Statics · Chapter 2

Force Vectors

Forces have both size and direction, so they must be handled as vectors rather than ordinary numbers. This chapter develops the tools for adding forces, breaking them into components, and describing them in two and three dimensions using Cartesian notation, position vectors, and the dot product.

Learning objectives

Distinguish a force magnitude from its directed vector representation.
Resolve and add coplanar forces using a stated coordinate convention.
Build a three-dimensional force vector from endpoint geometry and a unit vector.
Use a dot product to evaluate an angle or an axial projection.
Audit vector work with units, quadrants, and unit-vector magnitude checks.

SECTION 2.1

Scalars and Vectors

Engineering quantities fall into two groups depending on whether direction matters. Some are described fully by a single number; others need a direction to mean anything.

Scalar

A quantity fully described by a single magnitude — a positive or negative number with no associated direction. Length, mass, and time are scalars.

Vector

A quantity needing both a magnitude and a direction to be fully described. Force, position, and moment are vectors. A vector is drawn as an arrow: its length sets the magnitude, the angle to a reference axis sets the direction of its line of action, and the arrowhead sets the sense.

Notation

In print, a vector is shown in boldface (A) and its magnitude in italics (A). By hand, a vector is written with an arrow drawn above the symbol.

Fig. 2–1. A vector drawn as an arrow: length is magnitude, the angle θ to a fixed axis sets direction, and the arrowhead sets sense.

Course Note

Students lose marks here more than anywhere else: a magnitude is always positive. The sign you see in scalar notation (like −100 N) is bookkeeping for direction along an axis, not a "negative force." Keep magnitude and direction mentally separate from day one.

SECTION 2.2

Vector Operations

Scalar multiplication and division

Multiplying a vector by a positive scalar scales its magnitude only. A negative scalar also reverses its sense. Division is multiplication by the reciprocal.

Vector addition — the parallelogram law

Two vectors are added by accounting for both magnitude and direction, not by adding numbers. The parallelogram law gives the resultant.

LAW

Parallelogram Law of Addition

To add vectors A and B: join their tails so they are concurrent; from the head of each, draw a line parallel to the other, forming a parallelogram; the diagonal from the common tail is the resultant R = A + B.

Fig. 2–3. The resultant is the parallelogram diagonal extending from the common tail to point P.

Triangle rule

A special case of the parallelogram law: place B head-to-tail after A, and the resultant runs from the tail of A to the head of B. Vector addition is commutative, so R = A + B = B + A.

Collinear addition

When two vectors share the same line of action, the parallelogram law collapses to ordinary algebraic addition of magnitudes: R = A + B.

Vector subtraction

The difference is defined as addition of the negative: R′ = A − B = A + (−B). All rules of vector addition therefore apply to subtraction.

Important Points

A scalar is a positive or negative number.
A vector has magnitude, direction, and sense.
Scaling changes magnitude; a negative scalar also flips the sense.
Collinear vectors add algebraically.

SECTION 2.3

Vector Addition of Forces

Experiment confirms that force is a vector: it has magnitude, direction, and sense, and it combines by the parallelogram law. Two recurring tasks in statics are finding a resultant from known components, and resolving a known force into two components.

Finding a resultant force

Add the component forces by the parallelogram law (or the triangle rule), then apply the law of cosines for the resultant's magnitude and the law of sines for its direction.

Resolving a force into components

To study a force's effect along two chosen directions (axes u and v), draw lines from the tip of the force parallel to each axis. The intersections define the components, and the law of sines gives their magnitudes.

Adding several forces

For more than two forces, apply the parallelogram law successively — combine two, then add the next to that resultant. This is geometrically tedious, which motivates the rectangular-component method of Section 2.4.

Law of Cosines & Law of Sines

\[ C = \sqrt{A^2 + B^2 - 2AB\cos c} \qquad \frac{A}{\sin a} = \frac{B}{\sin b} = \frac{C}{\sin c} \]

Fig. 2–10. Triangle relating the resultant and its components; sides face their opposite interior angles.

PROCEDURE

Analysis for adding two forces

Parallelogram law: sketch the two components, complete the parallelogram, and label all known and unknown magnitudes and angles. Identify the two unknowns (magnitude and direction of the resultant, or the two component magnitudes).

Trigonometry: redraw half the parallelogram as a head-to-tail triangle; use the law of cosines for magnitude and the law of sines for angles.

