A boatman in Kolkata rows straight across the Hooghly at \(4\,\text{m/s}\) relative to the water. The river flows at \(3\,\text{m/s}\). The boat does not reach the point opposite. It lands downstream, and a person on the bank sees it move faster than the boatman rows. The two velocities have combined into a third, with its own size and direction.
The triangle law
Two displacements show the rule most clearly. Walk along \(\vec A\), and then walk along \(\vec B\) from where you stopped. The single displacement from the first starting point to the last end point is the sum \(\vec A + \vec B\). On paper, draw \(\vec A\), then draw \(\vec B\) with its tail at the tip of \(\vec A\). The arrow from the tail of \(\vec A\) to the tip of \(\vec B\) is the resultant \(\vec R\). Every vector adds this way, velocities and forces included.
Let two vectors be drawn tip to tail, as two sides of a triangle taken in order. Their resultant is the third side of the triangle, drawn from the tail of the first vector to the tip of the second.
The size of the resultant depends on the angle between the two vectors. That angle \(\theta\) is measured with both tails together. When \(\theta = 0°\) the vectors point the same way and the magnitudes add. When \(\theta = 180°\) they point opposite ways and the magnitudes subtract. In between, the resultant closes a triangle, and its length lies between these two limits.
The magnitude and the direction of the resultant
The figure can be turned into a formula. Put \(\vec A\) along a line and split \(\vec B\) into a part along \(\vec A\) and a part across it. The two parts are \(B\cos\theta\) and \(B\sin\theta\). The resultant is then the hypotenuse of a right triangle. The derivation below does this in five short steps.
\[ R_{along} = A + B\cos\theta \]
Why. along the line of \(\vec A\), the resultant covers all of \(A\) and the part \(B\cos\theta\) of \(\vec B\).
\[ R_{across} = B\sin\theta \]
Why. across that line, only \(\vec B\) contributes, by its part \(B\sin\theta\).
\[ R^2 = (A + B\cos\theta)^2 + (B\sin\theta)^2 \]
Why. the two parts are at right angles, so the theorem of Pythagoras gives the length of the resultant.
\[ R^2 = A^2 + 2AB\cos\theta + B^2(\cos^2\theta + \sin^2\theta) \]
Why. expand the first bracket and collect the two terms that contain \(B^2\).
\[ R^2 = A^2 + B^2 + 2AB\cos\theta \]
Read it. \(\cos^2\theta + \sin^2\theta = 1\) for every angle.
The formula contains the special cases. At \(\theta = 90°\), \(\cos\theta = 0\) and \(R = \sqrt{A^2 + B^2}\), the theorem of Pythagoras. Two equal vectors at \(120°\) give \(R^2 = 2A^2(1 + \cos 120°) = A^2\), so the resultant equals each of them. And for any angle, \(|A - B| \le R \le A + B\). Forces of \(8\,\text{N}\) and \(6\,\text{N}\) can therefore add to anything from \(2\,\text{N}\) to \(14\,\text{N}\), and to nothing outside that range.
The angle in the formula is the angle between the two vectors with their tails together. Inside the triangle, the angle at the joint of the two arrows is \(180° - \theta\), not \(\theta\). The cosine rule of trigonometry, \(R^2 = A^2 + B^2 - 2AB\cos(180° - \theta)\), uses that inside angle and gives the same result. Mixing the two angles changes the plus sign to a minus and gives a wrong answer. The rule also adds only quantities of one kind: a velocity and a force have no sum.
Two forces of \(3\,\text{N}\) and \(4\,\text{N}\) act on a body, with an angle of \(60°\) between them. Find the magnitude of their resultant, in N.
Two forces, each of magnitude \(F\), act at a point with an angle of \(120°\) between them. What is the magnitude of their resultant?
The parallelogram law
Forces on one body are often drawn from one point, tail to tail. The parallelogram law adds them without moving them. Complete the parallelogram with the two vectors as its sides. The diagonal drawn from their common tail is the resultant. This is the triangle law again, because the far side of the parallelogram is \(\vec B\) slid to the tip of \(\vec A\).
The parallelogram also shows that the order of addition does not matter. Going along \(\vec A\) then \(\vec B\), or along \(\vec B\) then \(\vec A\), reaches the same far corner. So \(\vec A + \vec B = \vec B + \vec A\). Grouping does not matter either: \((\vec A + \vec B) + \vec C = \vec A + (\vec B + \vec C)\).
