Q sin θ
- AD = AC cos θ = Q cos θ
The whole base is therefore OD = OA + AD = P + Q cos θ.
Now apply Pythagoras to the right-angled triangle OCD, where OC = R:
R² = OD² + CD² = (P + Q cos θ)² + (Q sin θ)²
R² = P² + 2PQ cos θ + Q² cos²θ + Q² sin²θ
Since cos²θ + sin²θ = 1, the two Q² terms collapse into one:
R² = P² + Q² + 2PQ cos θ, so R = √(P² + Q² + 2PQ cos θ)
Direction of the resultant
Let α be the angle between the resultant R and the force P. From the same triangle OCD:
tan α = CD / OD, so tan α = Q sin θ / (P + Q cos θ)
By symmetry, the angle β between R and Q is given by tan β = P sin θ / (Q + P cos θ), and α + β = θ. That last relation is a quick check on any answer: if your two angles do not add up to the angle between the forces, one of them is wrong.
What the formula tells you before you put numbers in
| Angle θ between the forces | cos θ | Resultant R | Direction |
|---|---|---|---|
| 0° (same direction) | +1 | P + Q, the maximum possible | Along both forces, α = 0° |
| 60° | +0.5 | √(P² + Q² + PQ) | Between the two forces, nearer the bigger one |
| 90° (perpendicular) | 0 | √(P² + Q²) | tan α = Q / P |
| 120° | −0.5 | √(P² + Q² − PQ) | Between the two forces |
| 180° (opposite) | −1 | |P − Q|, the minimum possible | Along the larger force |
So the resultant of two forces always lies between |P − Q| and P + Q. If a calculation gives you a resultant of 130 N from forces of 60 N and 40 N, the answer is wrong before you check anything else, because 100 N is the ceiling.
Two special results worth memorising. For two equal forces P at an angle θ, R = 2P cos(θ/2) and the resultant bisects the angle, α = θ/2. And two equal forces at 120° have a resultant equal to P itself, which is the geometry behind three-legged sling arrangements.
Solved examples on the parallelogram law of forces
Example 1: two forces at 60°
Forces of P = 60 N and Q = 40 N act at a point with an angle of 60° between them. Find the magnitude and direction of the resultant.
Magnitude:
R² = 60² + 40² + 2(60)(40) cos 60° = 3600 + 1600 + 4800(0.5) = 3600 + 1600 + 2400 = 7600
R = √7600 = 87.18 N
Direction:
tan α = 40 sin 60° / (60 + 40 cos 60°) = 34.641 / (60 + 20) = 34.641 / 80 = 0.43301
α = 23.41° from the 60 N force, measured towards the 40 N force.
Check by components. Along the 60 N force: 60 + 40 cos 60° = 80 N. At right angles to it: 40 sin 60° = 34.64 N. Then R = √(80² + 34.64²) = √(6400 + 1200) = √7600 = 87.18 N, and tan α = 34.64 / 80, the same as before. The result also sits comfortably between the limits of 20 N and 100 N.
Example 2: two equal forces
Two forces of 50 N each act at a point with 120° between them. Find the resultant.
Using R = 2P cos(θ/2): R = 2 × 50 × cos 60° = 2 × 50 × 0.5 = 50 N, acting along the bisector at 60° from each force.
Confirming with the general formula: R² = 2500 + 2500 + 2(50)(50)(−0.5) = 5000 − 2500 = 2500, so R = 50 N. Two 50 N forces at 120° produce exactly 50 N, not 100 N and not zero. Turning the two forces further apart, to 180°, would cancel them completely.
Example 3: finding the angle when the resultant is known
Two forces of 5 N and 12 N acting at a point have a resultant of 13 N. What is the angle between them?
13² = 5² + 12² + 2(5)(12) cos θ
169 = 25 + 144 + 120 cos θ = 169 + 120 cos θ
120 cos θ = 0, so cos θ = 0 and θ = 90°. The 5-12-13 triple is the signature of perpendicular forces, and the resultant lies at tan α = 5 / 12, that is 22.62° from the 12 N force.
Triangle law and polygon law of forces
Triangle law of forces
If two forces acting at a point are represented in magnitude and direction by the two sides of a triangle taken in order, their resultant is represented in magnitude and direction by the third side of the triangle taken in the opposite order.
