A Door Handle Sits Far From the Hinges for One Very Good Reason
Define the moment of a force and calculate it in newton metre and dyne centimetre, balance a metre rule with the principle of moments, tell translational from rotational equilibrium and static from dynamic, and locate the centre of gravity of regular and irregular bodies.
Why is it so much harder to push a door open near its hinges?
Push a door at the handle and it swings easily. Now push it with the same force a few centimetres from the hinges. It barely moves.
The force is identical, so force alone is clearly not what turns things. What matters is the force multiplied by its distance from the pivot, and that product is called the moment of the force, or its turning effect.
That is why every tool designed to turn something is made long:
- A door handle is placed at the far edge, as far from the hinges as the door allows
- A spanner has a long arm, and a nut that will not shift is loosened with a longer one
- A bicycle pedal turns on a crank, and a longer crank gives more turning effect for the same push
- A steering wheel is large in diameter, so a light grip produces a large moment about the column
- A hand pump's handle is long for exactly the same reason
And the same idea run backwards explains balance. A see-saw balances not when the two children weigh the same but when their moments are equal — so a lighter child can balance a heavier one by sitting further out. That statement is the principle of moments, and it is the tool for every numerical in this part of the chapter.
From there the chapter asks a more general question: when does a body not move at all? It turns out that two separate conditions are needed:
- No net force, or the body would accelerate along a line
- No net moment, or the body would start to rotate
Both must hold, and a body satisfying both is in complete equilibrium. The chapter then sorts equilibrium into static and dynamic, and into stable, unstable and neutral — the last three depending entirely on what happens to the body's centre of gravity when you nudge it.
This page covers the first part of the ICSE Class 10 Physics chapter on force: the moment of a force, the principle of moments, equilibrium and its types, and the centre of gravity.
The force is identical, so force alone is clearly not what turns things. What matters is the force multiplied by its distance from the pivot, and that product is called the moment of the force, or its turning effect.
That is why every tool designed to turn something is made long:
- A door handle is placed at the far edge, as far from the hinges as the door allows
- A spanner has a long arm, and a nut that will not shift is loosened with a longer one
- A bicycle pedal turns on a crank, and a longer crank gives more turning effect for the same push
- A steering wheel is large in diameter, so a light grip produces a large moment about the column
- A hand pump's handle is long for exactly the same reason
And the same idea run backwards explains balance. A see-saw balances not when the two children weigh the same but when their moments are equal — so a lighter child can balance a heavier one by sitting further out. That statement is the principle of moments, and it is the tool for every numerical in this part of the chapter.
From there the chapter asks a more general question: when does a body not move at all? It turns out that two separate conditions are needed:
- No net force, or the body would accelerate along a line
- No net moment, or the body would start to rotate
Both must hold, and a body satisfying both is in complete equilibrium. The chapter then sorts equilibrium into static and dynamic, and into stable, unstable and neutral — the last three depending entirely on what happens to the body's centre of gravity when you nudge it.
This page covers the first part of the ICSE Class 10 Physics chapter on force: the moment of a force, the principle of moments, equilibrium and its types, and the centre of gravity.
Formula
What is the moment of a force and what units does it have?
The moment of a force about a point is the product of the force and the perpendicular distance of the line of action of the force from that point.
The word "perpendicular" is part of the definition, not an extra. It is the perpendicular distance from the pivot to the line of action of the force, and a force applied at an angle has a smaller perpendicular distance than the same force applied at right angles.
The units.
- SI unit: the newton metre (N m), obtained from a force in newtons and a distance in metres
- CGS unit: the dyne centimetre (dyne cm)
- Gravitational units: the kilogram-force metre (kgf m) and the gram-force centimetre (gf cm), which are convenient for laboratory work with slotted weights
The conversion between the two absolute units, which is a standard question:
And the gravitational conversion: kgf N, so kgf m N m.
Moment is a vector quantity, because it has a direction as well as a magnitude — the sense in which it turns the body. Two senses are possible:
- Anticlockwise, conventionally taken as positive
- Clockwise, conventionally taken as negative
Worked example 1 — a spanner. A force of N is applied at the end of a spanner of length cm, at right angles to it. Find the moment of the force about the nut.
Convert the length to metres first, since the force is in newtons:
Worked example 2 — finding the force. A door handle is fixed cm from the hinges. What force, applied perpendicular to the door, produces a moment of N m about the hinges?
And the point of the opening section, in numbers. The same N applied only cm from the hinges would give
a quarter of the turning effect — which is exactly why the handle is not put there.
