Moment: the turning effect of a force
Why does a stuck wheel nut come loose with a long spanner, when the same force on a short spanner cannot make it budge?
Prior knowledge
- A force has a magnitude, a direction and a point of applicationchapter Interaction and force: Newton's first and third laws
- The components of an oblique force are obtained with sine and cosinechapter Free-body diagrams and the combined effect of forces
Core path: follow the steps in order. Tasks marked Optional give extra practice. Optional tasks: 2.
Think about it
A wheel nut is stuck tight. You pull the handle of a short spanner with all your strength, and the nut does not budge. Your friend fetches a spanner whose handle is twice as long, and the nut comes loose, even though he does not pull any harder than you.
If the size of the force alone does not decide it, what decides whether the nut comes loose?
Optional Review: the component of an oblique force
Predict
Make your predictions before investigating 2 questions
Prediction 1: distance of the point of application
Prediction 2: angle of the pull
Investigate
In the simulation a stuck nut is loosened with a spanner. Adjust the force of the hand, the point of application and the angle of the pull, and try whether the nut comes loose.
Simulation
How to use
Adjust the force $F$, the distance $d$ of the point of application from the centre of the nut and the angle $\alpha$ that the force makes with the extension of the handle. Press Turn: either the nut comes loose and the spanner turns, or the spanner just gives a jerk. Each attempt is drawn on the graph at the point whose horizontal coordinate is the lever arm $r$ and whose vertical coordinate is the force $F$: a filled green dot when the nut came loose and an empty ring when it did not move. The red cross shows the current setting. The Nut selection switches between a tight nut and a rusted nut and clears the graph, and so does Reset. Speed changes only the pace of the turning. The checkbox Show the lever arm draws the segment $r$ perpendicular from the axis to the line of action. The readouts are below the picture.
Assumptions of the model
The values are model values calculated by the simulation. The nut can only rotate about the bolt, and the axis of rotation passes through the centre of the nut. The nut comes loose when the moment of the hand's force is at least the moment required by the friction of the nut. This value is not shown but is found by trying. All angles $0^\circ$–$180^\circ$ turn the nut in the loosening direction, that is anticlockwise. The weight of the spanner is ignored. The turning is only a picture of the result: the simulation does not calculate how fast the nut rotates.
How the calculation works
The lever arm of a force is the perpendicular distance from the axis to the line of action, $r = d\sin\alpha$. The moment is $M = Fr$ in the unit $\text{Nm}$. An attempt succeeds when $M$ is at least the moment required by the nut.
Experiment 1: the smallest force that loosens the nut
The initial state selects the tight nut and the angle $\alpha = 90^\circ$. Choose the point of application $d$, adjust the force $F$ and press Turn until you find the smallest force that loosens the nut. Record a measurement only then. Repeat at least at the distances $0.10\,\text{m}$, $0.15\,\text{m}$, $0.20\,\text{m}$ and $0.30\,\text{m}$.
Initial state: nut: tight, angle $\alpha$ = $90^\circ$, point of application $d$ = $0{.}1\,\text{m}$, force $F$ = $100\,\text{N}$. Kept constant: nut, angle $\alpha$. Varied: point of application $d$.
| point of application $d$ | force $F$ | $M$ $(\text{Nm})$ | nut |
|---|
From the smallest loosening force in the row $d = 0.20\,\text{m}$ of the table, calculate the moment that the nut requires. Answer in newton metres.
Hint 1: a guiding question
The force is perpendicular to the handle. How long is the lever arm of the force then?
Hint 2: what is the connection
When $\alpha = 90^\circ$, the lever arm is $r = d$ and the moment is $M = Fd$.
Hint 3: an intermediate step
$M = F\cdot 0.20\,\text{m}$, where $F$ is the smallest loosening force of the row.
Show the solution
Given
from the table, point of application $d = 0.20\,\text{m}$ and smallest loosening force $F = 150\,\text{N}$ (a model value); angle $\alpha = 90^\circ$
Find
moment $M$ required by the nut, in the unit $\text{Nm}$
Principle
The nut rotates about its axis, so whether it comes loose is decided by the moment of the force $M = Fr$. At the smallest loosening force the moment is exactly the one that the nut requires. The force is perpendicular to the handle, so the lever arm is the whole distance $d$.
Formula
$M = Fr$ and $r = d\sin 90^\circ = d$, so $M = Fd$
Conversions
The force is in newtons and the distance in metres, so no conversions are needed.
Substitution
Answer
$30\,\text{Nm}$
Two significant figures like the distance $0.20\,\text{m}$.
Reasonableness
The other rows give the same product: $300\,\text{N}\cdot 0.10\,\text{m} = 30\,\text{Nm}$ and $100\,\text{N}\cdot 0.30\,\text{m} = 30\,\text{Nm}$. So the nut really does require a certain moment. On the graph the green boundary points lie on the curve on which $Fr$ is constant.
Experiment 2: the angle of the pull changes
The initial state selects the rusted nut, the force $F = 200\,\text{N}$ and the point of application $d = 0.40\,\text{m}$. Change only the angle $\alpha$ and press Turn at each angle. Record a measurement at least at the angles $90^\circ$, $60^\circ$, $45^\circ$, $30^\circ$ and $150^\circ$.