This is the same law with half the drawing. Instead of completing a parallelogram, shift Q so that its tail sits on the head of P. The line closing the triangle from the tail of P to the head of Q is the resultant. Since the shifted side is just AC in the parallelogram construction, the algebra is identical and gives the same R = √(P² + Q² + 2PQ cos θ), with θ still the angle between the forces themselves, not the interior angle of the triangle. The interior angle at the joint of the triangle is (180° − θ), which is where sign errors creep in when students switch between the two constructions.
Polygon law of forces
If any number of concurrent forces are represented in magnitude and direction by the sides of a polygon taken in order, their resultant is represented by the closing side of the polygon taken in the opposite order.
The polygon law is nothing more than the parallelogram law applied repeatedly. Combine the first two forces, then combine that resultant with the third, and so on. If the polygon closes on itself, the resultant is zero and the body is in equilibrium.
Worked check: three forces by two methods
Take three concurrent forces: 40 N along the x-axis, 30 N at 90°, and 20 N at 210°.
Step by step, using the parallelogram law twice. Combining 40 N and 30 N at 90° gives R₁ = √(1600 + 900) = 50 N at tan⁻¹(30/40) = 36.87° from the x-axis. The angle between R₁ and the 20 N force is 210° − 36.87° = 173.13°. Applying the law again:
R² = 50² + 20² + 2(50)(20) cos 173.13° = 2500 + 400 + 2000(−0.99283) = 2900 − 1985.7 = 914.3
R = 30.24 N, and tan α = 20 sin 173.13° / (50 + 20 cos 173.13°) = 2.392 / 30.143 = 0.0794, so α = 4.54°, putting the resultant at 36.87° + 4.54° = 41.4° from the x-axis.
All at once, by resolution. ΣFx = 40 + 0 + 20 cos 210° = 40 − 17.32 = 22.68 N. ΣFy = 0 + 30 + 20 sin 210° = 30 − 10 = 20 N. So R = √(22.68² + 20²) = √(514.4 + 400) = √914.4 = 30.24 N at tan⁻¹(20 / 22.68) = 41.4°.
Both routes land on the same answer, which is the point of the polygon law. In practice nobody adds four or five forces two at a time; they resolve everything into x and y components and add the columns. Resolution is faster, but it is the parallelogram law underneath.
Resolution of a force: the inverse operation
Resolution is composition run backwards. Instead of replacing two forces by one, you replace one force by two components along chosen directions, normally at right angles. A force F making an angle φ with the x-axis resolves into:
- Fx = F cos φ
- Fy = F sin φ
These are just the two sides of a parallelogram whose diagonal is F, drawn with a right angle at the corner.
Example. A tow rope pulls a crate with 200 N at 35° above the horizontal. The useful horizontal pull is 200 cos 35° = 163.8 N, and the vertical part, 200 sin 35° = 114.7 N, lifts the crate slightly and reduces the friction under it rather than moving it forward. Checking: √(163.8² + 114.7²) = √(26831 + 13156) = √39987 ≈ 200 N, so the components rebuild the original force.
Example on an incline. A block weighing 800 N rests on a 20° slope. Resolving its weight along and perpendicular to the surface gives a component down the slope of 800 sin 20° = 273.6 N and a component pressing into the slope of 800 cos 20° = 751.8 N. The first drives sliding; the second sets the normal reaction and therefore the available friction. The axes here are not horizontal and vertical, and that is the whole trick: you resolve along whatever pair of perpendicular directions makes the problem simple.
Experimental verification on Gravesand’s apparatus
Gravesand’s apparatus is the standard school and first-year laboratory check on the parallelogram law. It is a vertical drawing board with two smooth pulleys clamped at its upper corners.
- Fix a sheet of white paper to the board. Tie three strings to a common knot O.
- Pass two of the strings over the pulleys and hang slotted weights P and Q on them. Hang a third weight R from the string that dangles freely between the pulleys.
- Disturb the knot gently and let it settle, so friction at the pulleys is not holding it in a false position. Repeat once or twice to confirm the knot returns to the same place.
- Mark two points under each string with a sharp pencil, well apart, and mark the position of the knot O.
- Remove the paper, join the marks to get the three lines of action, and pick a scale such as 1 cm = 50 gram-weight.
- From O, lay off OA along the first string equal to P and OB along the second equal to Q. Complete the parallelogram and draw the diagonal OC.
- Measure OC. It should equal the third weight R to scale, and it should lie exactly opposite the third string.
