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Forces and Moments: Forces

Definition of force

In general, force is perceived as push or pull. Forces arise from different sources/causes like gravity, contact forces between bodies, pressure force by fluids on surface of solids, friction forces, electrostatic forces on charged particles, etc.

Physically, force has a magnitude and a direction in the space, and therefore, it is a vector quantity called the force vector. In this regard, the terms force and force vector are equivalent. Calculations regarding forces follow the concepts and methods presented in Chapter 2.

The SI unit of force is Newton (N). A force vector has the dimension \frac{\text{mass}\cdot \text{length}}{\text{time}^2}, being equal to \frac{\text{kg}\cdot \text{m}}{\text{s}^2} in the SI unit system (see Section 1.2). The direction of a force vector can be expressed using a unit vector. This unit vector shows the direction in the (geometrical) space. We can also have a graphical representation of a force vector. This graphical representation is by the means of an arrow as discussed in Section 2.2. The length of an arrow representing a force vector can be proportional to the vector’s magnitude following some scale.

Denoting a force vector as \bold F, we can write,

(3.1)   \[\bold F=|\bold F|\bold u_F \]

where |\bold F| is the magnitude of the force and \bold u_F is the unit vector expressing the direction of \bold F. Note that F has the unit of Newton, and the unit vector has no dimension and unit. Equation 3.1 readily implies that,

(3.2)   \[\bold u_F = \frac{\bold F}{|\bold F|} \]

Line of action of a force. The line of action of a force or a force vector \bold F is a line that is collinear with the force vector (arrow representing the force).

Like any vector, a force vector \bold F can be resolved into its Cartesian components using the CVN,

(3.3)   \[\bold F = F_x\bold i + F_y\bold j + F_z\bold k  \]

where (F_x,F_y,F_z) are the scalar components.

A graphical representation of \bold F in the Cartesian frame is shown in Fig. 3.1.

Fig. 3.1. A graphical representation of a force vector and its rectangular and scalar components in a Cartesian frame.
Fig. 3.1 A graphical representation of a force vector and its rectangular and scalar components in a Cartesian frame.

We say a body is subjected to a force (or a system of forces), or a force acts on a body. This expresses the fact that a force is acting or exerted at a particular location of the body. Therefore, the location at which a force vector is acting is deterministically important and not arbitrary. This is graphically demonstrated by drawing the force vector with its tail or head touching the point at which the force acts (Fig. 3.2a). Depending on the tail or the head of the vector touching the point, this representation may also imply a sense of pull or push by the force (Fig 3.2a).

Parallel forces. Two or more forces are called parallel if their line of actions are parallel. For example, \bold F_1, \bold F_2, and \bold F_3 are parallel in Fig. 3.2b.

Collinear forces. Two or more forces are called collinear if they share the same line of action. Consequently, collinear forces are parallel. However, parallel forces may not be collinear. For example, \bold F_4, \bold F_5, and \bold F_6 are collinear in Fig. 3.2b.

Concurrent forces. Two or more forces are called concurrent if their lines of action intersect at a common point. For example, \bold F_7, \bold F_8, and \bold F_9 are concurrent in Fig. 3.2b.

Fig. 3.2 (a) Graphical representations of forces acting at different points of a body, (b) parallel, collinear, and concurrent forces.

Remark: a force always accompanies a point at which the force is acting.

Concentrated force. In reality, forces are continuously distributed over areas. For example, we perceive and feel that the ground reaction force due to our weight is distributed on the soles of our feet; we do not feel a sharp feeling of a force there. In fact, all forces in nature are distributed, and there is no force concentrated at (or acting on) a point. However, we idealize or approximate a distributed force by a concentrated force. This idealization is performed when the area on which the load is distributed is relatively small compared to the size of the body.

Resultant force

Let \bold F_1, \bold F_2\dots ,\bold F_n be a system of n forces (set of forces), their resultant or the resultant force \bold F_R is a vector sum over the forces (see Section 2.4),

(3.4)   \[ \bold F_R = \bold F_1 + \bold F_2 + \dots + \bold F_n=\sum_{i=1}^n\bold F_i \]

Remark: vector addition of forces does not depend on the location of forces on a body.