Initial state: nut: rusted, force $F$ = $200\,\text{N}$, point of application $d$ = $0{.}4\,\text{m}$, angle $\alpha$ = $90^\circ$. Kept constant: nut, force $F$, point of application $d$. Varied: angle $\alpha$.
| angle $\alpha$ | $r$ $(\text{m})$ | $M$ $(\text{Nm})$ | nut |
|---|
Conclusion
Using your measurements, explain what decides whether the nut comes loose. Use the words force, lever arm and moment, and explain why the angle of the pull affects the result.
Model answer
In experiment 1 the smallest loosening force in the values of the model was $300\,\text{N}$ at the point $0.10\,\text{m}$, $150\,\text{N}$ at $0.20\,\text{m}$ and $100\,\text{N}$ at $0.30\,\text{m}$. So the force is not the same, but the product of the force and the distance is $30\,\text{Nm}$ every time: what decides whether the nut comes loose is the moment of the force $M = Fr$. In experiment 2 the force and the point of application stayed the same, but as the angle decreased, the line of action moved closer to the axis. The lever arm $r = d\sin\alpha$ shortened from $0.40\,\text{m}$ to $0.20\,\text{m}$ at the angle $30^\circ$, and the moment dropped from $80\,\text{Nm}$ to $40\,\text{Nm}$. At the angle $150^\circ$ the moment was the same as at the angle $30^\circ$, because the lever arm is equally long. On the graph the attempts with the same nut are divided by the same boundary, because loosening depends only on the product $Fr$.
Assess your own answer
Tick the points your own answer already contains.
Review the theory summary and practise with the exercises.
Theory summary
Key idea
A force can change the motion of a body, but it can also rotate the body. When a body is fixed to an axis, like a door on its hinges, a nut on a bolt or a seesaw on its support, it cannot move away but can only rotate about the axis. This axis is called the axis of rotation. The ability of a force to rotate a body about the axis of rotation is called the turning effect, and its size is measured by the moment (the moment of a force). The turning effect does not depend only on the size of the force. It also depends on the direction of the force and on the point of the body where the force acts.
A door and the point of application of a force
A door is easy to open by pushing at the handle, which is far from the hinges. If the same force pushes close to the hinge, the door turns much less easily. If the door is pushed at its edge in the direction of the door towards the hinge, the door does not turn at all, even though the force is equally large. The three forces $\overline{F}_1$, $\overline{F}_2$ and $\overline{F}_3$ in the figure are therefore equally large, but their turning effects about the hinge are completely different.
The lever arm of a force
The turning effect is decided by the line of action of the force, the straight line that passes through the point of application of the force (in a spanner, the point where the hand grips) in the direction of the force. The lever arm $r$ of a force is the perpendicular distance from the axis of rotation to the line of action. When the force is perpendicular to the handle of the spanner, the lever arm equals the distance $d$ of the point of application from the axis. When the force is oblique, the lever arm is shorter than $d$. If the line of action passes through the axis of rotation, the lever arm is zero and the force does not rotate the body at all. This is why the force $\overline{F}_3$ along the door does not open the door.
Moment of a force
The moment of a force about the axis of rotation is the product of the magnitude of the force and its lever arm:
$$M = F r$$
where $M$ is the moment of the force ($\text{Nm}$), $F$ the magnitude of the force ($\text{N}$) and $r$ the lever arm of the force, that is the perpendicular distance from the axis of rotation to the line of action ($\text{m}$). The unit of moment is the newton metre $\text{Nm}$. It is not the same thing as the unit of energy, the joule, even though both are a product of a newton and a metre: in a moment the distance is perpendicular to the force, in work it is a displacement in the direction of the force.
Direction of rotation and sign
In addition to its magnitude, a moment has a direction of rotation. In this book a moment that rotates anticlockwise is positive and a moment that rotates clockwise is negative. When several forces act on a body, their moments are added together with their signs. The result is the net moment:
$$\Sigma M = M_1 + M_2 + M_3 + \ldots$$
where $\Sigma M$ is the net moment ($\text{Nm}$) and $M_1$, $M_2$, … are the moments of the individual forces about the same chosen axis of rotation ($\text{Nm}$). If the net moment on a body at rest that can rotate about a fixed axis is not zero, the body starts to rotate in the direction of the net moment. When a body stays at rest is studied more closely in the chapter Equilibrium of a body: forces and moments.
The lever arm of an oblique force
Often the distance $d$ of the point of application from the axis of rotation is known, together with the angle $\alpha$ that the force makes with the line from the axis to the point of application (for example, the extension of the handle of the spanner). The axis, the point of application and the foot of the lever arm form a right-angled triangle whose hypotenuse is $d$. The lever arm is the side opposite the angle $\alpha$:
$$r = d\sin\alpha$$
where $r$ is the lever arm of the force ($\text{m}$), $d$ the distance of the point of application from the axis of rotation ($\text{m}$) and $\alpha$ the angle between the force and the extension of the arm. When $\alpha = 90^\circ$, $r = d$ and the moment is greatest. When $\alpha = 0^\circ$ or $\alpha = 180^\circ$, $r = 0$.
