Scalars and Vectors in Physics

Some physical quantities are fully described by a number and a unit: a mass of $3.2~\text{kg}$, a temperature of $290~\text{K}$, an energy of $18~\text{J}$. Others are not. Telling someone a plane flew $400~\text{km}$ leaves out the fact that determines where it landed. Quantities of the second kind carry a direction as an inseparable part of their identity, and they obey a different arithmetic — one in which $3 + 4$ can legitimately equal $5$, or $7$, or $1$, depending on the angle between the two quantities being added.

This chapter builds that arithmetic. It is short on new physics and long on technique, and it repays the effort many times over: forces, velocities, momenta, electric and magnetic fields are all vectors, and every mistake made here propagates into every one of those topics. The two operations that matter — resolving a vector into perpendicular components, and adding vectors by combining those components — are the workhorses of the entire mechanics course.

Why perpendicular components? Because motion along two perpendicular directions is independent. A projectile's horizontal motion does not know that gravity is acting vertically. Resolving is not a calculational trick; it is the statement that a two-dimensional problem is two one-dimensional problems that happen to share a clock.

1. The Distinction

A scalar is a physical quantity specified completely by a magnitude (a number with a unit). Scalars combine by ordinary arithmetic.

A vector is a physical quantity specified by a magnitude and a direction. Vectors combine by the geometric rules of this chapter.

The quantities you will meet in this course divide as follows:

• Scalars: distance, speed, mass, time, energy, work, power, temperature, charge, density, pressure, potential difference, electric potential, gravitational potential.
• Vectors: displacement, velocity, acceleration, force, weight, momentum, impulse, electric field strength, magnetic field strength, gravitational field strength, torque.

Three entries deserve comment. Work is a scalar even though it is computed from two vectors, force and displacement — the operation that combines them (the scalar product) deliberately destroys the directional information. Pressure is a scalar even though the force producing it has a direction, because pressure acts equally in all directions within a fluid. And note the systematic pairing: potential is a scalar, field strength is a vector; energy is a scalar, force is a vector. In each case the vector is a spatial rate of change of the scalar, a relationship developed fully in the gravitational and electric field chapters.

Notation matters in written answers. A vector is written $\vec{F}$ or in bold $\mathbf{F}$; its magnitude is written $|\vec{F}|$ or plain $F$, is never negative, and carries the unit. In one dimension the direction collapses to a sign relative to a stated positive direction, which is why a velocity of $-5~\dfrac{\text{m}}{\text{s}}$ is a perfectly good vector while a speed of $-5~\dfrac{\text{m}}{\text{s}}$ is meaningless.

2. Adding Vectors Geometrically

Two vectors are added tip-to-tail: draw $\vec{a}$, then draw $\vec{b}$ starting from the tip of $\vec{a}$; the sum $\vec{a}+\vec{b}$, called the resultant, runs from the tail of $\vec{a}$ to the tip of $\vec{b}$. Equivalently, drawing both from a common origin, the resultant is the diagonal of the parallelogram they span. Both constructions give the same answer, and vector addition is commutative: $\vec{a}+\vec{b} = \vec{b}+\vec{a}$.

Subtraction follows from addition once $-\vec{b}$ is defined as the vector with the same magnitude as $\vec{b}$ and the opposite direction:

This definition is what makes change in velocity computable. If a body's velocity changes from $\vec{u}$ to $\vec{v}$, then $\Delta\vec{v} = \vec{v} - \vec{u}$, and this is a genuine vector subtraction, not a subtraction of speeds. A ball thrown at $8~\dfrac{\text{m}}{\text{s}}$ at a wall and rebounding at $8~\dfrac{\text{m}}{\text{s}}$ has undergone no change of speed but a change of velocity of magnitude $16~\dfrac{\text{m}}{\text{s}}$ — which is why it exerted a force on the wall.

For two vectors at right angles the magnitude of the resultant follows from Pythagoras and its direction from a tangent:

For two vectors at a general angle $\phi$ between them, the cosine rule gives

Setting $\phi = 0$ recovers $|\vec{R}| = a+b$ (parallel vectors), and $\phi = 180^\circ$ gives $|\vec{R}| = |a-b|$ (antiparallel). The magnitude of a resultant therefore always lies between $|a-b|$ and $a+b$ — a bound worth checking every answer against.

3. Resolving a Vector into Components

The inverse operation is more useful still. Any vector $\vec{A}$ may be replaced by two perpendicular vectors that together have exactly the same effect. Taking the standard axes and letting $\theta$ be the angle measured from the $x$-axis:

The magnitude and direction are recovered by

Which one takes the cosine? Not the $x$-component — the component adjacent to the given angle. The safe habit is to draw the right-angled triangle with the vector as the hypotenuse and read off which side is adjacent to the marked angle. If the angle is measured from the vertical, the vertical component takes the cosine and the horizontal takes the sine. Blindly writing "horizontal $=A\cos\theta$" is the single most common error in this chapter.

Once every vector in a problem has been resolved, addition becomes pure arithmetic, done separately along each axis:

This is the component method, and it is the only method worth using for three or more vectors.

Worked example. Three coplanar forces act on a point: $F_1 = 5.0~\text{N}$ due east, $F_2 = 8.0~\text{N}$ at $60^\circ$ north of east, and $F_3 = 4.0~\text{N}$ due north. Find the resultant.

Resolving, with east as $+x$ and north as $+y$:

Summing each column:

A direction quoted without a reference — "$50.5^\circ$" alone — earns no mark. State the angle and what it is measured from.

4. Choosing the Axes: The Inclined Plane

Nothing obliges the axes to be horizontal and vertical. The right choice is the one that puts as many vectors as possible along an axis, and for a body on a slope that means axes parallel and perpendicular to the surface.

