Moments and couples
Introduit| English | Sens |
|---|---|
| moment/ˈməʊmənt/ | donné |
| moment arm/ˈməʊmənt ɑːm/ | bras de levier |
| line of action/laɪn ɒv ˈækʃn/ | ligne d'action |
| couple/ˈkʌpl/ | couple |
Why does a longer spanner win?
- Push a spanner $0.25\ \text{m}$ long with $40\ \text{N}$ and you turn the nut with $10\ \text{N}\cdot\text{m}$. A $0.50\ \text{m}$ spanner and the same push give $20\ \text{N}\cdot\text{m}$.
- The turning effect grows with the distance. It needs its own measure, and its own equation.
Moment = force × perpendicular arm
- The moment 力矩 of a force about a point O is $M = F d$. The moment arm 力臂 $d$ is the perpendicular distance from O to the force's line of action 作用线.
- Units are $\text{N}\cdot\text{m}$. Counterclockwise is positive, clockwise negative.
A $60\ \text{N}$ push acts at right angles to a spanner, $0.30\ \text{m}$ from the nut. What moment does it apply, in $\text{N}\cdot\text{m}$?
Difficulté 2/5 · 140 pièces + un petit bonus lors de la première réussite aujourd'hui
The push is perpendicular, so the arm is the full $0.30\ \text{m}$: $M = 60 \times 0.30 = 18\ \text{N}\cdot\text{m}$.
What is the moment arm of a force about a point O?
Difficulté 1/5 · 100 pièces + un petit bonus lors de la première réussite aujourd'hui
Only the perpendicular distance stays the same when the force slides along its line, so only it can measure turning.
A slanted force turns less
- If the force makes angle $\theta$ with the bar, only its perpendicular part $F\sin\theta$ turns it: $M = F r\sin\theta$, where $r$ is the distance along the bar.
- A force at $0^{\circ}$ (along the bar) has its line through O and no moment at all, however large it is.
- $300\ \text{N}$ pulls at $40^{\circ}$ to a bar, $1.2\ \text{m}$ from the pivot O.
- $M = 300 \times 1.2 \times \sin 40^{\circ} = 231.4\ \text{N}\cdot\text{m}$.
- Same answer the other way: the arm is $1.2\sin 40^{\circ} = 0.771\ \text{m}$, and $300 \times 0.771 = 231.4\ \text{N}\cdot\text{m}$.
The angle sets the moment arm
Turn the force from $90^{\circ}$ towards the bar. The amber perpendicular arm shrinks as $r\sin\theta$, and so does the moment; at $0^{\circ}$ the line passes through the axis and the moment is zero.
A $200\ \text{N}$ force pulls at $30^{\circ}$ to a bar, $0.50\ \text{m}$ from the pivot. What is its moment about the pivot, in $\text{N}\cdot\text{m}$?
Difficulté 2/5 · 140 pièces + un petit bonus lors de la première réussite aujourd'hui
$M = Fr\sin\theta = 200 \times 0.50 \times \sin 30^{\circ} = 50\ \text{N}\cdot\text{m}$. Half the force is wasted pulling along the bar.
Moments from components
- When the arm is awkward to find, split the force. For a force $(F_x, F_y)$ acting at point $(x, y)$: $M_O = xF_y - yF_x$.
- Each component has an easy arm, and the signs come out right on their own.
- $\vec F = (60,\ 80)\ \text{N}$ acts at $(2,\ 1)\ \text{m}$. Find its moment about the origin.
- $M_O = 2 \times 80 - 1 \times 60 = 100\ \text{N}\cdot\text{m}$, counterclockwise.
- Check: $|\vec F| = 100\ \text{N}$, so the arm must be $100/100 = 1\ \text{m}$. Measure it on a sketch: it is.
A force $\vec F = (30,\ -40)\ \text{N}$ acts at the point $(3,\ 2)\ \text{m}$. What is its moment about the origin, in $\text{N}\cdot\text{m}$, counterclockwise positive?
Difficulté 2/5 · 140 pièces + un petit bonus lors de la première réussite aujourd'hui
$M_O = xF_y - yF_x = 3 \times (-40) - 2 \times 30 = -180\ \text{N}\cdot\text{m}$: $180\ \text{N}\cdot\text{m}$ clockwise. Both parts turn the same way here.
A couple turns without pushing
- A couple 力偶 is two equal, opposite, parallel forces. Its moment is $M = F d$, with $d$ the distance between the two lines, and it is the same about every point.
- Why: about any point at distance $a$ from one line, $M = F(a + d) - Fa = Fd$. The $a$ cancels. A couple has no resultant force, only turning.
A couple of two $150\ \text{N}$ forces, $0.8\ \text{m}$ apart, has a moment of $120\ \text{N}\cdot\text{m}$ about every point in its plane.
Difficulté 1/5 · 100 pièces + un petit bonus lors de la première réussite aujourd'hui
$M = Fd = 150 \times 0.8 = 120\ \text{N}\cdot\text{m}$. The distance to the reference point cancels, so a couple's moment does not depend on the point.
An off-centre force is a force plus a couple
- A force $F$ acting at distance $e$ from a point is equal to the same force at the point plus a couple $M = Fe$. The swap is exact.
- A $500\ \text{N}$ load hung $0.3\ \text{m}$ off a column's centre line equals $500\ \text{N}$ down the centre line plus a $150\ \text{N}\cdot\text{m}$ couple. The couple is what bends the column.
- A bracket carries $300\ \text{N}$ down, $0.4\ \text{m}$ to the right of bolt O, and a clockwise couple of $60\ \text{N}\cdot\text{m}$.
- About O: $-300 \times 0.4 - 60 = -180\ \text{N}\cdot\text{m}$, so $180\ \text{N}\cdot\text{m}$ clockwise.
- About any other point the force's part changes, but the $60\ \text{N}\cdot\text{m}$ never does.
A bracket carries $400\ \text{N}$ straight down, $0.25\ \text{m}$ to the right of bolt O, and a counterclockwise couple of $50\ \text{N}\cdot\text{m}$. What is the total moment about O, in $\text{N}\cdot\text{m}$, counterclockwise positive?
Difficulté 2/5 · 140 pièces + un petit bonus lors de la première réussite aujourd'hui
The force gives $-400 \times 0.25 = -100\ \text{N}\cdot\text{m}$ (clockwise); the couple adds $+50$. Total $-50\ \text{N}\cdot\text{m}$, so $50\ \text{N}\cdot\text{m}$ clockwise.
Marks that slip away
- Using the distance to the point of application instead of the perpendicular distance to the line of action.
- Dropping a couple from a moment sum "because it has no force". A couple enters every moment equation, in full.
- Mixing senses. Give clockwise moments a minus sign before you add.
Match each idea to its formula.
Difficulté 2/5 · 140 pièces + un petit bonus lors de la première réussite aujourd'hui
All four are the same idea, force times perpendicular distance, written for different situations.
You've got it
- Moment $M = Fd$, with $d$ the perpendicular moment arm; or $M = Fr\sin\theta$; or $M_O = xF_y - yF_x$.
- A couple has moment $Fd$ about every point and no resultant force.
- An off-centre force = the same force at the point + a couple $Fe$.