Statics · Chapter 4

Force System Resultants

A force applied off the line of action of a point produces a turning effect — a moment. This chapter builds the tools to compute moments in two and three dimensions, define couples, and reduce any system of forces and couples to its simplest equivalent.

Learning objectives

Compute the moment of a force in two and three dimensions.
Find the moment of a force about a specified axis.
Define and evaluate the moment of a couple.
Reduce a nonconcurrent force system to a resultant force and couple moment.
Replace a simple distributed loading with an equivalent resultant force.

SECTION 4.1

Moment of a Force — Scalar Formulation

When a force acts on a body along a line that misses a given point, it tends to rotate the body about that point. Tightening a bolt with a wrench is the everyday case: the harder you push, or the farther out you push, the stronger the turn.

Moment of a force

The tendency of a force to rotate a body about a point or axis. Also called torque. Its magnitude grows with both the force and the perpendicular distance from the point to the force's line of action.

Moment arm

The perpendicular (shortest) distance d from the reference point to the line of action of the force. If the force points straight at the point, the moment arm is zero and no turning occurs.

Fig. 4–1A force F applied to a wrench turns it about point O; the moment arm d is the perpendicular distance from O to the line of action. (Original diagram.)

Moment magnitude (scalar)

\[ M_O = F\,d \]

Units are force times distance — \(\text{N}\cdot\text{m}\) in SI or \(\text{lb}\cdot\text{ft}\) in U.S. customary units. The direction follows the right-hand rule: curl the right-hand fingers in the sense of rotation and the thumb points along the moment axis, perpendicular to the plane of F and d.

Sign convention (2-D)

Counterclockwise moments are positive (directed out of the page, along +z).
Clockwise moments are negative (into the page).
The resultant moment about a point is the algebraic sum of the individual moments.

Resultant moment (2-D)

\[ (M_R)_O = \sum F\,d \]

Instructor Note

Students lose marks most often on the moment arm, not the force. Always extend the line of action as a dashed line first, then drop a perpendicular to it from the point. If you find yourself multiplying by a slanted distance, stop — that is not d.

SECTION 4.2

The Cross Product

Before formulating moments as vectors, we need the cross product — a vector multiplication that yields a third vector perpendicular to the first two.

Cross product

For vectors A and B, the product \(\mathbf{C}=\mathbf{A}\times\mathbf{B}\) is a vector whose magnitude is \(AB\sin\theta\) and whose direction is perpendicular to the plane of A and B, set by the right-hand rule (fingers sweep from A to B, thumb gives C).

Magnitude and direction

\[ \mathbf{C} = \mathbf{A}\times\mathbf{B} = (AB\sin\theta)\,\mathbf{u}_C \]

Fig. 4–2The cross product C is perpendicular to the plane containing A and B. (Original diagram.)

Laws of operation

Not commutative: \(\mathbf{A}\times\mathbf{B} = -\,\mathbf{B}\times\mathbf{A}\). Order matters.
Associative with a scalar: \(a(\mathbf{A}\times\mathbf{B}) = (a\mathbf{A})\times\mathbf{B} = \mathbf{A}\times(a\mathbf{B})\).
Distributive over addition: \(\mathbf{A}\times(\mathbf{B}+\mathbf{D}) = \mathbf{A}\times\mathbf{B} + \mathbf{A}\times\mathbf{D}\).

Applying the definition to the Cartesian unit vectors gives a set of results best recovered with the right-hand rule rather than memorized: \(\mathbf{i}\times\mathbf{j}=\mathbf{k}\), \(\mathbf{j}\times\mathbf{k}=\mathbf{i}\), \(\mathbf{k}\times\mathbf{i}=\mathbf{j}\), while reversing any pair flips the sign and crossing a vector with itself gives zero.

Cartesian cross product (determinant form)

\[ \mathbf{A}\times\mathbf{B} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{vmatrix} \]

Expanded form

\[ \mathbf{A}\times\mathbf{B} = (A_yB_z - A_zB_y)\mathbf{i} - (A_xB_z - A_zB_x)\mathbf{j} + (A_xB_y - A_yB_x)\mathbf{k} \]

Instructor Note

Watch the middle term: the j component carries a leading minus sign when you expand the determinant. Forgetting it is the single most common arithmetic error I see in 3-D moment problems.

SECTION 4.3

Moment of a Force — Vector Formulation

In three dimensions, finding the perpendicular distance is awkward. The cross product sidesteps it entirely.

Moment about a point (vector)

\[ \mathbf{M}_O = \mathbf{r}\times\mathbf{F} \]

Position vector r

A vector drawn from the moment point O to any point on the line of action of F. Because any such point works, the choice of r can be made to simplify the arithmetic.

