Moments and couples
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| moment/ˈməʊmənt/ | mô-men lực |
| moment arm/ˈməʊmənt ɑːm/ | cánh tay mô-men |
| line of action/laɪn ɒv ˈækʃn/ | trục tác dụng |
| couple/ˈkʌpl/ | cặp lực |
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 $\text{N}\cdot\text{m}$?
Độ khó 2/5 · 140 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
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?
Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
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 $\text{N}\cdot\text{m}$?
Độ khó 2/5 · 140 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
$M = Fr\sin\theta = 200 \times 0.50 \times \sin 30^{\circ} = 50\ \text{N}\cdot\text{m}$. 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?
Độ khó 2/5 · 140 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
$M_O = xF_y - yF_x = 3 \times (-40) - 2 \times 30 = -180\ \text{N}\cdot\text{m}$: $180\ \text{N}\cdot\text{m}$ 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.
Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
$M = Fd = 150 \times 0.8 = 120\ \text{N}\cdot\text{m}$. 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?
Độ khó 2/5 · 140 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
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.
Độ khó 2/5 · 140 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
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 $Fe$.