Common misconception
The moment is not obtained by multiplying the force by the distance to the point of application, but by the distance to the line of action. The distance $d$ and the lever arm $r$ are equal only when the force is perpendicular to the line from the axis to the point of application. Another mistake is to think that a long spanner adds force. The force of the hand is the same, but with a long spanner its lever arm is longer, so the moment is greater. Loosening a nut does not require a certain force but a certain moment.
Summary
- The turning effect of a force about an axis of rotation is measured by the moment $M = Fr$, whose unit is the newton metre $\text{Nm}$.
- The lever arm $r$ of a force is the perpendicular distance from the axis of rotation to the line of action of the force, not the distance to the point of application.
- The lever arm of an oblique force is $r = d\sin\alpha$. A force whose line of action passes through the axis does not rotate the body.
- A moment that rotates anticlockwise is positive and one that rotates clockwise is negative. The net moment is the sum of the moments with their signs.
- If the net moment on a body at rest that can rotate about an axis is not zero, the body starts to rotate.
Optional Return to your predictions
Compare your predictions with the measurements. What did you learn about what makes a force an effective rotator? Return also to the nut at the beginning: why does a long spanner help, even though the force of the hand does not grow?
Would you change your prediction?
Apply
Worked example
Problem
A stuck nut is being loosened with a spanner. The hand grips the handle at a distance of $0.35\,\text{m}$ from the centre of the nut, which is the axis of rotation. The hand pulls the handle with a force of $240\,\text{N}$ that makes an angle of $65^\circ$ with the extension of the handle and turns the nut in the loosening direction. The nut comes loose only when the moment is at least $80\,\text{Nm}$. What is the moment of the hand's force about the axis of the nut, and does the nut come loose?
Step 1/7
Given and find. $F = 240\,\text{N}$, $d = 0.35\,\text{m}$, $\alpha = 65^\circ$ and the moment required by the nut $80\,\text{Nm}$. We are asked for the moment $M$ of the force and whether it is enough.
Step 2/7
Principle. The nut can only rotate about its axis, so the turning effect of the force is given by its moment $M = Fr$. The force is oblique to the handle, so the lever arm $r$ is shorter than the distance $d$: it is obtained from the right-angled triangle, $r = d\sin\alpha$.
Step 3/7
Formula solved. Substitute the expression for the lever arm into the formula for the moment:
$$\begin{aligned} M &= F r \\ M &= F d\sin\alpha \end{aligned}$$
Step 4/7
Unit conversions. Not needed: the force is in newtons and the distance in metres, so the moment comes in newton metres.
Step 5/7
Substitution. Substitute the numerical values with their units:
$$\begin{aligned} M &= F d\sin\alpha \\ &= 240\,\text{N}\cdot 0.35\,\text{m}\cdot\sin 65^\circ \\ &\approx 76.1\,\text{Nm} \end{aligned}$$
Step 6/7
Result. The moment of the force is $M \approx 76\,\text{Nm}$, to two significant figures like the distance $0.35\,\text{m}$. It is smaller than the $80\,\text{Nm}$ required by the nut, so the nut does not come loose.
Step 7/7
Reasonableness. If the same force acted perpendicular to the handle, the lever arm would be the whole $0.35\,\text{m}$ and the moment $240\,\text{N}\cdot 0.35\,\text{m} = 84\,\text{Nm}$. An oblique pull gives less, because $\sin 65^\circ \approx 0.91 < 1$. So the nut would come loose with the same force if the pull were perpendicular to the handle.
Find the error
Apply to a new situation
The barrier of a car park is $3.00\,\text{m}$ long and has a mass of $12.0\,\text{kg}$. The barrier turns about an axis at one end, and its weight acts at the middle of the barrier. The barrier has been raised to an angle of $40^\circ$ to the horizontal.
What is the moment of the barrier's weight about the axis? Answer in newton metres.
Hint 1: a guiding question
The weight points straight down. How far from the axis does the vertical line of action pass?
Hint 2: what is the connection
The lever arm is the horizontal distance of the point of application of the weight from the axis: $r = 1.50\,\text{m}\cdot\cos 40^\circ$.
Hint 3: an intermediate step
Show the solution
Given
length of the barrier $3.00\,\text{m}$, mass $m = 12.0\,\text{kg}$, the weight acts at the middle, $d = 1.50\,\text{m}$ from the axis, angle of the barrier to the horizontal $40^\circ$; $g = 9.81\,\text{m/s}^2$
Find
moment $M$ of the weight about the axis in the unit $\text{Nm}$
Principle
The barrier can only turn about its axis, so the turning effect of the weight is given by its moment $M = Fr$. The weight points straight down, so its lever arm is the perpendicular distance from the axis to the vertical line of action, that is the horizontal distance of the point of application from the axis. In the right-angled triangle it is the side adjacent to the angle $40^\circ$.
Formula
$M = Gr$, where $G = mg$ and $r = d\cos 40^\circ$, so $M = mgd\cos 40^\circ$
Conversions
The mass is in kilograms and the length in metres, so no conversions are needed.
Substitution
Answer
$135\,\text{Nm}$
Three significant figures like the mass $12.0\,\text{kg}$ and the length $3.00\,\text{m}$.