The magnitude reduces correctly to the scalar result: since \(d = r\sin\theta\), we get \(M_O = rF\sin\theta = F(r\sin\theta) = Fd\), matching the scalar definition. The right-hand rule applied to \(\mathbf{r}\times\mathbf{F}\) fixes the direction, so the order r then F must be kept.

Principle of Transmissibility

A force may be applied at any point along its line of action without changing its moment about a point — so \(\mathbf{M}_O = \mathbf{r}_1\times\mathbf{F} = \mathbf{r}_2\times\mathbf{F} = \mathbf{r}_3\times\mathbf{F}\) for any position vectors reaching the line of action. A force is therefore a sliding vector.

Cartesian moment

\[ \mathbf{M}_O = \mathbf{r}\times\mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ r_x & r_y & r_z \\ F_x & F_y & F_z \end{vmatrix} \]

Resultant moment of a force system

\[ (\mathbf{M}_R)_O = \sum (\mathbf{r}\times\mathbf{F}) \]

SECTION 4.4

Principle of Moments

It is often easier to resolve a force into components and sum their moments than to find the moment of the whole force directly.

Varignon's Theorem (Principle of Moments)

Condition: a force F resolved into components \(\mathbf{F}_1 + \mathbf{F}_2\).

Conclusion: the moment of the force about a point equals the sum of the moments of its components about that same point:

\[ \mathbf{M}_O = \mathbf{r}\times\mathbf{F} = \mathbf{r}\times(\mathbf{F}_1+\mathbf{F}_2) = \mathbf{r}\times\mathbf{F}_1 + \mathbf{r}\times\mathbf{F}_2 \]

It follows directly from the distributive law of the cross product. Named for the French mathematician Pierre Varignon (1654–1722).

Scalar form (2-D)

\[ M_O = xF_y - yF_x \]

Fig. 4–3Resolving F into rectangular components lets each component's moment be summed using its own coordinate arm. (Original diagram.)

Important points

A moment is the tendency to turn about an axis through a chosen point O.
The right-hand rule gives the sense: curled fingers show rotation, thumb shows the moment axis.
Scalar magnitude is \(M_O = Fd\); in 3-D use \(\mathbf{M}_O = \mathbf{r}\times\mathbf{F}\) with r reaching the line of action.

SECTION 4.5

Moment of a Force About a Specified Axis

Sometimes only the turning effect about one particular axis matters — a lug nut can only spin about the axle, regardless of the total moment the wrench produces.

Moment about an axis

The component of the moment that acts along a chosen axis. It is the projection of \(\mathbf{M}_O\) onto that axis, capturing only the part of the turning effect the axis is free to rotate about.

Scalar form

\[ M_a = F\,d_a \]

Vector form — scalar triple product

\[ M_a = \mathbf{u}_a\cdot(\mathbf{r}\times\mathbf{F}) = \begin{vmatrix} u_{ax} & u_{ay} & u_{az} \\ r_x & r_y & r_z \\ F_x & F_y & F_z \end{vmatrix} \]

Here \(\mathbf{u}_a\) is the unit vector along the axis and r runs from any point on the axis to any point on the line of action of F. A positive result means \(M_a\) acts in the sense of \(\mathbf{u}_a\); a negative result means it acts opposite.

Fig. 4–4The moment about axis a is the projection of \(\mathbf{M}_O = \mathbf{r}\times\mathbf{F}\) onto the axis. (Original diagram.)

Important points

A force whose line of action intersects the axis, or runs parallel to it, produces zero moment about that axis.
If \(M_a\) comes out negative, its sense is opposite to \(\mathbf{u}_a\).
As a Cartesian vector: \(\mathbf{M}_a = M_a\,\mathbf{u}_a\).

SECTION 4.6

Moment of a Couple

Couple

Two parallel forces equal in magnitude, opposite in direction, separated by a perpendicular distance d. The net force is zero, so a couple produces pure rotation (or the tendency for it) with no translation.

Couple moment

The turning effect of a couple. It is a free vector: it depends only on the separation between the two forces, not on where it is applied, so it produces the same rotation anywhere on the body.

Fig. 4–5A couple consists of two equal, opposite, non-collinear forces a distance d apart. (Original diagram.)

Scalar magnitude

\[ M = F\,d \]

Vector form

\[ \mathbf{M} = \mathbf{r}\times\mathbf{F} \]

In the vector form, r runs from any point on the line of action of one force to any point on the line of action of the other. Taking moments about a point on one force makes that force's contribution zero, simplifying the work.