Course Note

For any "minimum force" problem, the answer is almost always a perpendicular. Don't grind through calculus — recognize that the shortest distance from a point to a line is the perpendicular, set \(\theta = 90^\circ\), and the rest is a right triangle. This trick reappears throughout statics.

SECTION 2.4

Addition of a System of Coplanar Forces

When a force is resolved along the x and y axes, the pieces are its rectangular components. These can be written with scalar notation or with Cartesian unit vectors, and either way the resultant of many coplanar forces is found by adding components.

Scalar notation

For a force at angle \(\theta\) from the x axis, \(F_x = F\cos\theta\) and \(F_y = F\sin\theta\). When direction is given by a slope triangle with sides \(a\), \(b\), hypotenuse \(c\), use proportions: \(F_x = F(a/c)\) and \(F_y = F(b/c)\). Signs encode sense and are for computation only, never for figures.

Cartesian vector notation

Unit vectors i and j point along the positive x and y axes and have magnitude 1. A force is written F = F_xi + F_yj, separating magnitude from direction.

Coplanar resultant

Resolve every force into x and y components, add the collinear components algebraically to get the resultant components, then recombine. The orientation of the axes is arbitrary.

Resultant of Coplanar Forces

\[ (F_R)_x = \sum F_x \qquad (F_R)_y = \sum F_y \] \[ F_R = \sqrt{(F_R)_x^{\,2} + (F_R)_y^{\,2}} \qquad \theta = \tan^{-1}\left|\frac{(F_R)_y}{(F_R)_x}\right| \]

Fig. 2–15. A force resolved into rectangular components \(F_x\) and \(F_y\).

Important Points

Set up an x, y frame and resolve every force along the axes.
Each force's direction comes from an angle to an axis or from a slope triangle.
The resultant components are the algebraic sums of all force components.
Magnitude follows from the Pythagorean theorem; direction from trigonometry.

Course Note

Pick one sign convention (right = +x, up = +y) and write it at the top of your page. Most sign errors in coplanar problems come from switching conventions mid-solution. Let the math return the negative sign — don't "force" the direction by inspection and then second-guess it.

SECTION 2.5

Cartesian Vectors

Three-dimensional vector algebra is far simpler when vectors are written in Cartesian form. This section sets up that representation and the angles that describe a vector's direction in space.

Right-handed coordinate system

A coordinate system is right-handed if curling the right-hand fingers from the +x axis toward the +y axis points the thumb along the +z axis. All theory here uses this convention.

Rectangular components & unit vectors

A vector in space resolves into three rectangular components, A = A_x + A_y + A_z. Using the unit vectors i, j, k (magnitude 1, along +x, +y, +z), this becomes the Cartesian form.

Cartesian Form & Magnitude

\[ \mathbf{A} = A_x\mathbf{i} + A_y\mathbf{j} + A_z\mathbf{k} \qquad A = \sqrt{A_x^2 + A_y^2 + A_z^2} \]

Coordinate direction angles

The angles \(\alpha\), \(\beta\), \(\gamma\) measured from the tail of A to the +x, +y, +z axes describe its direction. Each lies between \(0^\circ\) and \(180^\circ\). Their cosines are the direction cosines and equal the components of the unit vector along A.

Direction Cosines & Unit Vector

\[ \cos\alpha = \frac{A_x}{A}, \quad \cos\beta = \frac{A_y}{A}, \quad \cos\gamma = \frac{A_z}{A} \] \[ \mathbf{u}_A = \frac{\mathbf{A}}{A} = \cos\alpha\,\mathbf{i} + \cos\beta\,\mathbf{j} + \cos\gamma\,\mathbf{k} \]

IDENTITY

Direction Cosine Constraint

Because the unit vector has magnitude one, the three direction cosines are not independent:

\[ \cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1 \]

Knowing any two coordinate angles fixes the third.

Fig. 2–26. Coordinate direction angles \(\alpha\), \(\beta\), \(\gamma\) measured from the positive axes to the tail of A.

Alternatively, a direction can be set with a transverse angle \(\theta\) and an azimuth angle \(\phi\); the components follow from successive applications of trigonometry rather than from a memorized formula.

SECTION 2.6

Addition of Cartesian Vectors

Adding vectors in three dimensions reduces to adding like components — the chief reason Cartesian form is preferred for spatial problems.

