Scalars and vectors

AQA·Physics·3.4.1.1

Scalars and vectors

Some physical quantities need only a size and a unit to describe them. Others also need a direction. The size of a quantity is called its magnitude.

A scalar has magnitude only. For example, a mass of $$5\ \text{kg}$$ does not need a direction. A vector has both magnitude and direction. For example, a force of $$5\ \text{N}$$ east is different from a force of $$5\ \text{N}$$ west.

Speed, mass and distance are scalars. Velocity, force, weight, acceleration and displacement are vectors.

Representing vectors with arrows

A vector can be represented by an arrow. The arrow’s length shows its magnitude, and the direction in which it points shows the vector’s direction. When arrows are drawn using the same scale, a longer arrow represents a greater magnitude.

Distance and displacement

Distance is the total length of the path travelled. Displacement is the change in position from the starting point to the finishing point, including its direction.

For example, if a person walks $$10\ \text{m}$$ east and then $$10\ \text{m}$$ west, the distance travelled is $$20\ \text{m}$$. The displacement is zero because the person finishes at the starting point.

Speed and velocity

Speed is the rate at which distance is travelled. Velocity is the rate at which displacement changes, so it also has a direction. A speed might be $$5\ \text{m/s}$$, whereas a velocity could be $$5\ \text{m/s}$$ east.

Mass and weight

Mass is a scalar measure of an object’s inertia: its resistance to a change in motion. Weight is the gravitational force acting on the object. Weight is a vector because the force has both a magnitude and a direction.

Vector addition

Vector addition combines two or more vectors into a single vector called the resultant. The resultant has the same overall effect as the original vectors acting together.

A vector has both magnitude and direction, so both must be considered when finding the resultant. Adding magnitudes alone is not generally sufficient: vectors in different directions can partly or completely oppose one another.

Vector addition by scale drawing

A scale drawing represents each vector with an arrow. Its length represents the magnitude, and its orientation represents the direction.

  1. Choose a scale and draw the first vector as an arrow.
  2. Draw the next vector with its tail at the head of the first. Keep its length consistent with the scale and its direction unchanged. Continue head to tail for any further vectors.
  3. Draw the resultant arrow from the tail of the first vector to the head of the last vector.
  4. Measure the resultant’s length and use the scale to find its magnitude. Measure its direction relative to a stated reference direction.

Moving a vector to a new position does not change it, provided its magnitude and direction stay the same. The scale-drawing method can be used for vectors at any angle to one another, not just at $$90^\circ$$.

Adding perpendicular vectors

When two vectors are perpendicular, placing them head to tail forms two sides of a right-angled triangle. The resultant is the third side, opposite the right angle.

For perpendicular vectors with magnitudes $$A$$ and $$B$$, the resultant magnitude $$R$$ is found using Pythagoras’ theorem:

R=A2+B2R = \sqrt{A^2 + B^2}

If the angle $$\theta$$ is measured from the direction of $$A$$ towards $$B$$, its direction is found using:

tan⁡θ=BA\tan\theta = \frac{B}{A}

For example, consider a force of $$3,\mathrm{N}$$ east and a force of $$4,\mathrm{N}$$ north. Taking east as the direction of $$A$$, the resultant’s magnitude and angle north of east are:

R=32+42=5 Nθ=tan⁡−1(43)≈53∘\begin{aligned} R &= \sqrt{3^2 + 4^2} = 5\,\mathrm{N} \\ \theta &= \tan^{-1}\left(\frac{4}{3}\right) \approx 53^\circ \end{aligned}

The resultant is therefore $$5,\mathrm{N}$$ at approximately $$53^\circ$$ north of east. Stating the reference direction makes the angle unambiguous.

Vector resolution

A vector can be expressed as components along chosen directions. The components are vectors whose vector sum equals the original vector. Together, they represent the same combined effect as the original vector; they are not additional effects.

Components at right angles can be treated independently: a component along one axis does not contribute along the perpendicular axis. Choosing convenient axes can therefore make a vector easier to analyse.

Components along perpendicular axes

Consider a vector of magnitude $$F$$ making an angle $$\theta$$ with one axis. Its components form the perpendicular sides of a right-angled triangle, with the original vector as the hypotenuse. The component adjacent to $$\theta$$ uses cosine; the opposite component uses sine.

Falong axis=Fcos⁡θ,Fperpendicular=Fsin⁡θF_{\text{along axis}} = F\cos\theta, \qquad F_{\text{perpendicular}} = F\sin\theta

For these signed equations, take $$\theta$$ as the angle measured anticlockwise from the positive direction of the first axis, with the positive perpendicular axis a further $$90^\circ$$ anticlockwise. A positive component points along its positive axis; a negative component points in the opposite direction. If an angle or axes are chosen differently, identify the directions and assign the signs accordingly.

Resolving weight on an inclined plane

For a plane inclined at an angle $$\alpha$$ to the horizontal, it is often useful to choose axes parallel and perpendicular to the plane. Weight has magnitude $$mg$$ and acts vertically downwards, where $$m$$ is the object's mass and $$g$$ is gravitational field strength.

The inward perpendicular to the plane is tilted by $$\alpha$$ from the vertical. The angle between the weight and this inward direction is therefore also $$\alpha$$. The component perpendicular to the plane is adjacent to this angle, while the component parallel to the plane is opposite it:

Parallel magnitude=mgsin⁡α,perpendicular magnitude=mgcos⁡α\text{Parallel magnitude} = mg\sin\alpha, \qquad \text{perpendicular magnitude} = mg\cos\alpha

The parallel component points down the slope; the perpendicular component points into the plane. For example, if down the slope and away from the plane are chosen as positive, the signed components are $$+mg\sin\alpha$$ and $$-mg\cos\alpha$$ respectively.

Equilibrium of concurrent forces

Forces are concurrent when their lines of action meet at a point. They are in translational equilibrium when their vector sum, or resultant force, is zero.

∑F⃗=0⃗\sum \vec{F} = \vec{0}

With zero resultant force, the object has zero acceleration. It therefore remains at rest if it was already at rest, or continues to move with constant velocity if it was already moving.

Equilibrium of two or three forces

For two forces acting at a point to balance, they must be equal in magnitude and opposite in direction. Their vector sum is then zero.

For three coplanar forces—forces acting in the same plane—the vector sum must also be zero. To represent this, draw the force vectors head to tail. The third arrow must return to the starting point, forming a closed triangle of forces. Each side represents one force in magnitude and direction. If the triangle does not close, the forces have a non-zero resultant.

Resolving concurrent forces into components

Equilibrium can also be checked by resolving each force into components along two perpendicular axes. Choose a positive direction on each axis, then give each component a positive or negative sign according to its direction.

For equilibrium, the algebraic sum of the components must be zero along each axis:

∑Fx=0,∑Fy=0\sum F_x = 0, \qquad \sum F_y = 0

For example, if a force of magnitude $$F$$ makes an angle $$\theta$$ with the positive $$x$$-axis, its components are $$F\cos\theta$$ along the $$x$$-axis and $$F\sin\theta$$ along the $$y$$-axis. The signs show whether each component acts in the positive or negative direction. Both component sums must be zero for the resultant force to be zero.