The calculation procedure of resultants of forces can be presented separately for coplanar and spatial force systems.

Coplanar force systems (Two-dimensions)

A coplanar force system or a system of coplanar forces is a set of forces with their lines of action all in the same plane. Therefore, each force in a coplanar force system has two Cartesian components in a plane Cartesian coordinate system as presented in Section 2.4.2. With x and y being the axes of a Cartesian frame, a planar \bold F is written in CVN as,

(3.5a)   \[\bold F = \bold F_x + \bold F_y = F_x\bold i + F_y\bold j \]

The the following equations hold for the magnitude, and scalar components of planar forces,

(3.5b)   \[F=|\bold F| = \sqrt {F_x^2 + F_y^2},\quad \tan \theta=\frac{|F_y|}{|F_x|}\]

where 0\le\theta<90^\circ is the angle of \bold F with the x axis as shown in Fig. 3.3.

Fig. 3.3 A planar force.

Note that a new notation for the magnitude of \bold F (not its components) is defined as F being a non-bold capital letter. We will used F instead of |\bold F| when denoting the magnitude of a planar force. This labeling approach applies to the two-dimensional calculation of any vectors for simplicity.

To determine the resultant force of a coplanar force system, the equations presented in Sections 2.4.2 and 2.4.3 are implemented. Here, we repeat the equations for a force system.

For a system of coplanar forces \bold F_1, \bold F_2, \dots, \bold F_n, if each force is expressed in CVN as,

    \[\begin{split} \bold F_1&=F_{1x}\bold i + F_{1y}\bold j\\ \bold F_2&=F_{2x}\bold i + F_{2y}\bold j\\ &\vdots\\ \bold F_n&=F_{nx}\bold i + F_{ny}\bold j \end{split} \]

the resultant force \bold F_R is determined as,

(3.6a)   \[\bold F_R = \sum_{i=1}^n\bold F_i = \left(\sum_{i=1}^n F_{ix}\right)\bold i + \left(\sum_{i=1}^n F_{iy}\right )\bold j=(\bold F_R)_x + (\bold F_R)_y\]

which is written in the following concise way,

(3.6b)   \[ \bold F_R = \sum \bold F = (\sum F_x)\bold i +(\sum F_y)\bold j  \]


Determine the resultant force of the cable forces shown in the figure. Obtain the force vector, its magnitude, and its direction measured from the positive x-axis shown.


    \[\begin{split}\bold F_1&=\{20\cos 30^\circ \bold i+20\sin 30^\circ\bold j\}=\{17.3\bold i+10.0\bold j\}\text{ kN}\\\bold F_2&=\{0\bold i + 10\bold j\}\text{ kN}\\\bold F_3&=\{5\cos 45^\circ\bold i+5\sin 45^\circ\bold j\}=\{3.5\bold i - 3.5\bold j\}\text{ kN}\\\therefore \bold F_R&=\bold F_1 + \bold F_2 + \bold F_3= \{17.3+0+3.5\}\bold i+\{10.0+10.0-3.5\}\bold j\\&=\{20.8\bold i+16.5\bold j\}\text{ kN}\\\implies F_R&=|\bold F_R|=\sqrt{20.8^2+16.5^2}=26.6 \text{ kN}\\\\ \theta&=\arctan\frac{16.5}{20.8}=38.4^\circ\end{split}\]

Spatial force systems (three-dimensions)

A spatial system of forces, is a general force system with forces not necessarily lying in one plane. A spatial force \bold F can be represented by three Cartesian components as formulated in Section 2.4.4. With x, y and z being the axes of a Cartesian frame, \bold F is written in CVN as,

(3.7a)   \[\bold F = \bold F_x + \bold F_y + \bold F_z= F_x\bold i + F_y\bold j + F_z\bold k \]

The following equations hold for the magnitude, and scalar components of the force,

(3.7b)   \[\begin{split}|\bold F|&=\sqrt{F_x^2 +F_y^2 +F_z^2}\\\cos \alpha=\frac{F_x}{|\bold F|} \quad &\cos \beta=\frac{F_y}{|\bold F|} \quad \cos \gamma=\frac{F_z}{|\bold F|} \end{split}\]

where 0\le\alpha,\beta,\gamma \le 180^\circ are the coordinate direction angles as shown in Fig. 3.4.

