Lesson 1.7 · Unit 1 · CED topic 1.5 (vectors)
Vectors and components in two dimensions
Most real motion is not along a single line. A thrown ball rises and travels forward at the same time. A boat crossing a river is carried downstream while it heads for the other bank. A skier moves down a slope that is neither horizontal nor vertical. The central idea of two-dimensional mechanics is a trick that turns every such problem back into one-dimensional problems: split each vector into two perpendicular components, solve the two straight-line problems separately, and put the answers back together. This lesson develops the trick; the next one uses it on projectiles, and it returns in Unit 2 with forces on inclines.
Describe the perpendicular components of a vector.
- Vectors can be mathematically modeled as the resultant of two perpendicular components.
- Vectors can be resolved into components using a chosen coordinate system.
- Vectors can be resolved into perpendicular components using trigonometric functions and relationships: \(\sin\theta\), \(\cos\theta\), \(\tan\theta\), and \(a^2 + b^2 = c^2\).
Resolving a vector into components
Any vector in a plane can be replaced by two perpendicular vectors, usually along the horizontal and vertical axes, that together have exactly the same effect. These are its components. If a vector \(\vec{A}\) has magnitude \(A\) and makes an angle \(\theta\) with the positive \(x\)-axis, the right triangle in Figure 1.7a gives \[ A_x = A\cos\theta, \qquad A_y = A\sin\theta. \] The component along an axis is the “shadow” of the vector on that axis. When the vector lies along the axis (\(\theta = 0\)) its component is the whole vector, \(A_x = A\) and \(A_y = 0\); when it lies at right angles to the axis the shadow vanishes.
Going the other way, if you know the components, the magnitude of the vector and its direction follow from the Pythagorean relationship and the tangent: \[ A = \sqrt{A_x^2 + A_y^2}, \qquad \tan\theta = \frac{A_y}{A_x}. \] A reliable habit is to compute the angle from the absolute values of the components, then place it in the correct quadrant by looking at the signs. A calculator’s inverse tangent cannot distinguish \((+3, +4)\) from \((-3, -4)\), so always check the sketch.
| Direction of the vector | sign of \(A_x\) | sign of \(A_y\) |
|---|---|---|
| up and to the right (first quadrant) | + | + |
| up and to the left (second) | \(-\) | + |
| down and to the left (third) | \(-\) | \(-\) |
| down and to the right (fourth) | + | \(-\) |
Choosing the axes
The axes do not have to be horizontal and vertical. In Unit 2 you will often tilt them so that one axis runs parallel to the surface of an inclined plane, because that makes one of the two component problems trivial. The components depend on the axes you choose, but the vector itself does not. The skill is to choose axes that make the problem easiest, state them, and then resolve every vector in the problem onto those same axes.
Adding vectors by components
To add vectors in two dimensions, add their \(x\)-components to get the \(x\)-component of the resultant, and do the same for \(y\). The sum is independent of order, and it can be pictured by placing the vectors tip to tail: the resultant runs from the start of the first vector to the tip of the last (Figure 1.7b). A displacement of 30 m east followed by one of 40 m north gives a resultant with \(R_x = 30\ \text{m}\), \(R_y = 40\ \text{m}\), a magnitude of 50 m and a direction of \(\tan^{-1}(40/30) = 53^\circ\) north of east. Notice that the magnitude of the resultant, 50 m, is less than the sum of the magnitudes, 70 m: magnitudes add only when the vectors point the same way.
The same procedure applies to every vector quantity you meet: displacement, velocity, acceleration, and later force and momentum. It is the reason AP Physics 1 problems with several forces or velocities at angles are never harder than a table with two columns, one for each component.
A ball leaves a bat with a speed of 25 m/s at an angle of 37\(^\circ\) above the horizontal. Find the horizontal and vertical components of its velocity.
Step 1. \(v_x = v\cos\theta = 25\cos 37^\circ = 25(0.799) = 20.0\ \text{m/s}\).
Step 2. \(v_y = v\sin\theta = 25\sin 37^\circ = 25(0.602) = 15.0\ \text{m/s}\).