Worked example 3 — a wheel. A wheel of radius m is turned by a force of N applied tangentially at its rim. Find the moment of the force about the axle.
A tangential force is perpendicular to the radius, so the perpendicular distance is the radius itself:
Worked example 4 — converting units. Express a moment of N m in dyne centimetre.
Two cases where the moment is zero, and both are examinable:
- When the force is zero — nothing to turn with
- When the line of action of the force passes through the pivot, so the perpendicular distance is zero
That second case explains something you have done without thinking. Pushing a door directly toward its hinges, along the line of the hinges, does not turn it at all however hard you push. The force is not zero and the distance from the hinge to your hand is not zero — but the perpendicular distance to the line of action is, and that is the quantity in the formula.
The word "perpendicular" is part of the definition, not an extra. It is the perpendicular distance from the pivot to the line of action of the force, and a force applied at an angle has a smaller perpendicular distance than the same force applied at right angles.
The units.
- SI unit: the newton metre (N m), obtained from a force in newtons and a distance in metres
- CGS unit: the dyne centimetre (dyne cm)
- Gravitational units: the kilogram-force metre (kgf m) and the gram-force centimetre (gf cm), which are convenient for laboratory work with slotted weights
The conversion between the two absolute units, which is a standard question:
And the gravitational conversion: kgf N, so kgf m N m.
Moment is a vector quantity, because it has a direction as well as a magnitude — the sense in which it turns the body. Two senses are possible:
- Anticlockwise, conventionally taken as positive
- Clockwise, conventionally taken as negative
Worked example 1 — a spanner. A force of N is applied at the end of a spanner of length cm, at right angles to it. Find the moment of the force about the nut.
Convert the length to metres first, since the force is in newtons:
Worked example 2 — finding the force. A door handle is fixed cm from the hinges. What force, applied perpendicular to the door, produces a moment of N m about the hinges?
And the point of the opening section, in numbers. The same N applied only cm from the hinges would give
a quarter of the turning effect — which is exactly why the handle is not put there.
Worked example 3 — a wheel. A wheel of radius m is turned by a force of N applied tangentially at its rim. Find the moment of the force about the axle.
A tangential force is perpendicular to the radius, so the perpendicular distance is the radius itself:
Worked example 4 — converting units. Express a moment of N m in dyne centimetre.
Two cases where the moment is zero, and both are examinable:
- When the force is zero — nothing to turn with
- When the line of action of the force passes through the pivot, so the perpendicular distance is zero
That second case explains something you have done without thinking. Pushing a door directly toward its hinges, along the line of the hinges, does not turn it at all however hard you push. The force is not zero and the distance from the hinge to your hand is not zero — but the perpendicular distance to the line of action is, and that is the quantity in the formula.
What does the principle of moments say, and how do you balance a metre rule?
When a body is in rotational equilibrium, the sum of the anticlockwise moments about any point equals the sum of the clockwise moments about that point.
Which is the same as saying the net moment is zero, once the two senses are given opposite signs.
How the experiment is set up. A uniform metre rule is supported at its mid-point on a wedge, so that it balances horizontally with no weights on it — **that position tells you the rule's own weight acts at the cm mark. Weights are then hung on either side with thread loops, and their distances from the pivot are read off the scale.
Worked example 1 — one weight on each side.** A uniform metre rule is pivoted at its mid-point. A weight of gf is suspended at the cm mark. Where must a weight of gf be suspended to balance it?
**The pivot is at the cm mark**, so the gf weight is
from the pivot, on the left, giving an anticlockwise moment of
The gf weight must give an equal clockwise moment at a distance on the right:
so it hangs at the mark
Check the sense of the answer. The gf weight is lighter, so it must sit further from the pivot than the gf weight — and . Correct. If a lighter weight comes out nearer the pivot, the two sides have been exchanged.
Note that the rule's own weight was ignored, and legitimately so: the rule is uniform and the pivot is at its mid-point, so its weight acts through the pivot and its moment is zero. That is the case-two zero moment of the previous section doing useful work.
Worked example 2 — where the rule's own weight does count. A uniform metre rule of mass g is pivoted at the cm mark. Find the weight that must be suspended at the cm mark to keep the rule horizontal.
**The rule's weight is gf and acts at the cm mark**, which is
to the right of the pivot, giving a clockwise moment of
The unknown weight hangs at the cm mark, which is cm to the left, giving an anticlockwise moment of . Equating:
As soon as the pivot is moved off the mid-point the rule's own weight must be included, and forgetting it is the most common error in this section. **The test is simple: is the pivot at the cm mark? If not, the rule's weight has a moment.
