Turning effects of forces
| English | Chinese | Pinyin |
|---|---|---|
| pivot | 支点 | zhī diǎn |
| centre of gravity | 重心 | zhòng xīn |
| moment | 力矩 | lì jǔ |
| perpendicular distance | 垂直距离 | chuí zhí jù lí |
| line of action | 作用线 | zuò yòng xiàn |
| couple | 力偶 | lì ǒu |
| torque | 力偶矩 | lì ǒu jǔ |
Why the handle is far from the hinge
- A door handle sits as far from the hinges as possible.
- The same push, applied further from the pivot 支点, turns the door more easily.
- Turning effect depends on force and distance.
Centre of gravity 重心
- The whole weight of an object can be treated as acting at one point: its centre of gravity.
- For a uniform, regular shape it sits at the middle.
The moment 力矩 depends on the perpendicular distance 垂直距离 $d$ from the pivot to the line of action 作用线 of the force
Balance the see-saw
Put a weight on each side and slide it in or out. A small weight far from the pivot can balance a big weight close in — the beam is level when force × distance matches on both sides.
Moment of a force
- The moment is $M = F \times d$, where $d$ is the perpendicular distance from the pivot to the force's line of action.
- Unit: $\text{N}\cdot\text{m}$. A moment is clockwise or anticlockwise.
A force of $20\ \text{N}$ acts at a perpendicular distance of $0.50\ \text{m}$ from a pivot. What is the moment?
$M = F \times d = 20 \times 0.50 = 10\ \text{N}\cdot\text{m}$.
Why is a door handle placed far from the hinges?
Moment $= F \times d$. A bigger $d$ means a bigger turning effect for the same force, so the door opens more easily.
Angle matters
- Only the part of the force perpendicular to the arm turns it.
- For a force at angle $\theta$ to an arm of length $r$: $M = Fr\sin\theta$.
Weights on a rule balanced at a pivot — used to test the principle of moments
Only the part of a force perpendicular to the arm contributes to the turning.
Yes — that is why $M = Fr\sin\theta$: $\sin\theta$ picks out the perpendicular part of the force.
Match each term to the definition the examiner marks.
Two of these hinge on the phrase line of action, which is what makes the distance perpendicular rather than measured along the rod.
A couple 力偶
- A couple is two forces: equal in size, opposite in direction, a distance apart.
- It makes the body turn only — the resultant force is zero, so no straight-line push.
Select all the features of a couple.
A couple is equal, opposite forces a distance apart, giving zero resultant force but a pure turning effect. Same-direction forces are not a couple.
Torque 力偶矩 of a couple
- The torque is $\tau = F \times d$, where $d$ is the distance between the two lines of action.
- It is the same about any point — a special property of couples.
A couple has forces of $5.0\ \text{N}$ with their lines $0.40\ \text{m}$ apart. What is the torque?
$\tau = F \times d = 5.0 \times 0.40 = 2.0\ \text{N}\cdot\text{m}$.
Two equal forces in the same direction are not a couple — they have a resultant force and push the body along.
Two equal forces pointing in the same direction form a couple.
No — a couple needs equal and opposite forces. Same-direction forces have a resultant and push the body along.
Worked example: a beam that is not weightless
- A uniform beam of length $4.0\ \text{m}$ and weight $120\ \text{N}$ rests on a pivot $1.5\ \text{m}$ from its left end. A $200\ \text{N}$ load hangs from the left end. What force at the right end holds it in equilibrium?
- The beam is uniform, so its own weight acts at its centre, $2.0\ \text{m}$ from the left, which is $0.5\ \text{m}$ to the right of the pivot.
- Take moments about the pivot. Anticlockwise: $200 \times 1.5 = 300\ \text{N m}$.
- Clockwise: the beam's weight $120 \times 0.5 = 60\ \text{N m}$, plus the unknown $F$ at $2.5\ \text{m}$.
- $300 = 60 + 2.5F$, so $F = 96\ \text{N}$.
- Leaving out the beam's own weight is the standard lost mark, and it only vanishes when the pivot is at the centre.
Marks that slip away
- A moment uses the perpendicular distance from the point to the line of action of the force, not the distance along the rod.
- The weight of a uniform beam acts at its centre and has a moment about any pivot that is not there. Do not leave it out.
- The torque of a couple uses the distance between the two lines of action, so a symmetric couple gives $2Fd$, not $Fd$ measured from the middle.
- Check whether the geometry needs $\sin\theta$ or $\cos\theta$ by drawing the triangle. Do not guess.
- Equilibrium means the resultant force and the resultant torque are both zero. Two conditions, and a question can test either.
A uniform 4.0 m beam of weight 120 N is pivoted 1.5 m from its left end, with a 200 N load at that end. What upward force at the right end gives equilibrium, in N?
Moments about the pivot: 200 x 1.5 = 120 x 0.5 + 2.5F, so F = 96 N. Forgetting the beam's own 120 N at its centre gives 120 N, the classic wrong answer.
Put the method for a principle-of-moments problem in order.
Taking moments about a point an unknown passes through removes it from the equation, which is why the choice of point is worth a moment's thought.
A body is in equilibrium. Which must be true? Select all that apply.
Equilibrium permits constant velocity as well as rest. Both conditions are needed, and a question can test either one alone.
You've got it
- moment $M = F \times d$ — use the perpendicular distance to the line of action
- a couple is equal, opposite forces a distance apart: pure turning, torque $\tau = Fd$
- weight acts at the centre of gravity