Resultant of Concurrent Forces in 3-D

\[ \mathbf{R} = \mathbf{A} + \mathbf{B} = (A_x+B_x)\mathbf{i} + (A_y+B_y)\mathbf{j} + (A_z+B_z)\mathbf{k} \] \[ \mathbf{F}_R = \sum \mathbf{F} = \sum F_x\,\mathbf{i} + \sum F_y\,\mathbf{j} + \sum F_z\,\mathbf{k} \]

Important Points

A Cartesian vector has i, j, k components; its magnitude is \(\sqrt{A_x^2+A_y^2+A_z^2}\).
Direction is set by \(\alpha\), \(\beta\), \(\gamma\) via the unit vector \(\mathbf{u}_A = \mathbf{A}/A\); only two angles are independent.
Direction can also be given by a transverse angle \(\theta\) and azimuth angle \(\phi\).

SECTION 2.7

Position Vectors

A position vector is the bridge between geometry and force in three dimensions: it lets a force directed between two known points be written in Cartesian form.

Position vector

A fixed vector r locating one point relative to another. From the origin to \(P(x,y,z)\), r = xi + yj + zk — the distances traveled along each axis to reach the point.

Position Vector From A to B

\[ \mathbf{r} = (x_B - x_A)\mathbf{i} + (y_B - y_A)\mathbf{j} + (z_B - z_A)\mathbf{k} \]

The components are formed by subtracting the tail coordinates \(A(x_A,y_A,z_A)\) from the head coordinates \(B(x_B,y_B,z_B)\) — equivalently, the signed distance traveled along each axis going from A to B.

Fig. 2–34. Position vector r directed from A to B; components are head-minus-tail coordinates.

SECTION 2.8

Force Vector Directed Along a Line

In space, a force's direction is often given by two points on its line of action. The force shares the direction of the position vector between those points, so it can be built as a magnitude times a unit vector.

Force Along a Line

\[ \mathbf{F} = F\,\mathbf{u} = F\left(\frac{\mathbf{r}}{r}\right) = F\left(\frac{(x_B-x_A)\mathbf{i} + (y_B-y_A)\mathbf{j} + (z_B-z_A)\mathbf{k}}{\sqrt{(x_B-x_A)^2 + (y_B-y_A)^2 + (z_B-z_A)^2}}\right) \]

Important Points

A position vector locates one point relative to another.
Its components are the distances traveled along x, y, z from tail to head.
A force along r is \(\mathbf{F} = F\mathbf{u} = F(\mathbf{r}/r)\): multiply the magnitude by the unit vector.

Course Note

The whole game in 3-D force problems is: find r, normalize to u, multiply by the magnitude. If you can do those three steps cleanly, most of this chapter's spatial problems collapse into routine arithmetic. Practice reading coordinates straight off the figure — that's where the real difficulty hides, not in the algebra.

SECTION 2.9

Dot Product

Finding the angle between two lines, or the component of a force along a direction, is awkward in 3-D by geometry alone. The dot product handles both cleanly.

Dot product (scalar product)

The dot product of A and B is the product of their magnitudes and the cosine of the angle between their tails. The result is a scalar, which may be positive, negative, or zero.

Definition & Cartesian Form

\[ \mathbf{A}\cdot\mathbf{B} = AB\cos\theta, \quad 0^\circ \le \theta \le 180^\circ \] \[ \mathbf{A}\cdot\mathbf{B} = A_xB_x + A_yB_y + A_zB_z \]

Laws of operation

The dot product is commutative (\(\mathbf{A}\cdot\mathbf{B} = \mathbf{B}\cdot\mathbf{A}\)), compatible with scalar multiplication, and distributive over addition.

Angle Between Vectors & Projection Onto a Line

\[ \theta = \cos^{-1}\!\left(\frac{\mathbf{A}\cdot\mathbf{B}}{AB}\right) \qquad A_a = A\cos\theta = \mathbf{A}\cdot\mathbf{u}_a \]

Two key applications

Angle: the angle between two vectors or intersecting lines follows from \(\theta = \cos^{-1}(\mathbf{A}\cdot\mathbf{B}/AB)\). If the dot product is zero, the vectors are perpendicular.

Projection: the component of A along a line with unit vector \(\mathbf{u}_a\) is the scalar \(A_a = \mathbf{A}\cdot\mathbf{u}_a\). A positive value points along \(\mathbf{u}_a\); a negative value points opposite. The perpendicular component follows from \(A_\perp = \sqrt{A^2 - A_a^2}\).

Fig. 2–41. Projection of A onto line aa: the parallel component \(A_a\) and the perpendicular component form a right triangle.