Worked example 3 — deciding whether a rule is balanced.** A metre rule is pivoted at the cm mark. Weights of gf and gf are suspended at the cm and cm marks. Is the rule balanced? If not, where must a gf weight be placed to balance it?
Anticlockwise, from the gf weight cm to the left:
Clockwise, from the gf weight cm to the right:
The two are unequal, so the rule is not balanced — and since the clockwise moment is larger by gf cm, the right-hand side goes down.
To restore balance, the gf weight must supply gf cm anticlockwise, so it goes on the left at
from the pivot, that is at the ** cm mark.
Check by recomputing both sides.** Anticlockwise: gf cm. Clockwise: gf cm. Equal, so the rule balances.
Worked example 4 — a see-saw. Two children weighing kgf and kgf sit on a see-saw. The heavier child sits m from the pivot. Where must the lighter child sit?
The lighter child sits further out, at m against m — which is the everyday form of the principle and the reason a small child can lift a large one.
How the principle is verified in the laboratory with spring balances. The metre rule is suspended horizontally from two spring balances, one near each end, and slotted weights are hung at known positions. Then:
- The readings of the two balances are noted, giving the upward forces
- Moments are taken about the point where one balance is attached, so that balance's own force has zero moment
- The sum of the anticlockwise moments is compared with the sum of the clockwise moments, and the two are found equal within experimental error
- The sum of the two balance readings is also compared with the total downward weight, which verifies the force condition as well
Taking moments about the point where one unknown force acts is the technique to notice, because it removes that unknown from the equation entirely. It is the single most useful trick in the whole topic, and the next section uses it again.
Worked example 5 — a loaded beam, using both conditions. A uniform beam m long and weighing kgf rests on supports at its two ends. A load of kgf is placed m from the left-hand support. Find the force each support exerts.
Take moments about the left support, which removes its unknown force from the equation. The beam's weight acts at its mid-point, m along:
Now use the force condition, since the beam does not move up or down:
Check by taking moments about the right support instead:
The same value by an independent route. And notice that the nearer support carries the larger share of the load, which is what you would expect — a plausibility check that costs nothing.
Which is the same as saying the net moment is zero, once the two senses are given opposite signs.
How the experiment is set up. A uniform metre rule is supported at its mid-point on a wedge, so that it balances horizontally with no weights on it — **that position tells you the rule's own weight acts at the cm mark. Weights are then hung on either side with thread loops, and their distances from the pivot are read off the scale.
Worked example 1 — one weight on each side.** A uniform metre rule is pivoted at its mid-point. A weight of gf is suspended at the cm mark. Where must a weight of gf be suspended to balance it?
**The pivot is at the cm mark**, so the gf weight is
from the pivot, on the left, giving an anticlockwise moment of
The gf weight must give an equal clockwise moment at a distance on the right:
so it hangs at the mark
Check the sense of the answer. The gf weight is lighter, so it must sit further from the pivot than the gf weight — and . Correct. If a lighter weight comes out nearer the pivot, the two sides have been exchanged.
Note that the rule's own weight was ignored, and legitimately so: the rule is uniform and the pivot is at its mid-point, so its weight acts through the pivot and its moment is zero. That is the case-two zero moment of the previous section doing useful work.
Worked example 2 — where the rule's own weight does count. A uniform metre rule of mass g is pivoted at the cm mark. Find the weight that must be suspended at the cm mark to keep the rule horizontal.
**The rule's weight is gf and acts at the cm mark**, which is
to the right of the pivot, giving a clockwise moment of
The unknown weight hangs at the cm mark, which is cm to the left, giving an anticlockwise moment of . Equating:
As soon as the pivot is moved off the mid-point the rule's own weight must be included, and forgetting it is the most common error in this section. **The test is simple: is the pivot at the cm mark? If not, the rule's weight has a moment.
Worked example 3 — deciding whether a rule is balanced.** A metre rule is pivoted at the cm mark. Weights of gf and gf are suspended at the cm and cm marks. Is the rule balanced? If not, where must a gf weight be placed to balance it?
Anticlockwise, from the gf weight cm to the left:
Clockwise, from the gf weight cm to the right:
The two are unequal, so the rule is not balanced — and since the clockwise moment is larger by gf cm, the right-hand side goes down.
To restore balance, the gf weight must supply gf cm anticlockwise, so it goes on the left at
from the pivot, that is at the ** cm mark.
Check by recomputing both sides.** Anticlockwise: gf cm. Clockwise: gf cm. Equal, so the rule balances.
