2.2 Resolving and adding vectors — Learn

Prerequisites: scalars and vectors from Section 2.1; Pythagoras' theorem; the sine, cosine and tangent ratios; rearranging a formula; finding an angle from a tangent.

By the end of this section you can: resolve a vector into two perpendicular components; choose the component that is the cosine and the component that is the sine; add two perpendicular vectors to find the resultant; find the direction of a resultant with the tangent; and describe how a scale drawing adds vectors that are not at right angles.

1. Why resolve a vector

A vector at an angle is awkward to add to anything. Two perpendicular components are easier to work with, because each component acts along one axis.

Resolving a vector means replacing it by two perpendicular components that have the same combined effect. The vector is the resultant of its two components. Replacing the vector by the components changes nothing about the physics; it only changes how the vector is written.

The two components are found with the sine and cosine ratios. Take the vector as the hypotenuse of a right-angled triangle, with the components as the two shorter sides.

v_x = v\cos\theta \qquad v_y = v\sin\theta

Here $v$ is the magnitude of the vector and $\theta$ is the angle measured from the $x$ -axis.

2. Which component is the cosine

The rule follows from the triangle, and it is quick to apply once you know the pattern.

The component along the direction the angle is measured from is $v\cos\theta$ .
The component perpendicular to that direction is $v\sin\theta$ .

So if an angle is measured from the horizontal, the horizontal component uses the cosine. If the angle is measured from the vertical, the component along the vertical uses the cosine. The two components swap over when the angle is measured from the other axis.

Worked example 1. A velocity of $25\,\text{m}\,\text{s}^{-1}$ acts at 30° above the horizontal. Find its components.

\begin{aligned} v_x &= v\cos\theta = 25\cos 30^\circ = 21.7\,\text{m}\,\text{s}^{-1} \\ v_y &= v\sin\theta = 25\sin 30^\circ = 12.5\,\text{m}\,\text{s}^{-1} \end{aligned}

The angle is measured from the horizontal, so the horizontal component is the cosine.

Worked example 2. A force of 200 N acts at 120° from the positive $x$ -axis. Find its $x$ - and $y$ -components.

\begin{aligned} F_x &= 200\cos 120^\circ = -100\,\text{N} \\ F_y &= 200\sin 120^\circ = +173.2\,\text{N} \end{aligned}

The force points into the second quadrant. The negative sign on $F_x$ records that the component acts back along the $x$ -axis.

Watch out. The sine and cosine are often swapped. A student resolving a force at 40° above the horizontal might write $F_x = F\sin 40^\circ$ instead of $F\cos 40^\circ$ . Check with the two extreme angles. At 0° the vector lies along the $x$ -axis, so the $x$ -component must be the whole vector and the $y$ -component must be zero: $\cos 0^\circ = 1$ and $\sin 0^\circ = 0$ . At 90°, the reverse must hold. If your formula fails those cases, the sine and cosine are the wrong way round.

3. Try it: the two components of a vector

The figure shows a vector and its two perpendicular components. Change the magnitude and the angle and watch how the components depend on each.

Predict first. Do not move the sliders yet. Write down your answers.

At an angle of 0°, what is the $y$ -component?
At an angle of 90°, what is the $x$ -component?
At what angle are the two components equal?

Test. Set the magnitude to 10 and move the angle slider through 0°, 45° and 90°. Read the two components at each angle. Then set the angle back to 40° and change the magnitude.

Explain. Why is one component the cosine and the other the sine? Write two or three sentences. Use the idea that the vector is the hypotenuse of a right-angled triangle.

4. Adding two perpendicular vectors

Two perpendicular vectors are already two components of their resultant. Put the tail of the second vector at the tip of the first, and draw the resultant from the first tail to the second tip. The three vectors form a right-angled triangle.

The magnitude of the resultant comes from Pythagoras:

R = \sqrt{x^2 + y^2}

The direction comes from the tangent:

\tan\theta = \frac{\text{opposite}}{\text{adjacent}}

The angle $\theta$ is measured from the vector that you chose as the adjacent side.

