Parallelogram of force

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The parallelogram of forces is a method for solving (or visualizing) the results of applying two forces to an object.

Figure 1: Parallelogram construction for adding vectors

When more than two forces are involved, the geometry is no longer parallelogrammatic, but the same principles apply. Forces, being vectors are observed to obey the laws of vector addition, and so the overall (resultant) force due to the application of a number of forces can be found geometrically by drawing vector arrows for each force. For example, see Figure 1. This construction has the same result as moving F2 so its tail coincides with the head of F1, and taking the net force as the vector joining the tail of F1 to the head of F2. This procedure can be repeated to add F3 to the resultant F1 + F2, and so forth.

Newton's proof[edit]

Figure 2: Parallelogram of velocity

Preliminary: the parallelogram of velocity[edit]

Suppose a particle moves at a uniform rate along a line from A to B (Figure 2) in a given time (say, one second), while in the same time, the line AB moves uniformly from its position at AB to a position at DC, remaining parallel to its original orientation throughout. Accounting for both motions, the particle traces the line AC. Because a displacement in a given time is a measure of velocity, the length of AB is a measure of the particle's velocity along AB, the length of AD is a measure of the line's velocity along AD, and the length of AC is a measure of the particle's velocity along AC. The particle's motion is the same as if it had moved with a single velocity along AC.[1]

Newton's proof of the parallelogram of force[edit]

Suppose two forces act on a particle at the origin (the "tails" of the vectors) of Figure 1. Let the lengths of the vectors F1 and F2 represent the velocities the two forces could produce in the particle by acting for a given time, and let the direction of each represent the direction in which they act. Each force acts independently and will produce its particular velocity whether the other force acts or not. At the end of the given time, the particle has both velocities. By the above proof, they are equivalent to a single velocity, Fnet. By Newton's second law, this vector is also a measure of the force which would produce that velocity, thus the two forces are equivalent to a single force.[2]

Using a parallelogram to add the forces acting on a particle on a smooth slope. We find, as we'd expect, that the resultant (double headed arrow) force acts down the slope, which will cause the particle to accelerate in that direction.

Bernoulli's proof for perpendicular vectors[edit]

We model forces as Euclidean vectors or members of . Our first assumption is that the resultant of two forces is in fact another force, so that for any two forces there is another force . Our final assumption is that the resultant of two forces doesn't change when rotated. If is any rotation (any orthogonal map for the usual vector space structure of with ), then for all forces

Consider two perpendicular forces of length and of length , with being the length of . Let and , where is the rotation between and , so . Under the invariance of the rotation, we get

Similarly, consider two more forces and . Let be the rotation from to : , which by inspection makes .

Applying these two equations

Since and both lie along , their lengths are equal

which implies that has length , which is the length of . Thus for the case where and are perpendicular, . However, when combining our two sets of auxiliary forces we used the associativity of . Using this additional assumption, we will form an additional proof below.[3] [4]

Algebraic proof of the parallelogram of force[edit]

We model forces as Euclidean vectors or members of . Our first assumption is that the resultant of two forces is in fact another force, so that for any two forces there is another force . We assume commutativity, as these are forces being applied concurrently, so the order shouldn't matter .

Consider the map

If is associative, then this map will be linear. Since it also sends to and to , it must also be the identity map. Thus must be equivalent to the normal vector addition operator.[3][5]


The mathematical proof of the parallelogram of force is not generally accepted to be mathematically valid. Various proofs were developed (chiefly Duchayla's and Poisson's), and these also caused objections. That the parallelogram of force was true was not questioned; why it was true. Today the parallelogram of force is accepted as an empirical fact, non-reducible to Newton's first principles. [3] [6]

See also[edit]


  1. ^ Routh, Edward John (1896). A Treatise on Analytical Statics. Cambridge University Press. p. 6. , at Google books
  2. ^ Routh (1896), p. 14
  3. ^ a b c Spivak, Michael (2010). Mechanics I. Physics for Mathematicians. Publish or Perish, Inc. pp. 278–282. ISBN 0-914098-32-2. 
  4. ^ Bernoulli, Daniel (1728). Examen principiorum mechanicae et demonstrationes geometricae de compositione et resolutione virium. 
  5. ^ Mach, Ernest (1974). The Science of Mechanics. Open Court Publishing Co. pp. 55–57. 
  6. ^ Lange, Marc (2009). "A Tale of Two Vectors" (PDF). Dialectica, 63. pp. 397–431.