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2.1 · Calculate the moment of a force about a point

Learn to calculate the moment of a force about a point through clear examples and targeted practice.

University of Alberta ENGG 130: Engineering Mechanics: Statics

Force-System Resultants

ENGG 130 Engineering Mechanics: Statics — Study topic 2.1

A force can tend to turn a body about a point. The moment measures this turning effect about the chosen point. Its value depends not only on the force but also on where the force acts: the same force can have different moments about different points. This lesson focuses on calculating that moment for planar forces. We will define the reference point, choose a sign convention, and use two equivalent calculation methods. No equilibrium solution is needed to calculate a moment, but the moment calculation is one of the quantities used in planar equilibrium equations.

What you will learn

  • Explain what a force’s moment about a point represents and identify its sign in a plane.
  • Calculate a moment using either a perpendicular distance or force components.
  • Use the position-vector cross product to calculate a moment and check its units and direction.

1. Define the point and the force

Choose the body or system of interest and the point about which the moment is requested. Call the reference point OO. Identify the force’s point of application, AA, and the force vector, F\mathbf{F}. The position vector rOA\mathbf{r}_{OA} runs from OO to AA; its direction matters, so do not reverse its endpoints.
In a plane, use xx to the right and yy upward. Take counterclockwise moments as positive and clockwise moments as negative. This convention is a choice, but it must be used consistently. A force whose line of action passes through OO has zero moment about OO, because it has no perpendicular separation from that point.
The moment is a signed scalar in planar problems. Its SI unit is the newton-metre, written N·m. A moment is not a force: N·m and N describe different physical quantities.
MO=xFy−yFxM_O = xF_y-yF_x
  • Name the reference point before calculating.
  • Draw the position vector from the reference point to the force’s application point.
  • A line of action through the reference point gives zero moment.

2. Two equivalent ways to calculate a moment

The perpendicular-distance method is useful when the shortest distance from OO to the force’s line of action is clear. The moment magnitude is the force magnitude times that distance. Assign the sign by asking whether the force tends to turn the body clockwise or counterclockwise about OO. The distance must be perpendicular to the line of action, not merely the length from OO to the application point.
The component method works directly from coordinates. Write rOA=⟨x,y⟩\mathbf{r}_{OA}=\langle x,y\rangle and F=⟨Fx,Fy⟩\mathbf{F}=\langle F_x,F_y\rangle. The out-of-plane component of their cross product is xFy−yFxxF_y-yF_x. Thus, an upward component applied to the right of OO gives a positive contribution; a rightward component applied above OO gives a negative contribution. Add the contributions with their signs.
These methods agree. If the angle between the position vector and the force is θ\theta, the magnitude is also rFsin⁡θrF\sin\theta. This form makes clear that only the component perpendicular to the position vector contributes. In a calculation, use the angle between the actual vectors, not an angle measured from an unrelated reference direction.
MO=Fd⊥=rFsin⁡θM_O = Fd_\perp = rF\sin\theta
  • Perpendicular distance is measured from the point to the force’s line of action.
  • Component form automatically accounts for the force direction and its lever arm.
  • The moment’s sign records rotational sense under the chosen convention.

3. Calculation and checks

For several forces, calculate each force’s moment about the same point and add the signed values. If a force is given by its magnitude and direction, resolve it into horizontal and vertical components before using xFy−yFxxF_y-yF_x. Keep the distance and force units consistent; metres with newtons produce N·m.
A reliable check is to sketch the force’s tendency to turn the body and compare it with the sign of the calculated result. Then check the geometry: a force parallel to the position vector must give zero, while increasing the perpendicular separation at fixed force magnitude increases the moment magnitude.
In planar statics, equilibrium equations include ∑Fx=0\sum F_x=0, ∑Fy=0\sum F_y=0, and ∑MO=0\sum M_O=0. They express zero net force and zero net moment for an equilibrium body. This topic teaches how to calculate an individual force’s contribution to ∑MO\sum M_O; the examples below do not solve for unknown reactions or claim equilibrium.
∑Fx=0,∑Fy=0,∑MO=0\sum F_x=0,\quad \sum F_y=0,\quad \sum M_O=0
  • Add moments algebraically, preserving clockwise and counterclockwise signs.
  • Check rotational sense, perpendicular distance, and units.
  • Force equilibrium and moment equilibrium are separate checks when an equilibrium problem is being solved.

4. A repeatable setup

First state the point about which the moment is required. Identify the force’s application point and draw its line of action. Choose axes and the positive moment direction. Next choose the calculation method: use force times perpendicular distance when the geometry is direct, or resolve the force and use position and force components when coordinates are given.
Write one signed moment expression, substitute values with units, and state the direction represented by the sign. Finally, check the answer independently where possible—for example, calculate the perpendicular distance and compare its result with the component calculation. This is a check of the moment calculation, not a substitute for force and moment balance in a separate equilibrium problem.
  • Keep the reference point fixed throughout one calculation.
  • Write the sign convention before evaluating signed moments.
  • Use an independent geometry or component check when possible.

