2.2 Use the cross product for force moments | University of Alberta ENGG 130: Engineering Mechanics: Statics

University of Alberta ENGG 130: Engineering Mechanics: Statics

Force-System Resultants

ENGG 130 Engineering Mechanics: Statics — Study topic 2.2

A force can cause a body to tend to turn about a point. The moment of that force depends both on the force and on where it acts. The cross product combines these effects in one vector calculation. In planar statics, the moment vector points perpendicular to the plane, and its direction identifies a clockwise or counterclockwise turning tendency. This lesson develops the position-vector cross force rule, applies it to individual forces, and uses it to find reactions for a simple beam.

What you will learn

Represent a force and its position using planar vectors.
Calculate a force moment from the cross product of position and force vectors.
Interpret the sign of a moment using a consistent counterclockwise-positive convention.
Use cross-product moments with equilibrium equations to solve a simple planar statics problem.

1. Position vectors and the force moment

Choose the body or point of interest and identify the point about which you want the moment. Call that reference point

O

. The position vector

\mathbf{r}

starts at

O

and ends at the force's point of application. The force vector is

\mathbf{F}

. The moment of that force about

O

is defined by the cross product

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

.

For planar problems, take

x

to the right,

y

upward, and

z

perpendicular to the page. Choose counterclockwise as positive, corresponding to the

+z

direction. With

\mathbf{r}=x\mathbf{i}+y\mathbf{j}

and

\mathbf{F}=F_x\mathbf{i}+F_y\mathbf{j}

, the only nonzero moment component is the

z

component. Expanding the cross product gives

M_{O,z}=xF_y-yF_x

.

The coordinates in

\mathbf{r}

are measured from the chosen moment point, not from an unrelated origin. Use consistent units for position and force. If position is in metres and force is in newtons, moment is in newton-metres.

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

The moment depends on the force and its point of application relative to the reference point.
In a planar problem, use the signed scalar $M_{O,z}=xF_y-yF_x$ .
A positive moment is counterclockwise; a negative moment is clockwise.

2. Finding and interpreting a planar moment

If a force direction is given by an angle, resolve it into horizontal and vertical components first. For a force of magnitude

F

at angle

\theta

counterclockwise from

+x

, use

F_x=F\cos\theta

and

F_y=F\sin\theta

. Then substitute the position coordinates and force components into

xF_y-yF_x

.

This component calculation is the planar form of the cross product, not a different moment rule. The term

xF_y

is the contribution from the vertical force component, while

-yF_x

is the contribution from the horizontal component. Depending on their signs and sizes, these contributions may reinforce or oppose one another.

A useful check is to consider the force's line of action. If it passes through the reference point, the force has zero moment about that point. Changing the reference point generally changes the position vector and therefore changes the moment.

M_{O,z}=xF_y-yF_x

Resolve angled forces using the selected axes before calculating the moment.
Keep component signs; do not replace them with magnitudes too early.
A zero moment means no turning effect about the chosen point.

3. Using moments in planar equilibrium

For a body at rest in planar statics, the resultant force and the resultant moment are zero. In Cartesian components, write

\sum F_x=0

,

\sum F_y=0

, and

\sum M_O=0

. The moment equation may be taken about any convenient point on the body.

To use the cross product in an equilibrium problem, isolate the body with a free-body diagram. Show every external force, support reaction, and applied couple acting on that body. For each force, identify its position vector from the chosen moment point and calculate its moment with the cross product. An applied couple is already a moment; it is not multiplied by a position vector.

A convenient moment point can eliminate unknown reactions whose lines of action pass through that point, because those reactions have zero moment there. After solving, check both force-component equations and the moment equation. If a calculated force is negative, it acts opposite to the direction initially assumed.

\sum F_x=0,\quad \sum F_y=0,\quad \sum M_O=0

Planar equilibrium requires two force equations and one moment equation.
Choose a moment point that simplifies the calculation.
Check force balance as well as moment balance.

Worked example

1. Moment from a horizontal force

A force of

50\,\mathrm{N}

acts horizontally to the right at position

\mathbf{r}=(0.30\mathbf{i}+0.40\mathbf{j})\,\mathrm{m}

from point

O

. Find its moment about

O

and state the turning direction.

Set the axes and vectors
Take right as $+x$ , up as $+y$ , and counterclockwise moment as positive. The force has components $F_x=50\,\mathrm{N}$ and $F_y=0$ .
$\mathbf{r}=(0.30,0.40)\,\mathrm{m},\quad \mathbf{F}=(50,0)\,\mathrm{N}$
Apply the cross product
Use the planar component rule. The negative sign indicates a clockwise turning effect about $O$ .
$M_{O,z}=(0.30)(0)-(0.40)(50)=-20\,\mathrm{N\cdot m}$

Answer: The force moment is

20\,\mathrm{N\cdot m}

clockwise about

O

.

Check: The force points right while its point of application is above

O

, so it tends to turn the body clockwise, consistent with the negative result.

Worked example

2. An angled force with two nonzero components

A force of

100\,\mathrm{N}

acts at

30^\circ

above the positive horizontal at position

\mathbf{r}=(0.60\mathbf{i}+0.20\mathbf{j})\,\mathrm{m}

from

O

. Find the moment about

O

.

Resolve the force
Use trigonometry to find the Cartesian components. The angle is measured from $+x$ , so the horizontal and vertical components are both positive.
$F_x=100\cos30^\circ=86.6\,\mathrm{N},\quad F_y=100\sin30^\circ=50.0\,\mathrm{N}$
Calculate the moment
Use the position coordinates measured from $O$ . The two component contributions have opposite signs, so retain their signs in the calculation.
$M_{O,z}=(0.60)(50.0)-(0.20)(86.6)=12.7\,\mathrm{N\cdot m}$

Answer: The moment is

12.7\,\mathrm{N\cdot m}

counterclockwise about

O

.