Equivalent couples

Two couples are equivalent if their moments have the same magnitude and direction — regardless of the individual force magnitudes or spacing. A larger force over a shorter arm can match a smaller force over a longer arm.

Resultant couple moment

\[ \mathbf{M}_R = \sum (\mathbf{r}\times\mathbf{F}) \]

Instructor Note

Because a couple moment is a free vector, you can slide it anywhere on the body in a free-body diagram. This is the property that makes the "force + couple" reduction in §4.7 possible — keep it in mind as you move into the next section.

SECTION 4.7

Simplification of a Force and Couple System

Any collection of forces and couples on a body can be replaced by a single resultant force at a chosen point plus a single resultant couple moment, producing identical external effects.

Equivalent system

A simpler force–couple system that produces the same external effects — the same translation and rotation of a free body, or the same support reactions of a fixed body — as the original system.

Moving a force off its line of action to a new point P requires adding a couple moment equal to the moment of that force about P, keeping the system equivalent. Couple moments, being free vectors, simply move to the new point unchanged.

Equivalent force and couple at point O

\[ \mathbf{F}_R = \sum \mathbf{F} \qquad (\mathbf{M}_R)_O = \sum \mathbf{M}_O + \sum \mathbf{M} \]

Scalar equations (coplanar system)

\[ (F_R)_x = \sum F_x \qquad (F_R)_y = \sum F_y \qquad (M_R)_O = \sum M_O + \sum M \]

Important points

Force is a sliding vector (transmissibility): same external effect anywhere on its line of action.
A couple moment is a free vector: same external effect anywhere on the body.
Moving a force to a point off its line of action requires adding the couple moment of that force about the point.

SECTION 4.8

Further Simplification of a Force and Couple System

When the resultant force and resultant couple moment happen to be perpendicular, the couple can be eliminated by moving the force — reducing the whole system to a single resultant force.

Concurrent force system

All lines of action meet at a common point, so the system produces no moment there and reduces to a single resultant force \(\mathbf{F}_R = \sum\mathbf{F}\) acting at that point.

Coplanar force system

All forces lie in one plane; the resultant moment is perpendicular to it. The couple is absorbed by shifting \(\mathbf{F}_R\) a distance \(d = (M_R)_O / F_R\) from the point.

Parallel force system

All forces are parallel (say, along z). The resultant force is parallel to them and acts through a point located so its moment matches the system's: \(d = M_O / F_R\).

Wrench (screw)

When \(\mathbf{F}_R\) and \((\mathbf{M}_R)_O\) are not perpendicular, the system cannot reduce to a single force. The simplest equivalent is a wrench: a resultant force plus a collinear couple moment, which together translate and twist the body about one axis.

Fig. 4–6A coplanar resultant force and couple (left) reduce to a single force shifted a distance d (right). (Original diagram.)

Location of single resultant (coplanar / parallel)

\[ (M_R)_O = F_R\,d \quad\Rightarrow\quad d = \frac{(M_R)_O}{F_R} \]

Important point

Concurrent, coplanar, and parallel systems always reduce to a single resultant force at a specific point.
Every other system reduces to a wrench — a force plus a collinear couple moment.

SECTION 4.9

Reduction of a Simple Distributed Loading

Loads spread over a surface — wind on a sign, water in a tank, sand on a floor — can often be described along a single axis and replaced by one equivalent force.

Distributed loading

A load spread over a length or area rather than applied at a point. Along a single axis it is described by an intensity function \(w = w(x)\), measured in N/m or lb/ft (obtained by multiplying a pressure \(p(x)\) by the member's width).

Fig. 4–7A distributed load is replaced by a resultant equal to the area under the curve, acting through its centroid C. (Original diagram.)

Key results

The resultant force equals the area under the loading diagram.
Its line of action passes through the centroid (geometric center) of that area.

Magnitude of resultant

\[ F_R = \int_L w(x)\,dx = \int_A dA = A \]

Location of resultant

\[ \bar{x} = \frac{\displaystyle\int_L x\,w(x)\,dx}{\displaystyle\int_L w(x)\,dx} = \frac{\displaystyle\int_A x\,dA}{\displaystyle\int_A dA} \]

For common shapes — rectangles, triangles, trapezoids — the area and centroid can be read from standard tables rather than integrated, and composite shapes can be split into simple parts.

Instructor Note

For a triangle the resultant is \(\tfrac{1}{2}(\text{base})(\text{height})\) acting one-third of the base from the tall end; for a rectangle it is \((\text{base})(\text{height})\) at the midpoint. Memorize these two and you can break almost any practical beam load into pieces without integrating.