Reasonableness
For a horizontal barrier the lever arm of the weight would be the whole $1.50\,\text{m}$ and the moment $177\,\text{Nm}$. When raised, the lever arm shortens, so the moment decreases, and when vertical it would be zero. The angle can also be checked with the formula $r = d\sin\alpha$: the angle between the weight and the extension of the barrier is $90^\circ + 40^\circ = 130^\circ$ and $\sin 130^\circ = \cos 40^\circ$.
Check your understanding
Exit ticket
Answer on your own first and only then open the answer.
- What is the unit of moment, and from which two quantities is the moment obtained?
Answer
The unit of moment is the newton metre $\text{Nm}$. The moment is the product of the magnitude of the force and the lever arm, $M = Fr$, so it is affected both by the force and by how far from the axis the line of action of the force passes.
- Which rotates a body about an axis more effectively: a force of $100\,\text{N}$ with a lever arm of $0.20\,\text{m}$, or a force of $50\,\text{N}$ with a lever arm of $0.50\,\text{m}$?
Answer
The latter. The moments are $100\,\text{N}\cdot 0.20\,\text{m} = 20\,\text{Nm}$ and $50\,\text{N}\cdot 0.50\,\text{m} = 25\,\text{Nm}$. The smaller force rotates more effectively, because its lever arm is longer.
- The left end of a spanner is around a nut, and the right end of the handle is pressed straight down. Does the force rotate the spanner clockwise or anticlockwise, and what is the sign of the moment by the convention of this chapter?
Answer
The right end of the handle moves down, so the spanner rotates clockwise. A clockwise moment is negative.
1. Concept
2. Calculation
A driver turns a steering wheel whose radius is $0.19\,\text{m}$. The left hand pulls the rim along the rim with a force of $25\,\text{N}$, which rotates the wheel anticlockwise. The right hand acts on the rim along the rim with a force of $15\,\text{N}$, which rotates the wheel clockwise.
What is the net moment of the hands' forces about the axis of the steering wheel? Answer in newton metres, with an anticlockwise moment positive.
Hint 1: a guiding question
A force along the rim is perpendicular to the radius. How long is the lever arm of the force then?
Hint 2: what is the connection
The lever arm of both forces is the radius $0.19\,\text{m}$. An anticlockwise moment is positive and a clockwise moment negative.
Hint 3: an intermediate step
Show the solution
Given
radius of the steering wheel $R = 0.19\,\text{m}$, force of the left hand $F_1 = 25\,\text{N}$ anticlockwise, force of the right hand $F_2 = 15\,\text{N}$ clockwise, both along the rim
Find
net moment $\Sigma M$ about the axis in the unit $\text{Nm}$, anticlockwise positive
Principle
The steering wheel rotates about its axis, so the turning effect of each force is given by its moment. A force along the rim is perpendicular to the radius, so its lever arm is the radius. The net moment is the sum of the moments with their signs.
Formula
$M_1 = F_1 R$ and $M_2 = -F_2 R$, so $\Sigma M = F_1 R - F_2 R$
Conversions
The forces are in newtons and the radius in metres, so no conversions are needed.
Substitution
Answer
$1.9\,\text{Nm}$
Anticlockwise, two significant figures like the radius $0.19\,\text{m}$ and the forces.
Reasonableness
Both forces have the same lever arm, so the net moment is the product of the difference of the forces, $10\,\text{N}$, and the radius: $10\,\text{N}\cdot 0.19\,\text{m} = 1.9\,\text{Nm}$. The larger force rotates anticlockwise, so the sign is positive: together the hands' forces rotate the wheel anticlockwise.
Practise the net moment in the level 2 tasks.
3. Explain
Using the moment, explain why the handle of a door is placed at the edge of the door, far from the hinges. Also explain why the door does not open if it is pushed at the edge in the plane of the door towards the hinges.
Model answer
A door can only rotate about the axis of rotation through the hinges, so its turning is decided by the moment of the force $M = Fr$, where $r$ is the perpendicular distance from the axis to the line of action of the force. When the handle is at the edge of the door, the lever arm of a perpendicular push is almost the whole width of the door, for example $0.8\,\text{m}$. Then even a small force gives a sufficient moment, and the door turns easily. Close to the hinges the lever arm is short, so the same force gives a much smaller moment. If the door is pushed at the edge in the plane of the door towards the hinges, the line of action of the force passes through the axis of rotation. The lever arm is zero, so the moment is zero too, and the door does not turn even if the force is large.
Assess your own answer
Tick the points your own answer already contains.
Review the theory summary and practise with the exercises.
Review the chapter's concepts
Cards you mark Review later collect in your review deck.
- Turning effect
- The effect of a force that tends to rotate an object.
- Axis of rotation
- The line about which an object rotates or could rotate.
- Line of action
- The line through the point of application of a force, in the direction of the force.
- Net moment
- The sum of the moments of the forces on an object about the chosen axis.
Exercises
Level 1 · basic practice 5 exercises
One formula or one step of reasoning. Core exercises first, the rest is extra practice.