Check. \(\sqrt{20.0^2 + 15.0^2} = \sqrt{400 + 225} = \sqrt{625} = 25\ \text{m/s}\). ✓ The numbers 15, 20, 25 form a 3-4-5 right triangle, which is why 37\(^\circ\) appears so often in problems.
A particle has velocity components \(v_x = -6.0\ \text{m/s}\) and \(v_y = -8.0\ \text{m/s}\). Find its speed and the direction of its motion.
Speed. \(v = \sqrt{(-6.0)^2 + (-8.0)^2} = \sqrt{36 + 64} = 10\ \text{m/s}\).
Angle. Using the absolute values, \(\tan\phi = 8.0/6.0\), so \(\phi = 53^\circ\). Both components are negative, so the vector points down and to the left: 53\(^\circ\) below the negative \(x\)-axis. (Measured counterclockwise from \(+x\), that is \(180^\circ + 53^\circ = 233^\circ\).)
A hiker walks 8.0 m east, then 6.0 m west, then 9.0 m north. Find the resultant displacement.
Components. With east as \(+x\) and north as \(+y\): \(R_x = +8.0 - 6.0 + 0 = +2.0\ \text{m}\) and \(R_y = 0 + 0 + 9.0 = +9.0\ \text{m}\).
Magnitude. \(R = \sqrt{2.0^2 + 9.0^2} = \sqrt{85} = 9.2\ \text{m}\).
Direction. \(\theta = \tan^{-1}(9.0/2.0) = 77^\circ\) north of east. The hiker has walked 23 m but is only 9.2 m from the start.
- Always draw the triangle. Sketch the vector, mark the angle and say from which axis it is measured. Mixing up sine and cosine is the usual error, and the sketch prevents it.
- Measured from the horizontal. For an angle from the horizontal, the horizontal component uses the cosine. For an angle from the vertical, the roles swap.
- Use the same axes for every vector in one problem.
- Keep signs. A component pointing in the negative direction is negative. The sign tells you the quadrant.
- Calculator in degrees. Check the mode: \(\sin 30^\circ = 0.5\), not \(-0.988\).
- A plane climbs at 200 m/s at 30\(^\circ\) above the horizontal. Find its horizontal speed and its rate of climb.
Show answer
\(v_x = 200\cos 30^\circ = 173\ \text{m/s}\). \(v_y = 200\sin 30^\circ = 100\ \text{m/s}\).
- A vector has components 12 and 5. Find its magnitude and the angle above the \(x\)-axis.
Show answer
Magnitude \(\sqrt{144 + 25} = 13\). Angle \(\tan^{-1}(5/12) = 22.6^\circ\) above the \(x\)-axis.
- An object slides 10 m along an incline that makes 30\(^\circ\) with the horizontal. Find the horizontal and vertical parts of the displacement.
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Horizontal: \(10\cos 30^\circ = 8.66\ \text{m}\). Vertical: \(10\sin 30^\circ = 5.0\ \text{m}\) (a drop, if it slides downhill).
- A velocity has \(v_x = -3.0\ \text{m/s}\) and \(v_y = +4.0\ \text{m/s}\). Give its speed and direction.
Show answer
Speed \(= 5.0\ \text{m/s}\). \(\tan^{-1}(4.0/3.0) = 53^\circ\). The vector points up and to the left: 53\(^\circ\) above the negative \(x\)-axis, which is \(127^\circ\) from the positive \(x\)-axis.
- Three students walk the displacements 20 m east, 20 m north and 20 m west. Find the net displacement.
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\(R_x = 20 - 20 = 0\), \(R_y = +20\ \text{m}\). The displacement is 20 m north.
- A student claims a vector of magnitude 10 at 45\(^\circ\) has components of 10 and 10 because both axes see the whole arrow. Evaluate the claim.
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The claim is wrong. At 45\(^\circ\) the two components are equal, but each is \(10\cos 45^\circ = 7.1\), and \(\sqrt{7.1^2 + 7.1^2} = 10\). Components of 10 and 10 would give a vector of magnitude 14, which is larger than the original.