Fig. 3.4 A spatial force.

The resultant force \bold F_R of a system of spatial forces containing \bold F_1,\bold F_2,\dots,\bold F_n expressed in CVN is,

(3.8)   \[ \bold F_R = \sum \bold F = (\sum F_x)\bold i +(\sum F_y)\bold j +(\sum F_z)\bold k \]


A box beam is subjected to the forces as shown. Express each force in CVN and determine the magnitude and coordinate direction angles of the resultant force.


To express the \bold F_1, the scalar components of \bold F_1 are determined and we write \bold F=F_{1x}\bold i+F_{1y}\bold j + F_{1z}\bold k.

    \[\begin{split}F_{1x} &= 0 \rm{N}\\F_{1y} &= 500 \frac{3}{5} = 300 \rm{N}\\F_{1z} &= -500 \frac{4}{5} = -400 \rm{N}\\\implies \bold F_1&=\{0\bold i + 300\bold j - 400\bold k\}\rm N\end{split}\]

And for \bold F_2,

    \[\begin{split}F_{2x} &= 150\cos(60^\circ)= 75\rm{N}\\F_{2y} &= 150\cos(150^\circ)= -130\rm{N}\\F_{2z} &= 150\cos(60^\circ)= 75\rm{N}\\\implies \bold F_1&=\{75\bold i -130\bold j +75\bold k\}\rm N\end{split}\]


    \[\bold F_R=\bold F_1 + \bold F_2= (0+75)\bold i + (300-130)\bold j+ (- 400+75)\bold k= \{75\bold i + 170\bold j -325\bold k\}\rm N\]

Leading to,

    \[|\bold F_R|=\sqrt{75^2+170^2+325^2}= 374 \rm N\]

The direction angles of \bold F_R are,

    \[\begin{split}\cos \alpha &= \frac{75}{374.4} \Rightarrow \alpha = 78.5^\circ \\\cos \beta &= \frac{170}{374.4} \Rightarrow \beta = 63.0^\circ \\\cos \gamma &= \frac{-325}{374.4} \Rightarrow \gamma = 150.1^\circ\end{split}\]

Force directed along a line

In many problems, a force is directed along a line-shape physical object or structure like a cable or a bar. In such a case, the line of action of the force is collinear with the direction of the object (Fig. 3.5). The line of action of the force is, therefore, determined by two points along the structure.

To determine a force directed along a line, we initially choose (or consider a given) a Cartesian coordinate system. Choosing a point as the coordinate system origin (if not given) is arbitrary, however, we should try to wisely choose a point that will simplify the calculations. Then, two points, say A and B, on the line of action or the object are identified. These are points with known coordinates or points that their coordinates are easy enough to calculate (Fig. 3.5). Considering the position vectors of A and B, we find the vector \bold r_{AB} as,

(3.9)   \[\bold r_{AB}=\bold r_B-\bold r_A=(x_B-x_A)\bold i+(y_B-y_A)\bold j+(z_B-z_A)\bold k  \]

The unit vector \bold u_F is obtained by normalizing \bold r_{AB} by its magnitude (length), \bold u_F=\frac{\bold r_{AB}}{r_{AB}}, assuming AB is the direction of the force. Consequently, the force vector \bold F can be written as,

(3.10)   \[\bold F=|\bold F|\bold u_F=|\bold F|\frac{\bold r_{AB}}{|\bold r_{AB}|}=|\bold F|\frac{\bold (x_B-x_A)\bold i+(y_B-y_A)\bold j+(z_B-z_A)\bold k}{\sqrt{(x_B-x_A)^2+(y_B-y_A)^2+(z_B-z_A)^2}}\]

As shown in Fig. 3.5, the graphical representation of \bold F, \bold r_{AB} and \bold u_{AB} are collinear. Note that \bold F has the dimension of force, \bold r_{AB} has the dimension of length, and the unit vector \bold u_F has no dimension. Despite having different dimensions and units, all the vector quantities can be graphically shown in one Cartesian coordinate system (frame).