- Exercise 1-1Core
A nut is being tightened with a spanner. The hand acts on the handle at a distance of $0.25\,\text{m}$ from the centre of the nut with a force of $150\,\text{N}$ perpendicular to the handle.
What is the moment of the force about the axis of the nut?
Answer
$37.5\,\text{Nm} \approx 38\,\text{Nm}$
Show the solution
Given
force $F = 150\,\text{N}$ perpendicular to the handle, distance of the point of application from the centre of the nut $d = 0.25\,\text{m}$
Find
moment $M$ of the force about the axis of the nut in the unit $\text{Nm}$
Principle
The nut rotates about its axis, so the turning effect of the force is given by its moment $M = Fr$. The force is perpendicular to the handle, so the lever arm is the whole distance to the point of application, $r = d$.
Formula
$M = Fr$ and $r = d$, so $M = Fd$
Conversions
The force is in newtons and the distance in metres, so no conversions are needed.
Substitution
$M$$= 150\,\text{N}\cdot 0.25\,\text{m}$$= 37.5\,\text{Nm}$Answer
$37.5\,\text{Nm} \approx 38\,\text{Nm}$
Two significant figures like the distance $0.25\,\text{m}$.
Reasonableness
With a quarter-metre handle the moment is a quarter of the numerical value of the force: $150 : 4 \approx 38$. The tightening moment of an ordinary nut is of the order of tens of newton metres, so the result is believable.
- Exercise 1-2Core
A door is opened by pushing the handle with a force of $12\,\text{N}$ perpendicular to the door. The handle is $0.80\,\text{m}$ from the hinges.
What is the moment of the force about the hinges?
Answer
$9.6\,\text{Nm}$
Show the solution
Given
push $F = 12\,\text{N}$ perpendicular to the door, distance of the handle from the hinges $d = 0.80\,\text{m}$
Find
moment $M$ of the force about the hinges in the unit $\text{Nm}$
Principle
The door rotates about the axis of rotation through the hinges, so the turning effect of the push is given by its moment. The push is perpendicular to the door, so the lever arm is the distance of the handle from the hinges.
Formula
$M = Fr$ and $r = d$, so $M = Fd$
Conversions
The force is in newtons and the distance in metres, so no conversions are needed.
Substitution
$M$$= 12\,\text{N}\cdot 0.80\,\text{m}$$= 9.6\,\text{Nm}$Answer
$9.6\,\text{Nm}$
Two significant figures like the force $12\,\text{N}$ and the distance $0.80\,\text{m}$.
Reasonableness
The lever arm is a little less than a metre, so the numerical value of the moment is a little smaller than the numerical value of the force, $12$. If the same force acted $0.10\,\text{m}$ from the hinges, the moment would be only $1.2\,\text{Nm}$, which explains why the handle is at the edge of the door.
- Exercise 1-3
The axle nut of a bicycle must be tightened to a moment of $45\,\text{Nm}$. The point of application on the spanner is $0.30\,\text{m}$ from the centre of the nut, and the force is perpendicular to the handle.
How large a force is needed?
Answer
$150\,\text{N} = 1.5 \cdot 10^{2}\,\text{N}$
Show the solution
Given
required moment $M = 45\,\text{Nm}$, distance of the point of application $d = 0.30\,\text{m}$, force perpendicular to the handle
Find
required force $F$ in the unit $\text{N}$
Principle
The tightness of the nut is decided by the moment $M = Fr$. The force is perpendicular to the handle, so the lever arm is $r = d$, and the force is solved from the formula for the moment.
Formula
$M = Fd$, so $F = \dfrac{M}{d}$
Conversions
The moment is in newton metres and the distance in metres, so the force comes directly in newtons.
Substitution
$F$$= \dfrac{45\,\text{Nm}}{0.30\,\text{m}}$$= 150\,\text{N}$Answer
$150\,\text{N} = 1.5 \cdot 10^{2}\,\text{N}$
Two significant figures like the moment $45\,\text{Nm}$ and the distance $0.30\,\text{m}$.
Reasonableness
Check: $150\,\text{N}\cdot 0.30\,\text{m} = 45\,\text{Nm}$. The force corresponds to the weight of a body of about $15\,\text{kg}$, which an adult can pull by hand. With a spanner half as long, twice the force would be needed.
- Exercise 1-4
In the figure, a horizontal force of $40\,\text{N}$ acts at the point $P$ on a rod that turns about the axis of rotation $O$. One square of the grid corresponds to $0.10\,\text{m}$. Read the lever arm of the force from the figure.
What is the moment of the force about the axis $O$?
Answer
$12\,\text{Nm}$
Show the solution
Given
horizontal force $F = 40\,\text{N}$ at the point $P$; one square of the grid is $0.10\,\text{m}$
Find
moment $M$ of the force about the axis $O$ in the unit $\text{Nm}$
Principle
The rod rotates about the axis $O$, so the turning effect of the force is given by its moment $M = Fr$. The lever arm is the perpendicular distance of the axis from the line of action of the force, not the length $OP$ of the rod.
Reading the figure
The force is horizontal, so its line of action is the horizontal dashed line through the point $P$. The axis $O$ is three squares below this line, so the lever arm is three squares. The rod itself is slanted in the figure, four squares to the right and three up, so its length is five squares, but that is not the lever arm.
