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1.5 · Use position and unit vectors to describe force directions

Learn to use position and unit vectors to describe force directions through clear examples and targeted practice.

University of Alberta ENGG 130: Engineering Mechanics: Statics

Force Vectors and Mechanics Foundations

ENGG 130 Engineering Mechanics: Statics — Study topic 1.5

A force has both a magnitude and a direction. When its direction is given by a line between two points, coordinates provide a direct way to describe it. First form a position vector pointing from the force’s point of application toward the other point. Then divide by the position vector’s length to obtain a unit vector. Multiplying that unit vector by the force magnitude gives the force vector. This lesson develops that process and uses it in statics examples. Unless stated otherwise, use SI units and the usual Cartesian axes: positive xx to the right and positive yy upward.

What you will learn

  • Find a directed position vector between two points from their coordinates.
  • Convert a position vector into a unit vector that describes a force direction.
  • Use a unit vector and a force magnitude to write force components.
  • Apply vector components consistently in a planar equilibrium calculation.

1. From coordinates to a directed position vector

A position vector describes a displacement from one point to another. Let the starting point be A(xA,yA)A(x_A,y_A) and the ending point be B(xB,yB)B(x_B,y_B). The vector from AA to BB is found by subtracting the starting coordinates from the ending coordinates. Its components tell how far to move horizontally and vertically.
The order matters: the vector from AA to BB is generally not the vector from BB to AA. Reversing the order reverses both component signs. For a cable force acting at AA and pulling toward BB, use the vector from AA to BB. For a cable force acting at BB and pulling toward AA, use the vector from BB to A.
Coordinates may be given in metres, millimetres, or another length unit. Keep the coordinate units consistent. A unit vector has no units, so its direction remains the same whether the coordinate differences were calculated in metres or millimetres.
rAB=(xB−xA)i+(yB−yA)j\mathbf{r}_{AB}=(x_B-x_A)\mathbf{i}+(y_B-y_A)\mathbf{j}
  • A directed position vector starts at the point of force application and points along the force direction.
  • Subtract starting coordinates from ending coordinates.
  • Reversing the point order reverses the vector.

2. From a position vector to a force vector

The length of a planar position vector follows from the Pythagorean theorem. Dividing each component by that length gives a unit vector: a vector with length one that points in the same direction. This is useful because it separates a force’s direction from its magnitude.
If a force has magnitude FF and acts from AA toward BB, multiply its magnitude by the unit vector along ABAB. The resulting components are signed: a negative component means the force points opposite to the positive direction of that axis.
A quick check is to calculate the unit vector’s length. The sum of the squares of its components should equal one. Also check the signs against a sketch or the point coordinates. For example, if the destination lies left and above the starting point, the horizontal component is negative and the vertical component is positive.
uAB=rAB∥rAB∥,F=FuAB\mathbf{u}_{AB}=\frac{\mathbf{r}_{AB}}{\lVert\mathbf{r}_{AB}\rVert},\qquad \mathbf{F}=F\mathbf{u}_{AB}
  • The position-vector length is positive.
  • A unit vector has length one and no units.
  • Force components equal magnitude times the corresponding unit-vector components.

3. Using vector directions in statics

For a particle in planar equilibrium, draw the particle and all applied forces, then choose horizontal and vertical axes. Write each force as a vector before adding components. A force whose direction is known from two points can be written using the position-vector method, even when its angle is not specified.
Equilibrium requires the vector sum of forces to be zero. In component form, the sum of horizontal components and the sum of vertical components must each be zero. These equations can determine unknown force magnitudes or support reactions when the system is statically determinate.
For a rigid body, force balance alone is not sufficient: moments must also balance. Use the force’s point of application when calculating its moment. In the beam example below, the force direction comes from coordinates, and the independent moment check confirms the complete equilibrium calculation.
∑Fx=0,∑Fy=0,∑MA=0\sum F_x=0,\qquad \sum F_y=0,\qquad \sum M_A=0
  • Keep the assumed force direction, diagram, and component signs consistent.
  • For particle equilibrium, require zero resultant force.
  • For a rigid body, verify both force balance and moment balance.