Fig. 3.3. Obtaining the force vector of a force directed along a line.
Fig. 3.5 Obtaining the force vector of a force directed along a line.

The following interactive tool illustrates obtaining a force vector directed along a line with two known points.


Determine the magnitude and coordinate direction angles of the resultant force. |\bold F_1|= 400\rm N, |\bold F_2|= 200 \rm N and |\bold F_3|=300 \rm N.


The following position vectors are needed to determine the directions of the forces.

    \[\begin{split}\bold u_{DA} &= \frac{\bold r_A - \bold r_D}{| \bold r_{DA} |} = \frac{-3 \bold i + 2 \bold j - 20 \bold k}{\sqrt{3^2 + 2^2 + 20^2}}\\\bold u_{DB} &= \frac{\bold r_B - \bold r_D}{| \bold r_{DB} |} = \frac{4\bold i + 4 \bold j - 20 \bold k}{\sqrt{4^2 + 4^2 + 20^2}}\\\bold u_{DC} &= \frac{\bold r_C - \bold r_D}{| \bold r_{DC} |} = \frac{5 \bold i - 3 \bold j - 20\bold k}{\sqrt{5^2 + 3^2 + 20^2}}\end{split}\]

The forces are found as,

    \[\begin{split}\bold F_1 &= |\bold F_1|\bold u_{DA} = 400 \frac{-3 \bold i + 2 \bold j - 20 \bold k}{\sqrt{3^2 + 2^2 + 20^2}} \\&=\{-59.04 \bold i + 39.37 \bold j - 393.65 \bold k\}\rm N\\\bold F_2 &= |\bold F_2|\bold u_{DB} =  200 \frac{4\bold i + 4 \bold j - 20 \bold k}{\sqrt{4^2 + 4^2 + 20^2}} \\&=\{38.49 \bold i + 38.49 \bold j - 192.45 \bold k\} \rm N\\\bold F_3 &= |\bold F_3|\bold u_{DC}= 300 \frac{5 \bold i - 3 \bold j - 20\bold k}{\sqrt{5^2 + 3^2 + 20^2}}\\&=\{72 \bold i - 43.20 \bold j - 288 \bold k\} \rm N\end{split}\]

Therefore, the resultant force \bold F_R is,

    \[\begin{split}\bold F_R &= \bold F_1 + \bold F_2 + \bold F_3\\&=(-59.04 + 38.49 + 72 )\bold i + (39.37 + 38.49 - 43.2)\bold j + (-393.65 - 192.45 - 288) \bold k \\&=\{51.45 \bold i +34.66 \bold j - 874.10 \bold k\} \rm N\\|\bold F_R| &= 876.72 \rm N\end{split}\]

The coordinate direction angles \alpha, \beta, and \gamma (the angls of the resultant force with x, y, and z axes respectively) are, therefore, determined as,

    \[\begin{split}\alpha &= \arccos (\frac{51.45}{876.72}) = 86.64^{\circ} \\\beta &= \arccos (\frac{34.66}{876.72}) = 87.70^{\circ} \\\gamma &= \arccos (\frac{-874.10}{876.72}) = 175.94^{\circ}\end {split}\]

Context: Force directed along a Line
  • There is a relationship between position vectors (e.g. member length and angles) and force vectors (e.g. force carried by a member).
  • In two dimensional systems, simple trigonometric rules can relate vector components and the angle between vectors. This approach is much more complicated in 3-D systems, so another approach is used instead: dot products.
  • Force directed along a line is most easily visualized in cables but occurs in many other members (such as columns, struts, and truss members).
  • Engineers select member start and end points based on many factors (e.g. aesthetics, materials, ease of construction) but the most important factor is optimizing how force is carried through the member. For instance, in a system supported by many cables, engineers may proportion cable start and end points to ensure that each cable carries approximately the same forces under the expected loads.
Applications: Why is force along a line important?