Formula
$M = Fr$
Conversions
The squares are converted to metres:
$r$$= 3\cdot 0.10\,\text{m}$$= 0.30\,\text{m}$Substitution
$M$$= 40\,\text{N}\cdot 0.30\,\text{m}$$= 12\,\text{Nm}$Answer
$12\,\text{Nm}$
Two significant figures like the force $40\,\text{N}$; the lever arm is read from the grid exactly, in whole squares.
Reasonableness
If the length of the rod, $0.50\,\text{m}$, were taken as the lever arm, the moment would be $20\,\text{Nm}$, but that would hold only for a perpendicular force. The horizontal force pulls the rod partly along its own direction, so the moment is smaller. The same result is obtained with the formula $r = d\sin\alpha$: $0.50\,\text{m}\cdot\tfrac{3}{5} = 0.30\,\text{m}$.
- Exercise 1-5
The magnitude of a force is $60\,\text{N}$, and its moment about the axis of rotation is $18\,\text{Nm}$. What is the lever arm of the force?
Answer
$0.30\,\text{m}$
Show the solution
Given
force $F = 60\,\text{N}$, moment $M = 18\,\text{Nm}$
Find
lever arm $r$ of the force in the unit $\text{m}$
Principle
The moment is the product of the force and the lever arm, $M = Fr$. When the moment and the force are known, the lever arm is solved from the same formula.
Formula
$M = Fr$, so $r = \dfrac{M}{F}$
Conversions
The moment is in newton metres and the force in newtons, so the lever arm comes directly in metres: $\tfrac{\text{Nm}}{\text{N}} = \text{m}$.
Substitution
$r$$= \dfrac{18\,\text{Nm}}{60\,\text{N}}$$= 0.30\,\text{m}$Answer
$0.30\,\text{m}$
Two significant figures like the moment $18\,\text{Nm}$ and the force $60\,\text{N}$.
Reasonableness
Check: $60\,\text{N}\cdot 0.30\,\text{m} = 18\,\text{Nm}$. The numerical value of the moment is smaller than that of the force, so the lever arm must be less than a metre.
Level 2 · applied practice 4 exercises
Two steps, working with measured data, a unit conversion or rearranging a formula. Core exercises first, the rest is extra practice.
- Exercise 1-6Core
A door is opened by pulling a string tied to the handle. The handle is $0.75\,\text{m}$ from the hinges, and the tension of the string is $40\,\text{N}$. The string makes an angle of $35^\circ$ with the plane of the door.
What is the moment of the string's force about the hinges?
Answer
$17\,\text{Nm}$
Show the solution
Given
tension of the string $F = 40\,\text{N}$, distance of the handle from the hinges $d = 0.75\,\text{m}$, angle of the string to the plane of the door $\alpha = 35^\circ$
Find
moment $M$ of the string's force about the hinges in the unit $\text{Nm}$
Principle
The door rotates about the hinges, so the turning effect of the string is given by its moment $M = Fr$. The string is oblique to the door, so the lever arm is shorter than $d$. The plane of the door runs from the hinges to the handle, so the angle $35^\circ$ is the angle between the force and the extension of the arm, and $r = d\sin\alpha$.
Formula
$M = Fr$ and $r = d\sin\alpha$, so $M = Fd\sin\alpha$
Conversions
The force is in newtons and the distance in metres, so no conversions are needed.
Substitution
$M$$= 40\,\text{N}\cdot 0.75\,\text{m}\cdot\sin 35^\circ$$= 30\,\text{Nm}\cdot 0.5736$$= 17.21\,\text{Nm}$Answer
$17\,\text{Nm}$
Two significant figures like the force $40\,\text{N}$ and the distance $0.75\,\text{m}$.
Reasonableness
A perpendicular pull would give $40\,\text{N}\cdot 0.75\,\text{m} = 30\,\text{Nm}$. An oblique string gives less, because $\sin 35^\circ \approx 0.57$, a little more than half. The part of the string's force along the door pulls the door towards the hinges and does not rotate it.
- Exercise 1-7Core
A wheel nut is being loosened with a force of $85\,\text{N}$ perpendicular to the handle of the spanner. The point of application is $35\,\text{cm}$ from the centre of the nut.
What is the moment of the force about the axis of the nut?
Answer
$30\,\text{Nm}$
Show the solution
Given
force $F = 85\,\text{N}$ perpendicular to the handle, distance of the point of application $d = 35\,\text{cm}$
Find
moment $M$ of the force about the axis of the nut in the unit $\text{Nm}$
Principle
The nut rotates about its axis, so the turning effect of the force is given by its moment $M = Fr$. The force is perpendicular to the handle, so the lever arm is the whole distance to the point of application.
Formula
$M = Fr$ and $r = d$, so $M = Fd$
Conversions
The distance is converted to metres so that the moment comes in newton metres:
$d$$= 35\,\text{cm}$$= 35\cdot 0.01\,\text{m}$$= 0.35\,\text{m}$Substitution
$M$$= 85\,\text{N}\cdot 0.35\,\text{m}$$= 29.75\,\text{Nm}$Answer
$30\,\text{Nm}$
Two significant figures like the force $85\,\text{N}$ and the distance $35\,\text{cm}$.