Worked example

1. A force along a cable between two points

A cable pulls a ring at A(1,2)A(1,2) m toward an anchor at B(4,6)B(4,6) m. Find the unit vector along the cable and the force components if the cable tension is 250250 N.
  1. Set the direction
    The ring is the system. Since the cable pulls from A toward B, subtract A’s coordinates from B’s coordinates. Both coordinate differences are positive, matching the right-and-up direction shown.
    rAB=(4−1)i+(6−2)j=(3i+4j) m\mathbf{r}_{AB}=(4-1)\mathbf{i}+(6-2)\mathbf{j}=(3\mathbf{i}+4\mathbf{j})\,\mathrm{m}
  2. Normalize the direction
    Find the length of the position vector, then divide each component by that length. The metre units cancel, leaving a unit vector.
    ∥rAB∥=5 m,uAB=0.6i+0.8j\lVert\mathbf{r}_{AB}\rVert=5\,\mathrm{m},\qquad \mathbf{u}_{AB}=0.6\mathbf{i}+0.8\mathbf{j}
  3. Apply the tension
    Multiply the unit vector by the tension. The horizontal and vertical components are both positive because the cable pulls right and upward.
    T=250uAB=(150i+200j) N\mathbf{T}=250\mathbf{u}_{AB}=(150\mathbf{i}+200\mathbf{j})\,\mathrm{N}
Answer: The cable’s unit vector from A to B is 0.6i+0.8j0.6\mathbf{i}+0.8\mathbf{j}. Its force components on the ring are 150150 N right and 200200 N upward.
Check: The unit-vector length is 0.62+0.82=1\sqrt{0.6^2+0.8^2}=1. The force magnitude is 1502+2002=250\sqrt{150^2+200^2}=250 N, as required.

Worked example

2. A cable pulling down and left

A small ring at P(5,3)P(5,3) m is pulled by a cable connected to Q(1,0)Q(1,0) m. Write the force vector if the tension is 500500 N.
  1. Form the directed vector
    The force acts at P and points toward Q, so use Q minus P. The negative coordinate differences correctly indicate left and downward directions.
    rPQ=(1−5)i+(0−3)j=(−4i−3j) m\mathbf{r}_{PQ}=(1-5)\mathbf{i}+(0-3)\mathbf{j}=(-4\mathbf{i}-3\mathbf{j})\,\mathrm{m}
  2. Find the unit vector
    The vector length is 55 m. Dividing each signed component by this length preserves the direction while giving a vector of length one.
    uPQ=−0.8i−0.6j\mathbf{u}_{PQ}=-0.8\mathbf{i}-0.6\mathbf{j}
  3. Calculate force components
    Multiply by the tension magnitude. The negative components mean the force acts opposite to the positive xx and yy axes.
    F=500uPQ=(−400i−300j) N\mathbf{F}=500\mathbf{u}_{PQ}=(-400\mathbf{i}-300\mathbf{j})\,\mathrm{N}
Answer: The force vector is −400i−300j-400\mathbf{i}-300\mathbf{j} N: 400400 N left and 300300 N downward.
Check: Its magnitude is (−400)2+(−300)2=500\sqrt{(-400)^2+(-300)^2}=500 N. The signs agree with Q being left and below P.