The Port Mann Bridge between Surrey and Coquitlam (Vancouver area) opened in 2012 and has one of the longest spans of a cable stayed bridge in North America (470 m) (Fig. 3.6a). To support the roadway, dozens of steel cables connect to one of two large reinforced concrete towers.

Engineers need to consider how much force is carried in each cable to ensure that the tower is not excessively loaded on one side (which would cause the tower to bend or twist). Since each cable acts at a different angle, engineers used force along a line concepts to determine the force in each cable and evaluate if the tower is loaded evenly and ensure that no cables are overloaded (i.e. at risk of snapping). For added complexity, engineers also need to account for how the cables deform (i.e. stretch) when they are under load to ensure they share loads as designed. Deformations are discussed in future courses.

Note that cable stayed bridges are commonly mistaken as suspension bridges. In cable stayed bridges, cables are connected directly between the roadway and the supporting towers (Fig. 3.6b) while suspension bridges (e.g. Golden Gate Bridge) have relatively small vertical cables (‘suspenders’) that transfer loads (weight of road and the traffic) into large curved cables that rest on top of towers and are anchored (held) to the ground (usually via giant concrete anchorage blocks) at either end of the bridge.

Fig. 3.6 (a) Cables connecting to a tower as part of the Port Mann Bridge between Surrey and Coquitlam, (b) general layout of a cable stayed bridge, and (c) General layout of a suspension bridge. (Photo by D. Tomlinson)

Force component using dot product

The component of a force along an arbitrary axis can be determined using the dot product. Let \bold F be a force and l an axis with its positive direction defined by a unit vector \bold u as shown in Fig. 3.7. Then, the component of \bold F along l is denoted by \bold F_u and determined by the orthogonal projection of \bold F onto l. Using the dot product to formulate the orthogonal projection (see Chapter 2), \bold F_u is calculated as,

(3.11)   \[\bold F_u=(\bold F\cdot\bold u)\bold u \]

Fig. 3.7 Component of a force along an arbitrary axis.

Note that F_u = \bold F\cdot\bold u is the scalar component of \bold F along the l axis. Therefore, we can write,

    \[\bold F_u=F_u\bold u\]


Consider the flagpole shown in the figure. The pole is 15 meters high and stabilized by a cable/rope as shown. Due to the wind, a tension force \bold F with a magnitude of 520\rm N develops in the cable. Determine the force developed along the pole.


1- determine the position vector \bold r_{PG}.

    \[\bold r_{PG} = \bold r_{OG}-\bold r_{OP} = 8\bold i - 4\bold j -15 \bold k\]

2- Find \bold F in CVN.

    \[\begin{split}|\bold r_{PG}|&= \sqrt {8^2 + 4^2 +15^2} = 17.46 \text{ m}\\\bold F &= |\bold F|\frac{\bold r_{PG}}{|\bold r_{PG}|}=510 \frac{8\bold i - 4\bold j -15\bold k}{17.46}\text{N}\\\implies \bold F &=\{234\bold i -117\bold j -438\bold k\}\text {N}\end{split}\]

3- Project \bold F onto the z axis.

i) Find the unit vector on z axis.

    \[\bold u=0\bold i + 0\bold j + 1 \bold k\]


    \[\bold F_z = (\bold F\cdot\bold u)\bold u=\left( (234)(0)+(-117)(0)+(-438)(1)\right ) \bold u= -438\bold u \text { N}\]

where -438\bold u indicates that \bold F_u in in the opposite direction of \bold u.

As in this example \bold u = \bold k coincidentally, we can write \bold F_u=\bold F_z=\bold -438\bold k\text { N}.