Reasonableness
Without the conversion the result would be $2975$, which would be a hundred times too large: centimetres must always be converted to metres. With a handle of about a third of a metre, the moment is about a third of the numerical value of the force, $85 : 3 \approx 28$, which is close to the result.
- Exercise 1-8
A horizontal board turns about an axis at its left end. A force of $30\,\text{N}$ acts straight upwards at the right end of the board, $0.50\,\text{m}$ from the axis. In addition, a force of $40\,\text{N}$ acts straight downwards at $0.20\,\text{m}$ from the axis.
What is the net moment of the forces about the axis, when an anticlockwise moment is positive?
Answer
$7.0\,\text{Nm}$
Show the solution
Given
force $F_1 = 30\,\text{N}$ upwards at the distance $r_1 = 0.50\,\text{m}$ from the axis, force $F_2 = 40\,\text{N}$ downwards at the distance $r_2 = 0.20\,\text{m}$; the board is horizontal and the axis is at its left end
Find
net moment $\Sigma M$ about the axis in the unit $\text{Nm}$, anticlockwise positive
Principle
The board rotates about the axis, and its rotation is decided by the net moment, which is the sum of the moments of the forces with their signs. Both forces are vertical and the board is horizontal, so the lever arms are the distances from the axis. On the right-hand side of the axis, an upward force rotates the board anticlockwise and a downward force clockwise.
Formula
$M_1 = F_1 r_1$ and $M_2 = -F_2 r_2$, so $\Sigma M = F_1 r_1 - F_2 r_2$
Conversions
The forces are in newtons and the distances in metres, so no conversions are needed.
Substitution
$\Sigma M$$= 30\,\text{N}\cdot 0.50\,\text{m} - 40\,\text{N}\cdot 0.20\,\text{m}$$= 15\,\text{Nm} - 8.0\,\text{Nm}$$= 7.0\,\text{Nm}$Answer
$7.0\,\text{Nm}$
Anticlockwise, two significant figures like the data.
Reasonableness
The moment of the weight of the board itself is ignored: the board is assumed to be light compared with the forces. The larger force $40\,\text{N}$ does not decide the direction of rotation, because its lever arm is short. The moment of the smaller force, $15\,\text{Nm}$, is greater than $8.0\,\text{Nm}$, so the board starts to rotate anticlockwise. The net moment is positive, as it should be.
- Exercise 1-9
The lever of a water-pipe shut-off valve is $0.25\,\text{m}$ long, and the valve turns only with a moment of $50\,\text{Nm}$. The lever is pulled at its end so that the force makes an angle of $60^\circ$ with the extension of the lever.
How large a force is needed?
Answer
$231\,\text{N} \approx 2.3 \cdot 10^{2}\,\text{N}$
Show the solution
Given
required moment $M = 50\,\text{Nm}$, length of the lever, that is the distance of the point of application from the axis, $d = 0.25\,\text{m}$, angle between the force and the extension of the lever $\alpha = 60^\circ$
Find
required force $F$ in the unit $\text{N}$
Principle
Turning the valve is decided by the moment $M = Fr$. The force is oblique to the lever, so its lever arm is $r = d\sin\alpha$, shorter than the length of the lever. The force is solved from the formula for the moment.
Formula
$M = Fd\sin\alpha$, so $F = \dfrac{M}{d\sin\alpha}$
Conversions
The moment is in newton metres and the distance in metres, so no conversions are needed.
Substitution
$F$$= \dfrac{50\,\text{Nm}}{0.25\,\text{m}\cdot\sin 60^\circ}$$= \dfrac{50\,\text{Nm}}{0.2165\,\text{m}}$$= 230.9\,\text{N}$Answer
$231\,\text{N} \approx 2.3 \cdot 10^{2}\,\text{N}$
Two significant figures like the moment $50\,\text{Nm}$ and the length $0.25\,\text{m}$.
Reasonableness
With a perpendicular pull, $50\,\text{Nm} : 0.25\,\text{m} = 200\,\text{N}$ would be enough. An oblique pull needs more, because the lever arm is only $\sin 60^\circ \approx 0.87$ times the length of the lever: $200\,\text{N} : 0.87 \approx 230\,\text{N}$, the same as $231\,\text{N}$ to the accuracy of the roundings.
Level 3 · challenging 3 exercises
Optional: application, two ideas together or a new situation where you reason for yourself.
- Exercise 1-10
A cyclist's foot presses the pedal straight down with a force of $500\,\text{N}$. The distance of the pedal from the axis of the crank is $0.175\,\text{m}$. The crank points forwards and $45^\circ$ below the horizontal.
In which position of the crank would the same force give the greatest moment and in which zero? What is the moment of the foot's force about the axis of the crank when the crank is in this $45^\circ$ position?