Worked example

A vertical force with a direct lever arm

A 24 N downward force acts at a point 0.35 m to the right of point O. Calculate its moment about O.
  1. Set the reference and sign
    Take moments about O and choose counterclockwise as positive. The force is downward to the right of O, so it tends to turn the bar clockwise.
    clockwise<0\text{clockwise} < 0
  2. Use the perpendicular distance
    The horizontal separation is perpendicular to the vertical force, so it is the required lever arm.
    MO=−(24 N)(0.35 m)M_O=-(24\,\mathrm{N})(0.35\,\mathrm{m})
  3. Evaluate and check
    The negative sign matches the clockwise turning tendency. The product of newtons and metres has units of N·m.
    MO=−8.4 N⋅mM_O=-8.4\,\mathrm{N\cdot m}
Answer: The moment about O is 8.4 N·m clockwise.
Check: The force’s line of action is 0.35 m from O, so the magnitude is 24 × 0.35 = 8.4 N·m. Its downward direction on the right side gives clockwise rotation.

Worked example

An angled force resolved into components

A force of magnitude 50 N acts at point A, whose coordinates relative to O are (0.40 m, 0.20 m). Its direction is 30° above the positive x-axis. Calculate its moment about O.
  1. Resolve the force
    The angle is measured from the positive x-axis, so both components are positive.
    Fx=50cos⁡30∘=43.3 N,Fy=50sin⁡30∘=25.0 NF_x=50\cos 30^\circ=43.3\,\mathrm{N},\quad F_y=50\sin 30^\circ=25.0\,\mathrm{N}
  2. Apply the component formula
    Use the coordinates of A measured from O. The upward component at positive x contributes counterclockwise moment; the rightward component at positive y contributes clockwise moment.
    MO=(0.40)(25.0)−(0.20)(43.3)M_O=(0.40)(25.0)-(0.20)(43.3)
  3. Evaluate and interpret
    The result is positive, so the net turning tendency is counterclockwise. Both terms have N·m units.
    MO=1.34 N⋅mM_O=1.34\,\mathrm{N\cdot m}
Answer: The moment about O is 1.34 N·m counterclockwise.
Check: The component contributions are 10.0 N·m counterclockwise and 8.66 N·m clockwise. Their signed difference is 1.34 N·m counterclockwise.

Worked example

A force whose line of action is specified

A 16 N force acts along a straight line whose perpendicular distance from point P is 0.25 m. The force tends to rotate the body clockwise about P. Calculate its moment about P.
  1. Choose the sign
    Use counterclockwise as positive. The stated clockwise tendency means the moment is negative.
    MP<0M_P<0
  2. Multiply force by perpendicular distance
    The given 0.25 m is already the shortest distance from P to the force’s line of action, so no trigonometric resolution is needed.
    MP=−(16 N)(0.25 m)M_P=-(16\,\mathrm{N})(0.25\,\mathrm{m})
  3. Report the moment
    The magnitude is 4.0 N·m, and the negative sign denotes clockwise rotation under the selected convention.
    MP=−4.0 N⋅mM_P=-4.0\,\mathrm{N\cdot m}
Answer: The moment about P is 4.0 N·m clockwise.
Check: The perpendicular-distance method directly gives a magnitude of 16 × 0.25 = 4.0 N·m. The stated clockwise tendency confirms the negative sign.

Common mistakes and how to avoid them

Using the full distance from the reference point to the application point as the lever arm.
Correction: Use only the perpendicular distance from the reference point to the force’s line of action, or use the component formula.
Changing the sign convention partway through a calculation.
Correction: Choose a positive rotational direction first and apply it to every moment contribution.
Using coordinates measured from a different origin than the moment reference point.
Correction: Form the position vector from the chosen reference point to the force application point.
Reporting a moment in newtons instead of newton-metres.
Correction: A moment is force multiplied by distance, so its unit is N·m.

Lesson summary

  • A force’s moment about a point depends on the force and its perpendicular separation from that point.
  • For a planar force, use MO=Fd⊥M_O=F d_\perp with a signed direction, or use MO=xFy−yFxM_O=xF_y-yF_x.
  • With counterclockwise positive, a positive result is counterclockwise and a negative result is clockwise.
  • Check the force direction, perpendicular geometry, and moment units.

Check your understanding

Question 1

A 10 N downward force acts 0.60 m to the right of O. With counterclockwise positive, what is its moment about O?
  1. 6.0 N·m counterclockwise
  2. 6.0 N·m clockwise
  3. 0.060 N·m clockwise
  4. 10.6 N·m clockwise
Show answer and explanation
6.0 N·m clockwise
The perpendicular distance is 0.60 m, giving a magnitude of 10 × 0.60 = 6.0 N·m. A downward force on the right tends to rotate clockwise.

Question 2

A force’s line of action passes through the moment reference point. What is its moment about that point?
  1. Zero
  2. Equal to the force magnitude
  3. Always positive
  4. Equal to the force times its application-point distance
Show answer and explanation
Zero
The perpendicular distance from the point to the line of action is zero, so the moment is zero.

Key terms

Moment about a point
The signed measure of a force’s tendency to turn a body about a specified point.
Position vector
The vector from the chosen reference point to the force’s point of application.
Line of action
The straight line extending through a force in its direction.
Perpendicular distance
The shortest distance from the reference point to the force’s line of action.

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