Consider a block of weight $W = mg$ on a frictionless incline at angle $\theta$ to the horizontal. With the $x$-axis down the slope, the normal force $\vec{N}$ lies entirely along the $y$-axis and only the weight needs resolving. Geometry gives the angle between the weight and the slope-normal as $\theta$ as well, so

The component down the slope, $mg\sin\theta$, is the one that accelerates the block; the component into the slope, $mg\cos\theta$, is balanced by the normal force. Two checks confirm the assignment: at $\theta = 0$ (flat ground) the driving component $mg\sin 0 = 0$ vanishes and $N = mg$ as it must; at $\theta = 90^\circ$ (a vertical wall) the driving component is the full weight and $N = 0$. Any candidate who swaps the sine and cosine fails both checks instantly, which is why they are worth doing.

Had the axes been left horizontal and vertical, two vectors would have needed resolving and the acceleration would have had components along both axes — the same physics, three times the algebra.

5. Equilibrium and Closed Vector Polygons

A body is in translational equilibrium when the resultant of all forces on it is zero:

The two component equations are equivalent to the single vector equation and are how equilibrium problems are actually solved. Geometrically, zero resultant means the vectors drawn tip-to-tail form a closed polygon — for three forces, a closed triangle. This is the basis of the triangle of forces method and the reason a sine-rule calculation can sometimes replace a component calculation.

Worked example. A $2.0~\text{kg}$ lamp hangs from two wires making angles of $30^\circ$ and $60^\circ$ with the ceiling. Find the tensions.

Taking the horizontal and vertical components at the knot, with $T_1$ at $30^\circ$ and $T_2$ at $60^\circ$ to the horizontal:

From the first equation $T_2 = T_1\dfrac{\cos 30^\circ}{\cos 60^\circ} = T_1\sqrt{3}$. Substituting into the second, with $mg = 2.0\times 9.81 = 19.6~\text{N}$:

The steeper wire carries the larger tension, as intuition demands — and since the wires happen to be perpendicular here, $T_1^{2}+T_2^{2} = 96+289 = 385 \approx (19.6)^{2}$, a satisfying check.

6. Vector Multiplication: A Preview

Two vectors can be multiplied in two distinct ways, both of which appear later in the course.

The scalar (dot) product produces a scalar:

where $\phi$ is the angle between them. It selects the component of one vector along the other, which is exactly what work requires: $W = \vec{F}\cdot\vec{s} = Fs\cos\phi$, so a force perpendicular to the displacement does no work — the reason the tension in a string does no work on a body in circular motion.

The vector (cross) product produces a vector of magnitude

directed perpendicular to both, with its sense given by the right-hand rule. It selects the component of one vector perpendicular to the other, which is what torque and the magnetic force require: $F = qvB\sin\phi$, maximal when the velocity is perpendicular to the field and zero when it is parallel.

Notice the symmetry: $\cos\phi$ vanishes at $90^\circ$, $\sin\phi$ vanishes at $0^\circ$. Whenever a formula in later chapters contains a sine or a cosine of an angle between two vectors, this is why.

7. Common Exam Pitfalls

• Adding vector magnitudes arithmetically. Two $3~\text{N}$ forces give a resultant anywhere between $0$ and $6~\text{N}$, and only give $6~\text{N}$ when they are parallel.
• Assuming the horizontal component always takes the cosine. The cosine belongs to the component adjacent to the marked angle. Draw the triangle before writing anything.
• Quoting a direction without a reference. "$50^\circ$" is not an answer; "$50^\circ$ above the horizontal" or "$50^\circ$ north of east" is.
• Treating $\Delta\vec{v}$ as a difference of speeds. For a ball rebounding at the same speed, $|\Delta\vec{v}| = 2v$, not zero — this is the whole basis of the momentum chapters.
• Resolving on an incline with horizontal and vertical axes. Legal, but it doubles the work and introduces two acceleration components where one would do.
• Forgetting that a magnitude is never negative. A negative component is normal and informative; a negative magnitude signals an error in the algebra.
• Using the calculator's $\arctan$ output blindly. It returns angles between $-90^\circ$ and $+90^\circ$; if $R_x < 0$ the true direction lies in a different quadrant and $180^\circ$ must be added.

Summary

• A scalar has magnitude only; a vector has magnitude and direction and obeys geometric addition. Speed, mass, energy, work, power and potential are scalars; displacement, velocity, acceleration, force, momentum and field strength are vectors.
• Vectors add tip-to-tail; the resultant of two vectors at angle $\phi$ has magnitude $\sqrt{a^{2}+b^{2}+2ab\cos\phi}$ and always lies between $|a-b|$ and $a+b$.
• Subtraction is addition of the reversed vector: $\Delta\vec{v} = \vec{v}-\vec{u}$, which is why a rebound at equal speed gives $|\Delta\vec{v}| = 2v$.
• Resolving into perpendicular components gives $A_x = A\cos\theta$, $A_y = A\sin\theta$ with $\theta$ measured from the $x$-axis; the cosine always belongs to the component adjacent to the marked angle.
• The component method — resolve, sum each axis separately, recombine with $R = \sqrt{R_x^{2}+R_y^{2}}$ and $\theta = \arctan\dfrac{R_y}{R_x}$ — is the general method for any number of vectors.
• On an incline, axes parallel and perpendicular to the surface give $W_\parallel = mg\sin\theta$ and $W_\perp = mg\cos\theta$; check the assignment at $\theta = 0$ and $\theta = 90^\circ$.
• Equilibrium means $\sum F_x = 0$ and $\sum F_y = 0$, equivalently that the force vectors form a closed polygon.
• The scalar product $ab\cos\phi$ extracts parallel components (work); the vector product magnitude $ab\sin\phi$ extracts perpendicular components (torque, magnetic force).