The polygon law
Any number of vectors add in the same way. Draw them tip to tail in any order, and the resultant runs from the first tail to the last tip. If the last tip falls exactly on the first tail, the polygon is closed and the resultant is zero. A body at rest under several forces is exactly such a case. The forces on it, drawn tip to tail, must close.
A knot is held at rest by three strings. Two strings pull with \(50\,\text{N}\) and \(86.6\,\text{N}\), at right angles to each other. The third string carries a hanging load. Find the weight of the load, in N.
Two forces of \(8\,\text{N}\) and \(6\,\text{N}\) act at one point, at an angle you may choose. Which of these can never be their resultant?
Subtraction is the addition of a negative
To subtract \(\vec B\) from \(\vec A\), add the negative of \(\vec B\): \(\vec A - \vec B = \vec A + (-\vec B)\). Draw \(-\vec B\) from the tip of \(\vec A\) and close the triangle. The same arrow appears in the parallelogram as the second diagonal, from the tip of \(\vec B\) to the tip of \(\vec A\). The angle between \(\vec A\) and \(-\vec B\) is \(180° - \theta\), so the magnitude is \(|\vec A - \vec B|^2 = A^2 + B^2 - 2AB\cos\theta\).
Subtraction depends on the order. The vector \(\vec B - \vec A\) has the same length as \(\vec A - \vec B\) and points the opposite way: \(\vec B - \vec A = -(\vec A - \vec B)\). In physics the order is always "final minus initial". A car that turns from \(10\,\text{m/s}\) east to \(10\,\text{m/s}\) north has a change of velocity \(\Delta\vec v = \vec v_f - \vec v_i\). Here \(-\vec v_i\) points west, at right angles to \(\vec v_f\). So \(|\Delta\vec v| = \sqrt{10^2 + 10^2} = 14.1\,\text{m/s}\), towards the north-west.
If \(\vec A - \vec B = \vec D\), what is \(\vec B - \vec A\)?
This is the worked example "The scooter seen from the bus" of Relative Velocity in Two Dimensions, lesson 1, done with the tool. A bus moves north at \(12\,\text{m/s}\). A scooter moves east at \(5\,\text{m/s}\). Find the velocity of the scooter relative to the bus.
The velocity relative to the bus is \(\vec v_{SB} = \vec v_S - \vec v_B = \vec v_S + (-\vec v_B)\). Here \(\vec v_S\) is \(5\,\text{m/s}\) east and \(-\vec v_B\) is \(12\,\text{m/s}\) south.
East and south are at right angles, so \(\theta = 90°\) between \(\vec v_S\) and \(-\vec v_B\). Then \(R^2 = 5^2 + 12^2 + 2 \times 5 \times 12 \times \cos 90°\).
\(R^2 = 25 + 144 + 0 = 169\), so \(v_{SB} = 13\,\text{m/s}\). The direction from east is \(\tan\alpha = 12\sin 90°/(5 + 12\cos 90°) = 2.4\), so \(\alpha = 67°\) towards the south.
From the bus, the scooter moves at \(13\,\text{m/s}\), \(23°\) east of south, the same answer as the physics chapter.
A car moving at \(20\,\text{m/s}\) east turns a corner and then moves at \(20\,\text{m/s}\) north. Find the magnitude of its change of velocity, in m/s.
The boat on the Hooghly can now be answered. Its velocity relative to the bank is the sum of \(4\,\text{m/s}\) across and \(3\,\text{m/s}\) downstream, at right angles. The resultant is \(\sqrt{16 + 9} = 5\,\text{m/s}\). It points at \(\tan^{-1}(3/4) = 37°\) downstream from the straight line across. The person on the bank sees \(5\,\text{m/s}\), faster than the boatman rows.
Relative Velocity in Two Dimensions subtracts velocities tip to tail. It does so for a scooter seen from a bus, for rain on a moving person and for an aircraft in a wind. Objects in Equilibrium closes the triangle and the polygon of forces on a knot or a pulley. Kinematics of Circular Motion finds the centripetal acceleration from the change \(\Delta\vec v\). Gravitation and Electrostatics add the fields of many bodies by the principle of superposition. Momentum and Its Conservation adds the momenta of the pieces of a body that bursts.
- Adding the magnitudes of two vectors that are not parallel.Magnitudes add only at \(0°\). At any other angle use \(R^2 = A^2 + B^2 + 2AB\cos\theta\).
- Putting the angle inside the triangle into the formula.\(\theta\) is the angle between the vectors drawn tail to tail. The angle inside the triangle is \(180° - \theta\).
- Treating \(\vec A - \vec B\) and \(\vec B - \vec A\) as the same vector.They have equal lengths and opposite directions. A change is always final minus initial.