Answer
$61.9\,\text{Nm}$
Show the solution
Given
force of the foot $F = 500\,\text{N}$ straight down, distance of the pedal from the axis $d = 0.175\,\text{m}$, crank $45^\circ$ below the horizontal
Find
moment $M$ of the force about the axis of the crank in the unit $\text{Nm}$, and the positions in which the moment is greatest and zero
Principle
The crank rotates about its axis, so the turning effect of the foot's force is given by its moment $M = Fr$. The force is vertical, so its line of action is the vertical line through the pedal, and the lever arm is the horizontal distance of the pedal from the axis. It is greatest when the crank is horizontal ($r = d$) and zero when the crank is vertical, because then the line of action passes through the axis. In the position $45^\circ$ the horizontal distance is the side adjacent to the angle.
Formula
$M = Fr$ and $r = d\cos 45^\circ$, so $M = Fd\cos 45^\circ$
Conversions
The force is in newtons and the distance in metres, so no conversions are needed.
Substitution
$M$$= 500\,\text{N}\cdot 0.175\,\text{m}\cdot\cos 45^\circ$$= 87.5\,\text{Nm}\cdot 0.7071$$= 61.87\,\text{Nm}$Answer
$61.9\,\text{Nm}$
Three significant figures like the distance $0.175\,\text{m}$. The moment is greatest, $87.5\,\text{Nm}$, when the crank is horizontal pointing forwards, and zero when the crank points straight down or up.
Reasonableness
The result lies between the moment of a horizontal crank, $87.5\,\text{Nm}$, and zero, as it should. The formula $r = d\sin\alpha$ gives the same: the angle between the force and the extension of the crank is $45^\circ$, and $\sin 45^\circ = \cos 45^\circ$. This is why a cyclist presses the pedal hardest when the crank is horizontal at the front, that is, pointing in the direction of travel.
- Exercise 1-11
An adult pushes a garden gate at the handle, $0.40\,\text{m}$ from the hinges, with a force of $50\,\text{N}$ perpendicular to the gate. A child can push with a force of only $20\,\text{N}$. The child also pushes perpendicular to the gate.
How far from the hinges must the child push so that the turning effect of the child's force is as large as the adult's?
Answer
$1.0\,\text{m}$
Show the solution
Given
adult's force $F_1 = 50\,\text{N}$ and lever arm $r_1 = 0.40\,\text{m}$, child's force $F_2 = 20\,\text{N}$; both push perpendicular to the gate
Find
distance $r_2$ of the child's push from the hinges in the unit $\text{m}$
Principle
The turning effects are equal when the moments about the hinges are equal. The lever arm of a perpendicular push is the distance of the pushing point from the hinges, so a smaller force needs a longer lever arm.
Formula
$F_2 r_2 = F_1 r_1$, so $r_2 = \dfrac{F_1 r_1}{F_2}$
Conversions
The forces are in newtons and the lever arm in metres, so no conversions are needed.
Substitution
$r_2$$= \dfrac{50\,\text{N}\cdot 0.40\,\text{m}}{20\,\text{N}}$$= \dfrac{20\,\text{Nm}}{20\,\text{N}}$$= 1.0\,\text{m}$Answer
$1.0\,\text{m}$
Two significant figures like the forces and the distance $0.40\,\text{m}$.
Reasonableness
The child's force is $\tfrac{2}{5}$ of the adult's, so the lever arm must be $\tfrac{5}{2}$ times as long: $2.5\cdot 0.40\,\text{m} = 1.0\,\text{m}$. An ordinary gate is about a metre wide, so the child has to push right at the edge of the gate.
- Exercise 1-12
The arms of a wheel brace extend $0.20\,\text{m}$ from the centre of the nut on both sides. A mechanic presses the end of the left arm straight down with a force of $120\,\text{N}$ and pulls the end of the right arm straight up with a force of $120\,\text{N}$. The arms are horizontal.
What is the sum of the hands' forces? What is the net moment of the hands' forces about the axis of the nut?
Answer
$48\,\text{Nm}$
Show the solution
Given
force of each hand $F = 120\,\text{N}$, the left one downwards and the right one upwards, distance of each point of application from the centre of the nut $d = 0.20\,\text{m}$; the arms are horizontal
Find
sum of the hands' forces and net moment $\Sigma M$ about the axis of the nut in the unit $\text{Nm}$
Principle
The forces are equal and opposite, so their sum is zero and the wheel brace does not move sideways. The rotation, however, is decided by the net moment. Both forces are vertical and the arms horizontal, so the lever arm of each is $d$. On the left of the axis a downward force and on the right of the axis an upward force both rotate anticlockwise, so the moments have the same sign and add up.
Formula
$\Sigma F = F - F$ and $\Sigma M = Fd + Fd$, so $\Sigma M = 2Fd$
Conversions
The forces are in newtons and the distances in metres, so no conversions are needed.
Substitution
$\Sigma F$$= 120\,\text{N} - 120\,\text{N}$$= 0$$\Sigma M$$= 2\cdot 120\,\text{N}\cdot 0.20\,\text{m}$$= 48\,\text{Nm}$Answer
The sum of the forces is zero, and the net moment is $48\,\text{Nm}$ anticlockwise, two significant figures like the distance $0.20\,\text{m}$.
Reasonableness
One hand would produce $24\,\text{Nm}$, so two hands rotating in the same direction give twice the moment. Because the sum of the forces is zero, the wheel brace only rotates and does not push the bolt sideways, which is the advantage of a wheel brace over turning with one hand.