The reasoning is that the knot is in equilibrium under three forces. The resultant of P and Q must therefore be equal in magnitude to R and opposite in direction, so the diagonal of the parallelogram should mirror the third string. A worked instance: with P = 200 gram-weight, Q = 250 gram-weight and a measured angle of 70° between them, the predicted resultant is √(200² + 250² + 2 × 200 × 250 × cos 70°) = √(40000 + 62500 + 34202) = √136702 = 369.7 gram-weight, which in SI is about 3.63 N. The third weight should read close to 370 gram-weight.
A small gap between the predicted and the measured value is normal. The gaps come from pulley friction, the weight of the strings, the board not being truly vertical, and the thickness of the pencil marks. Because all three forces meet at one point and the system is in equilibrium, the same setup also verifies Lami’s theorem, which relates each of three concurrent forces in equilibrium to the sine of the angle between the other two, and is the faster route when you already know the system is balanced.
Where the parallelogram law is used
- Structural analysis. Resolving member forces at a joint of a truss, and combining wind and dead loads into a single design force.
- Lifting and rigging. Working out the tension in each leg of a two-leg sling. The wider the angle between the legs, the larger each leg tension is for the same load, which is why sling angle limits appear on lifting charts.
- Navigation and flight. Adding a boat’s own velocity to the current, or an aircraft’s airspeed to the wind, to get the track over the ground.
- Machine design. Combining radial and tangential gear tooth loads into a single bearing load.
- Everyday mechanics. Two people pulling a stuck vehicle at an angle, a kite held by string tension and wind pressure, or a cable-stayed sign carrying its own weight and a wind load at once.
Common mistakes
- Using the wrong θ. The formula needs the angle between the two forces at their common point. If a question gives the angle between one force and the diagonal, or the interior angle of the force triangle (180° − θ), convert it first.
- Adding magnitudes directly. 60 N and 40 N give 100 N only when they point the same way. At 60° the answer is 87.18 N.
- Dropping the 2PQ cos θ term. Writing R = √(P² + Q²) is correct only at 90°.
- Applying the law to forces that are not concurrent. Two parallel forces on a beam do not meet at a point, so this law does not apply directly; there the turning effect matters too. That is the subject of the page on the moment of a force, which covers moments, couples and the principle of moments.
- Quoting an angle without saying what it is measured from. Always write “23.41° from the 60 N force”, not just “23.41°”.
- Forgetting the calculator mode. Degrees, not radians. A cos 60° that returns −0.952 means the calculator is in radian mode.
References
- NCERT Physics – Motion in a Plane (Vectors), Class 11, Part 1.
- NPTEL – Engineering Mechanics lecture series, IIT.
FAQs
What is the parallelogram law of forces?
The parallelogram law of forces states that if two forces acting at a point are represented in magnitude and direction by the two adjacent sides of a parallelogram drawn from that point, then their resultant is represented in magnitude and direction by the diagonal of the parallelogram passing through that same point. It applies to any two concurrent forces, and to any other vector quantity such as velocity or acceleration.
What is the formula for the resultant of two forces?
For two forces P and Q acting at a point with an angle θ between them, the resultant is R = √(P² + Q² + 2PQ cos θ). The resultant acts at an angle α to P, where tan α = Q sin θ / (P + Q cos θ). The resultant is largest, P + Q, when θ = 0°, and smallest, |P − Q|, when θ = 180°.
How is the parallelogram law derived?
Draw P as OA and Q as OB with angle θ between them, complete the parallelogram OACB, extend OA and drop a perpendicular from C to meet it at D. Then CD = Q sin θ and AD = Q cos θ, so OD = P + Q cos θ. Pythagoras on triangle OCD gives R² = (P + Q cos θ)² + (Q sin θ)², which simplifies to R² = P² + Q² + 2PQ cos θ because cos²θ + sin²θ = 1. Dividing CD by OD gives tan α.
What is the difference between the parallelogram law and the triangle law of forces?
They give the same result from different drawings. The parallelogram law places both forces tail to tail from a common point and takes the diagonal as the resultant. The triangle law places them head to tail and takes the closing side, reversed, as the resultant. The triangle law extends naturally to any number of forces as the polygon law, while the parallelogram construction handles two at a time.
Can the parallelogram law be used for three or more forces?
Yes, but only two at a time. Combine the first two into a resultant, then combine that resultant with the third force, and continue. This repeated application is the polygon law of forces. In practice it is quicker to resolve every force into x and y components, add each column and recombine, which is the same mathematics in a faster form.