Worked example

3. Position-vector direction in beam equilibrium

A horizontal beam is pinned at A and extends to B, 44 m to the right. A cable at B pulls toward an anchor whose position relative to B gives the directed vector (−3i+4j)(-3\mathbf{i}+4\mathbf{j}) m. The cable tension is 1010 kN. A 1616 kN downward load acts 22 m from A. Find the pin reactions and verify equilibrium.
  1. Define the system and cable direction
    Isolate the beam as a rigid body. The pin can exert horizontal and vertical reactions, and the cable pulls at B toward its anchor. Normalize the given position vector before multiplying by the tension.
    u=−3i+4j5=−0.6i+0.8j,T=(−6i+8j) kN\mathbf{u}=\frac{-3\mathbf{i}+4\mathbf{j}}{5}=-0.6\mathbf{i}+0.8\mathbf{j},\qquad \mathbf{T}=(-6\mathbf{i}+8\mathbf{j})\,\mathrm{kN}
  2. Apply force equilibrium
    Take right and upward as positive. The cable contributes −6-6 kN horizontally and +8+8 kN vertically; the applied load contributes −16-16 kN vertically.
    ∑Fx=0: Ax−6=0,∑Fy=0: Ay+8−16=0\sum F_x=0:\ A_x-6=0,\qquad \sum F_y=0:\ A_y+8-16=0
  3. Apply moment equilibrium
    Take counterclockwise moments about A as positive. The pin reactions have zero moment about A. The cable’s vertical component acts 44 m from A, while the downward load acts 22 m from A.
    ∑MA=0: (8 kN)(4 m)−(16 kN)(2 m)=0\sum M_A=0:\ (8\,\mathrm{kN})(4\,\mathrm{m})-(16\,\mathrm{kN})(2\,\mathrm{m})=0
  4. Solve and verify
    The force equations give a rightward horizontal reaction and an upward vertical reaction. Substituting all forces confirms horizontal and vertical balance; the moment equation above independently confirms rotational balance.
    Ax=6 kN,Ay=8 kNA_x=6\,\mathrm{kN},\qquad A_y=8\,\mathrm{kN}
Answer: The pin reactions are Ax=6A_x=6 kN to the right and Ay=8A_y=8 kN upward.
Check: Force balance: 6−6=06-6=0 kN horizontally and 8+8−16=08+8-16=0 kN vertically. Moment balance about A: 32−32=032-32=0 kN·m.

Common mistakes and how to avoid them

Subtracting coordinates in the reverse order from the force direction.
Correction: Start at the point where the force acts and end at the point toward which it pulls.
Using the position vector directly as the force vector.
Correction: First divide by the position-vector length to obtain a unit vector, then multiply by the force magnitude.
Dropping negative signs when calculating components.
Correction: Keep the signs from the directed coordinate differences; they show whether a component points with or against an axis.
Treating the unit vector’s components as force components.
Correction: Unit-vector components describe direction only. Multiply them by the force magnitude to obtain force components with units.

Lesson summary

  • Construct the position vector by subtracting starting coordinates from ending coordinates.
  • Divide the position vector by its length to obtain a unit vector in the same direction.
  • Multiply a force magnitude by its unit vector to find signed force components.
  • Use component equilibrium equations and, for a rigid body, check moments as well as forces.

Check your understanding

Question 1

A force acts from C(2,1)C(2,1) m toward D(2,5)D(2,5) m. What is its unit vector?
  1. j\mathbf{j}
  2. −j-\mathbf{j}
  3. i\mathbf{i}
  4. −i-\mathbf{i}
Show answer and explanation
j\mathbf{j}
The directed vector is (0i+4j)(0\mathbf{i}+4\mathbf{j}) m, so its length is 44 m and its unit vector is j\mathbf{j}.

Question 2

A 300300 N force acts along the unit vector −0.6i+0.8j-0.6\mathbf{i}+0.8\mathbf{j}. What are its components?
  1. 180180 N right and 240240 N up
  2. 180180 N left and 240240 N up
  3. 240240 N left and 180180 N down
  4. 300300 N left and 300300 N up
Show answer and explanation
180180 N left and 240240 N up
Multiplying each unit-vector component by 300300 N gives −180-180 N horizontally and +240+240 N vertically.

Key terms

Position vector
A vector describing the displacement from one specified point to another.
Unit vector
A vector of length one that specifies a direction without specifying a physical magnitude.
Component
The signed contribution of a vector along one coordinate axis.
Force equilibrium
A condition in which the vector sum of forces on a system is zero.

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