Topic 4 of 4
Vectors and perpendicular components
A vector includes direction as well as magnitude. Account for both when combining forces, displacements or velocities.
| Scalars: magnitude only | Vectors: magnitude and direction |
|---|---|
| Mass, temperature, energy, time, distance and speed | Displacement, velocity, acceleration and force |
A temperature of -5 °C is still a scalar: it has no spatial direction. A signed coordinate is not, just because it has a minus sign, a complete vector description. For a one-dimensional velocity, a sign can encode direction once the positive axis has been stated.
Represent a vector by an arrow: length represents magnitude, and the arrowhead shows direction. Vectors in the same plane are coplanar. Use fixed axes when calculating their components; rotating the axes changes the component values, not the physical vector.
Add vectors head to tail
Draw the first vector, then place the tail of the second at the head of the first. Preserve the second vector's length and direction. The resultant points from the first tail to the final head.
Worked addition
East and north components
Take east as +x and north as +y. Let A = (3, 4) m and B = (2, -1) m. A points 3 m east and 4 m north; B points 2 m east and 1 m south.
Add corresponding components: A + B = (5, 3) m.
Magnitude = √(52 + 32) = √34 = 5.83 m.
Direction = tan-1(3/5) = 31.0° north of east.
A + B: join head to tail
A = (3, 4) m and B = (2, -1) m use east and north as positive. Move B without turning or stretching it, putting its tail at the tip of A. The purple resultant goes from the original start to the final tip.
A + B = (5, 3) m. Its magnitude is 5.83 m and its direction is 31.0° north of east. The component signs and the arrow's quadrant agree.
A paper scale drawing can also give the magnitude and direction: state a scale, draw the arrows head to tail, then measure the resultant and convert its length back. A resized screen diagram is not a physical centimetre ruler.
Subtract by reversing the second vector
A - B = A + (-B). The vector -B has the same magnitude as B and the opposite direction. Reverse B, then perform the same head-to-tail construction.
Worked subtraction
Use the signs of both components
For the same A and B, -B = (-2, +1) m. Therefore:
A - B = (3 - 2, 4 - (-1)) = (1, 5) m.
Magnitude = √(12 + 52) = √26 = 5.10 m.
Direction = tan-1(5/1) = 78.7° north of east.
A - B: add the reversed B
A = (3, 4) m and B = (2, -1) m use east and north as positive. Reverse B without changing its length, then put the tail of -B at the tip of A. The purple resultant goes from the original start to the final tip.
A - B = (1, 5) m. Its magnitude is 5.10 m and its direction is 78.7° north of east. The component signs and the arrow's quadrant agree.
Subtracting the magnitudes of A and B would ignore their directions. Coplanar vector subtraction is performed using the arrows or their signed components.
Resolve into perpendicular components
Perpendicular components are an equivalent way of representing one vector. The original vector is the hypotenuse of a right triangle. The component adjacent to the stated angle uses cosine; the opposite component uses sine.
Worked resolution
A 10.0 N force at 30.0°
Take right as +x and up as +y. The angle is measured above the positive horizontal.
Fx = 10.0 cos 30.0° = +8.66 N.
Fy = 10.0 sin 30.0° = +5.00 N.
The same force is at 60.0° to the positive vertical. Using that angle, Fx = 10.0 sin 60.0° and Fy = 10.0 cos 60.0°, giving the same components.
For a different 10.0 N force at 150° from +x, the arrow points left and up. Its components are Fx = -8.66 N and Fy = +5.00 N.
Resolve one force into perpendicular components
Purple is the original 10.0 N force. Green arrows represent its horizontal and vertical components placed head to tail. They are an equivalent description of that force, not two extra forces acting alongside it.
One direction, two angle references
Using the horizontal reference: Fx = 10.0 cos 30° and Fy = 10.0 sin 30°. From the vertical reference, use Fx = 10.0 sin 60° and Fy = 10.0 cos 60°. The vector and its components stay the same.
The identity sin2θ + cos2θ = 1 gives Fx2 + Fy2 = F2(cos2θ + sin2θ) = F2. Here (8.66 N)2 + (5.00 N)2 is approximately 100 N2, recovering the original 10.0 N magnitude.
The horizontal component is negative
The direction is 150° from +x, or 30° above -x. The components are (-8.66, +5.00) N: left and up. An inverse-tangent value must be interpreted in the correct quadrant.
Use the geometry of the component triangle
Angles and similar triangles
The interior angles of a triangle sum to 180°. The right triangle for the force at 30° therefore has third angle 180° - 90° - 30° = 60°. This is why the same force can be described as 30° above horizontal or 60° from vertical.
Triangles with equal corresponding angles are similar: corresponding side lengths have the same scale factor. A 5.00 N force at that same 30° has a component triangle similar to the 10.0 N triangle. The scale factor is 5.00/10.0 = 0.500, so its components are 0.500 × 8.66 = 4.33 N horizontally and 0.500 × 5.00 = 2.50 N vertically.
Both components scale together and the angle is unchanged. Similarity cannot be assumed for two triangles merely because they are both right-angled; their remaining corresponding angles must agree too.
The components replace the original force in the calculation. They are not two extra forces to add alongside it on the same free-body diagram.
The identity sin2θ + cos2θ = 1 explains why recombining the two perpendicular components recovers the original magnitude, as shown in the diagram.
To reconstruct a vector from perpendicular components, use magnitude √(Fx2 + Fy2) and determine the quadrant from the signs. An inverse tangent alone can return an angle in the wrong quadrant. For (-8.66, +5.00) N, the direction is 30° north of west, or 150° anticlockwise from +x.
A velocity change is a vector difference
Worked application
Turning from east to north
An initial velocity is 8.0 m/s east and a final velocity is 6.0 m/s north. Take east and north as positive.
Δv = vfinal - vinitial
= (0, 6.0) - (8.0, 0)
= (-8.0, +6.0) m/s.
Magnitude = √(8.02 + 6.02) = 10.0 m/s.
Direction = tan-1(6.0/8.0) = 36.9° north of west.
Velocity change runs from the initial tip to the final tip
Draw the initial and final velocities from the same origin. The purple arrow from the initial tip to the final tip is the change. It points 8.0 m/s west and 6.0 m/s north.
Change in velocity = (-8.0, +6.0) m/s, with magnitude 10.0 m/s and direction 36.9° north of west. A time interval would also be needed to calculate an average acceleration.
The speed decreases by 2.0 m/s, but the magnitude of the velocity change is 10.0 m/s because the direction changes too. The two velocity values alone do not establish an acceleration or show that it is constant; time information and a motion model are needed.