Important Points

The dot product gives the angle between two vectors or the projection along a direction.
In Cartesian form, multiply matching components and add: \(A_xB_x + A_yB_y + A_zB_z\).
Angle: \(\theta = \cos^{-1}(\mathbf{A}\cdot\mathbf{B}/AB)\).
Projection along \(\mathbf{u}_a\): \(A_a = \mathbf{A}\cdot\mathbf{u}_a\).

EXAMPLE 1Component load on a sensor bracket

A compact sensor bracket is pulled by a 520 N cable directed 35° below the positive x-axis. Determine the signed Cartesian components of the cable force acting on the bracket.

Recognition strategy

This is one known force and one clean planar angle, so rectangular components expose the horizontal mounting load and the downward load without needing a force triangle.

Given information

Cable magnitude: 520 N; direction: 35° clockwise from +x; requested components use +x right and +y upward.

Model and interpretation

Treat the cable force as a vector applied at the bracket. The angle is measured from x, making the x component adjacent; the cable points downward, so its y component is negative.

Step-by-step solution

1. Set signs. Adopt rightward +x and upward +y; the arrow lies in quadrant IV, so the horizontal component is positive and the vertical component is negative.

2. Formulate. Because 35° is measured from the x-axis, write \(F_x=F\cos35^\circ\) and \(F_y=-F\sin35^\circ\) before inserting numbers.

3. Calculate x. Substitution gives \(F_x=(520\,\text{N})\cos35^\circ=426\,\text{N}\), the portion that pulls the bracket toward +x.

4. Calculate y. The vertical component is \(F_y=-(520\,\text{N})\sin35^\circ=-298\,\text{N}\), where the sign records the downward sense.

Verification

Units: Sine and cosine are dimensionless, so both \(520\,\text{N}\cos35^\circ\) and \(520\,\text{N}\sin35^\circ\) retain the required unit of newtons.

Signs and direction: A force below +x must have \(F_x>0\) and \(F_y<0\); the calculated signs therefore reproduce the stated quadrant rather than reversing the cable.

Plausibility: Recombining gives \(\sqrt{426^2+(-298)^2}=520\,\text{N}\) to rounding, and the horizontal component is larger because 35° is closer to the x-axis.

Final result: \( \boxed{\mathbf F=(426\,\mathbf i-298\,\mathbf j)\,\text{N}} \).

Engineering interpretation

The fastener group must transmit both a 426 N lateral pull and a 298 N downward pull. Reporting components makes those separate load paths visible; it does not by itself establish bracket strength or bolt capacity.

EXAMPLE 2Spatial cable force from surveyed points

A 1.80 kN guy cable pulls a mast attachment at \(A(0,0,0)\,\text{m}\) toward ground anchor \(B(3,-4,12)\,\text{m}\). Express the force exerted on A as a Cartesian vector.

Recognition strategy

Endpoint coordinates specify a line of action in space. Forming and normalizing the displacement is safer than trying to infer three direction angles from a perspective drawing.

Given information

\(T=1.80\,\text{kN}\), \(A=(0,0,0)\,\text{m}\), and \(B=(3,-4,12)\,\text{m}\); the force on A points from A toward B.

Model and interpretation

An ideal cable carries tension only and pulls the attachment along AB. Coordinates are metres only for geometry; the unit vector is dimensionless before it is multiplied by tension.

Step-by-step solution

1. Build direction. Subtract tail from head: \(\mathbf r_{AB}=3\mathbf i-4\mathbf j+12\mathbf k\,\text{m}\), preserving the signs associated with the anchor location.

2. Normalize. Its length is \(|\mathbf r_{AB}|=\sqrt{3^2+(-4)^2+12^2}=13\,\text{m}\), so \(\mathbf u_{AB}=(3\mathbf i-4\mathbf j+12\mathbf k)/13\).

3. Scale by tension. Use \(\mathbf F_{AB}=T\mathbf u_{AB}\), which keeps the direction purely geometric while assigning the physical magnitude of 1.80 kN.

4. Resolve. Thus \(\mathbf F_{AB}=0.415\mathbf i-0.554\mathbf j+1.66\mathbf k\,\text{kN}\), rounded consistently from the exact component ratios.

Verification

Units: Dividing a metre displacement by its 13 m length cancels metres, so multiplying the dimensionless unit vector by 1.80 kN correctly leaves every force component in kN.

Signs and direction: The anchor is at positive x, negative y, and positive z from A, exactly matching the \(+,-,+\) signs of the force components for a cable that pulls toward B.

Plausibility: The reconstructed magnitude \(\sqrt{0.415^2+(-0.554)^2+1.66^2}=1.80\,\text{kN}\) to rounding, while the dominant upward component agrees with the 12 m vertical rise.