Check: The vertical component contributes

30.0\,\mathrm{N\cdot m}

counterclockwise, and the horizontal component contributes

17.3\,\mathrm{N\cdot m}

clockwise. Their difference is

12.7\,\mathrm{N\cdot m}

counterclockwise.

Worked example

3. Beam reactions using cross-product moments

A horizontal beam is supported by a pin at

A

and a vertical cable at

B

. The distance

AB

is

1.2\,\mathrm{m}

. A downward force of

200\,\mathrm{N}

acts

0.8\,\mathrm{m}

to the right of

A

. Find the cable tension and the pin reaction components. Assume the beam is in planar equilibrium and the cable pulls upward.

Take moments about A
The beam is the system. Take $+x$ along the beam, $+y$ upward, and counterclockwise moment as positive. The pin reactions pass through $A$ , so they have zero moment about $A$ . The cable force acts at $B$ and the downward force acts $0.8\,\mathrm{m}$ from $A$ .
$(1.2\,\mathrm{m})T-(0.8\,\mathrm{m})(200\,\mathrm{N})=0$
Solve for cable tension
Solve the moment equation for $T$ . The units reduce to newtons.
$T=\frac{(0.8\,\mathrm{m})(200\,\mathrm{N})}{1.2\,\mathrm{m}}=133.3\,\mathrm{N}$
Apply force equilibrium
There are no horizontal applied forces, so the horizontal pin reaction is zero. Vertical force balance gives the remaining pin reaction.
$\sum F_x=0:\ A_x=0,\qquad \sum F_y=0:\ A_y+133.3-200=0$
Verify equilibrium
Check the vertical and horizontal force sums, then recalculate the moments about $A$ using the solved reactions. The vertical check includes the pin reaction as well as the cable tension and downward load.
$\sum F_x=0,\quad \sum F_y=66.7+133.3-200=0\,\mathrm{N},\quad \sum M_A=(1.2)(133.3)-(0.8)(200)\approx0\,\mathrm{N\cdot m}$

Answer: The cable tension is

133.3\,\mathrm{N}

upward. The pin reaction is

A_x=0

and

A_y=66.7\,\mathrm{N}

upward.

Check: The vertical reaction is

200-133.3=66.7\,\mathrm{N}

. The vertical forces sum to

66.7+133.3-200=0\,\mathrm{N}

, and the moments about

A

balance to approximately zero. The small rounding difference is from reporting tension to one decimal place.

Common mistakes and how to avoid them

Using the force magnitude and a distance without checking the force direction or moment sign.

Correction: Resolve the force into signed components and use

M_{O,z}=xF_y-yF_x

. Then interpret the sign using the stated convention.

Measuring the position vector from the force point back to the moment point.

Correction: For

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

, draw

\mathbf{r}

from the moment point

O

to the force's point of application.

Adding an applied couple to the position-vector calculation as if it were a force.

Correction: A couple is already a moment. Include it directly, with its signed moment, in the moment equilibrium equation.

Checking moments but not checking the resultant force.

Correction: For planar equilibrium, verify both force-component sums and the moment sum.

Lesson summary

The force moment about a point is the cross product of the position vector from that point to the force application point and the force vector.
In planar coordinates, the signed moment is $M_{O,z}=xF_y-yF_x$ .
A positive result is counterclockwise and a negative result is clockwise when that convention is chosen.
Use moment and force equilibrium together to solve statically determinate planar problems, then check all three equilibrium equations.

Check your understanding

Question 1

A force has components

\mathbf{F}=(0,-40)\,\mathrm{N}

and acts at

\mathbf{r}=(0.5,0)\,\mathrm{m}

from

O

. What is its moment about

O

?

$20\,\mathrm{N\cdot m}$ counterclockwise
$20\,\mathrm{N\cdot m}$ clockwise
$0\,\mathrm{N\cdot m}$
$40\,\mathrm{N\cdot m}$ counterclockwise

Show answer and explanation

20\,\mathrm{N\cdot m}

clockwise

The cross product gives

M_{O,z}=(0.5)(-40)-(0)(0)=-20\,\mathrm{N\cdot m}

, which is clockwise.

Question 2

Which position vector is used to calculate a force moment about point

O

?

A vector from $O$ to the force's point of application
A vector from the force's point of application to $O$
A vector parallel to the force with length equal to the force magnitude
A vector from the origin of the page, regardless of the chosen moment point

Show answer and explanation

A vector from

O

to the force's point of application

The defining expression

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

uses the position vector from the reference point

O

to the point where the force acts.

Key terms

Position vector: A vector from a chosen reference point to a force's point of application.
Cross product: A vector operation that combines two vectors and produces a vector perpendicular to their plane; for a force and position vector, it gives the moment.
Moment: The turning effect of a force about a chosen point, represented in planar statics by a signed value perpendicular to the plane.
Counterclockwise-positive convention: A chosen sign rule in which counterclockwise moments are positive and clockwise moments are negative.

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2.2 · Use the cross product for force moments

University of Alberta ENGG 130: Engineering Mechanics: Statics

ENGG 130 Engineering Mechanics: Statics — Study topic 2.2

What you will learn

1. Position vectors and the force moment

2. Finding and interpreting a planar moment

3. Using moments in planar equilibrium

1. Moment from a horizontal force

2. An angled force with two nonzero components

3. Beam reactions using cross-product moments

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Key terms

Continue through ENGG 130

About this lesson and its review