Worked example 3. Two forces of 6.0 N and 8.0 N act at right angles. Find the magnitude and direction of the resultant.

\begin{aligned} R &= \sqrt{6.0^2 + 8.0^2} \\ &= \sqrt{36 + 64} \\ &= \sqrt{100} = 10\,\text{N} \end{aligned}

For the direction, take the 6.0 N force as the adjacent side.

\begin{aligned} \tan\theta &= \frac{8.0}{6.0} \\ \theta &= \tan^{-1}\!\left(\frac{8.0}{6.0}\right) = 53.1^\circ \end{aligned}

The resultant is 10 N at 53.1° to the 6.0 N force, on the side of the 8.0 N force.

5. Resolving on a slope

A weight on a slope is a common resolution. The weight acts vertically down. The slope makes an angle $\theta$ with the horizontal, so the weight makes the same angle with the perpendicular to the slope.

The component of the weight along the slope is

W_{\parallel} = W\sin\theta

and the component perpendicular to the slope is

W_{\perp} = W\cos\theta

Worked example 4. A block of weight 500 N rests on a slope at 25° to the horizontal. Find the two components of the weight.

\begin{aligned} W_{\parallel} &= 500\sin 25^\circ = 211.3\,\text{N} \\ W_{\perp} &= 500\cos 25^\circ = 453.2\,\text{N} \end{aligned}

The perpendicular component pushes the block into the slope. The along-slope component pulls it down the slope. On a shallow slope the perpendicular component is the larger of the two.

6. Adding vectors that are not at right angles

When two vectors are not at right angles, the triangle they form has no right angle, so Pythagoras does not apply. Two methods work.

A scale drawing. Choose a scale such as 1 cm to 10 N. Draw the first vector as an arrow to scale in its direction. Draw the second vector head to tail from the tip of the first. Draw the resultant from the tail of the first to the tip of the second. Measure the resultant with a ruler, convert the length with the scale, and measure its angle with a protractor.
Resolving into components. Put both vectors along the same pair of axes, add the components in each direction, and combine the two totals. When the vectors are already perpendicular, their components are the vectors themselves, so this reduces to the Pythagoras method of Section 4.

Worked example 5. A displacement of 120 m east is followed by a displacement of 50 m north. Find the resultant displacement.

The two displacements are already perpendicular components.

\begin{aligned} s &= \sqrt{120^2 + 50^2} \\ &= \sqrt{14400 + 2500} \\ &= \sqrt{16900} = 130\,\text{m} \end{aligned}

\tan\theta = \frac{50}{120}, \qquad \theta = 22.6^\circ

The resultant displacement is 130 m at 22.6° north of east.

Progress check

Return to the chapter problem. The ferry points straight across at $4.0\,\text{m}\,\text{s}^{-1}$ and the current is $3.0\,\text{m}\,\text{s}^{-1}$ downstream. Their resultant is $5.0\,\text{m}\,\text{s}^{-1}$ at 36.9° to the straight-across direction.

Resolve the $5.0\,\text{m}\,\text{s}^{-1}$ resultant into a component straight across and a component downstream. Check that you recover 4.0 and 3.0.
The river is 120 m wide. Use the across component to find the crossing time.
Use the downstream component and the crossing time to find how far downstream the ferry lands.

If question 1 goes wrong, the angle is measured from the straight-across direction, so that component uses the cosine. If question 3 goes wrong, the downstream distance is the downstream speed multiplied by the crossing time.

You should now be able to

resolve a vector into two perpendicular components using the sine and cosine;
decide which component uses the cosine from the direction the angle is measured from;
add two perpendicular vectors with Pythagoras and find the direction with the tangent;
resolve a weight into components along and perpendicular to a slope;
describe how a scale drawing adds two vectors that are not at right angles;
explain why two perpendicular vectors are already the components of their resultant.