Final result: \( \boxed{\mathbf F_{AB}=(0.415\mathbf i-0.554\mathbf j+1.66\mathbf k)\,\text{kN}} \).

Engineering interpretation

The cable supplies a substantial upward component but also a 0.554 kN transverse pull that the mast and anchor system must resist. A real guy-system review also checks pretension, anchor soil capacity, fittings, and wind combinations.

EXAMPLE 3Actuator load along a guide rail

A positioning actuator applies \(\mathbf F=(4\mathbf i-3\mathbf j+12\mathbf k)\,\text{kN}\) to a carriage. Its guide rail has direction from one mount to the next of \(2\mathbf i+\mathbf j+2\mathbf k\). Find the signed force along the rail and the magnitude perpendicular to it.

Recognition strategy

The requested axial load is a scalar projection, not a nonperpendicular component resolution. A dot product with a unit rail direction gives the signed load, then the remaining magnitude follows from orthogonal decomposition.

Given information

Applied force components are in kN; rail direction ratio is \(2:1:2\); positive rail sense is \(2\mathbf i+\mathbf j+2\mathbf k\).

Model and interpretation

Model the rail as a straight direction only; no friction or bearing compliance is included. The projection is positive when the actuator tends to load the carriage in the specified rail sense.

Step-by-step solution

1. Normalize rail direction. The direction-vector magnitude is \(\sqrt{2^2+1^2+2^2}=3\), hence \(\mathbf u_r=(2\mathbf i+\mathbf j+2\mathbf k)/3\).

2. Project. Compute \(F_{\parallel}=\mathbf F\cdot\mathbf u_r=[4(2)-3(1)+12(2)]/3=9.67\,\text{kN}\).

3. Find total force. The actuator magnitude is \(|\mathbf F|=\sqrt{4^2+(-3)^2+12^2}=13.0\,\text{kN}\), needed for the perpendicular remainder.

4. Resolve remainder. Orthogonality gives \(F_\perp=\sqrt{13.0^2-9.67^2}=8.69\,\text{kN}\), an unsigned magnitude normal to the rail.

Verification

Units: The rail unit vector has no units, so each dot-product term remains kN; the squared magnitudes in the perpendicular calculation consistently carry \(\text{kN}^2\).

Signs and direction: The positive 9.67 kN projection means the applied force has a component in the selected \(+2,+1,+2\) rail sense, not opposite the stated mounting direction.

Plausibility: \(9.67\,\text{kN}\) is less than the 13.0 kN total and \(\sqrt{9.67^2+8.69^2}=13.0\,\text{kN}\), confirming a valid orthogonal split.

Final result: \( \boxed{F_{\parallel}=+9.67\,\text{kN},\quad F_\perp=8.69\,\text{kN}} \).

Engineering interpretation

The rail sees a 9.67 kN driving component, while bearings and structure must react an 8.69 kN side load. Service-life decisions require manufacturer load ratings, stiffness, alignment tolerance, duty cycle, and safety-factor review beyond this ideal vector calculation.

CHAPTER SUMMARY

Key Takeaways

The thread of the chapter

Forces are vectors, so they combine by the parallelogram law, not by adding numbers. In two dimensions, resolving into x–y components turns force addition into algebra. In three dimensions, Cartesian vectors, position vectors, and unit vectors carry the same idea into space, and the dot product extracts angles and projections that geometry alone makes difficult.

Course Note

Master Chapter 2 before moving on — every later topic (moments, equilibrium, trusses) assumes you can resolve and add force vectors without thinking. If a problem feels hard later, it's usually a Chapter 2 skill that wasn't solid. Time spent here pays back tenfold.

Physical meaning

Components are not extra forces added to a body: they are perpendicular descriptions of the same force. Their signed sums tell where the combined pull acts; a zero component means cancellation along that chosen axis.

Problem-solving strategy

Identify whether the requested quantity is a resultant, a component, a line direction, or a projection.
Draw axes and record where each given angle is measured.
Use coordinates to form a direction vector, then normalize it before multiplying by force magnitude.
Add only like Cartesian components and use \(\operatorname{atan2}(R_y,R_x)\) for a quadrant-safe planar direction.
Audit signs, force units, and whether every unit vector has magnitude one.

Before moving on

Can you state a force vector with unambiguous units and signs?
Can you normalize a position vector and verify the result?
Can you explain whether a dot-product projection is along or opposite a selected member?