Worked example 4 — a see-saw. Two children weighing kgf and kgf sit on a see-saw. The heavier child sits m from the pivot. Where must the lighter child sit?
The lighter child sits further out, at m against m — which is the everyday form of the principle and the reason a small child can lift a large one.
How the principle is verified in the laboratory with spring balances. The metre rule is suspended horizontally from two spring balances, one near each end, and slotted weights are hung at known positions. Then:
- The readings of the two balances are noted, giving the upward forces
- Moments are taken about the point where one balance is attached, so that balance's own force has zero moment
- The sum of the anticlockwise moments is compared with the sum of the clockwise moments, and the two are found equal within experimental error
- The sum of the two balance readings is also compared with the total downward weight, which verifies the force condition as well
Taking moments about the point where one unknown force acts is the technique to notice, because it removes that unknown from the equation entirely. It is the single most useful trick in the whole topic, and the next section uses it again.
Worked example 5 — a loaded beam, using both conditions. A uniform beam m long and weighing kgf rests on supports at its two ends. A load of kgf is placed m from the left-hand support. Find the force each support exerts.
Take moments about the left support, which removes its unknown force from the equation. The beam's weight acts at its mid-point, m along:
Now use the force condition, since the beam does not move up or down:
Check by taking moments about the right support instead:
The same value by an independent route. And notice that the nearer support carries the larger share of the load, which is what you would expect — a plausibility check that costs nothing.
What are the conditions for equilibrium, and what is dynamic equilibrium?
A body is in equilibrium when the resultant force on it is zero and the resultant moment on it is zero. The first prevents it accelerating along a line; the second prevents it starting to spin.
Translational equilibrium is the force condition:
which, resolved into directions, means the forces in each direction balance — up with down, left with right.
Rotational equilibrium is the moment condition:
which is the principle of moments of the previous section.
Both are needed, and neither implies the other. Two equal and opposite forces applied at different points on a rod have zero resultant force, so the rod will not move off; but they form a couple and the rod will rotate. So the rod is in translational equilibrium and not in rotational equilibrium, and it is not in complete equilibrium.
And now the distinction the syllabus asks about: static and dynamic equilibrium.
Static equilibrium — the body is at rest, and remains at rest. Examples:
- A book lying on a table, where its weight is balanced by the table's normal reaction
- A ladder leaning against a wall and not slipping
- A beam balance at rest with equal weights in its pans
- A hanging picture frame, still on its hook
Dynamic equilibrium — the body is moving with uniform velocity, in a straight line at a constant speed, and the forces on it still balance. Examples:
- A raindrop falling at its terminal velocity, where its weight is balanced by the air's viscous drag and upthrust
- A parachutist descending at a steady speed
- An aeroplane cruising at a constant height and a constant speed
- A car travelling at a constant speed on a level road, where the engine's driving force equals the total resistance
The single fact that makes dynamic equilibrium make sense. A balanced set of forces produces zero acceleration, and zero acceleration means the velocity does not change. It does not mean the velocity is zero. So a body moving steadily is just as much in equilibrium as one at rest, and this follows directly from the first law of motion.
The misconception this is set against. "No net force" does not mean "no motion". **It means no change of motion.** A car cruising at a steady km/h on a level road has balanced forces; the moment it speeds up or slows down or turns, it does not.
Worked example — a body under three forces. A body is acted on by a force of N toward the east and a force of N toward the north. What single additional force would put it in translational equilibrium?
The resultant of the two given forces, being at right angles, has magnitude
**So the third force must be N, directed exactly opposite to that resultant — that is, toward the south-west side, along the line of the resultant but reversed. Then all three forces sum to zero and the body is in translational equilibrium.
But note what this does not guarantee. If the three forces act at different points on an extended body, their moments need not cancel. Translational equilibrium alone is not complete equilibrium**, and a question asking for complete equilibrium wants both conditions stated.
Translational equilibrium is the force condition:
which, resolved into directions, means the forces in each direction balance — up with down, left with right.
Rotational equilibrium is the moment condition:
which is the principle of moments of the previous section.
Both are needed, and neither implies the other. Two equal and opposite forces applied at different points on a rod have zero resultant force, so the rod will not move off; but they form a couple and the rod will rotate. So the rod is in translational equilibrium and not in rotational equilibrium, and it is not in complete equilibrium.
And now the distinction the syllabus asks about: static and dynamic equilibrium.
Static equilibrium — the body is at rest, and remains at rest. Examples:
- A book lying on a table, where its weight is balanced by the table's normal reaction
- A ladder leaning against a wall and not slipping
- A beam balance at rest with equal weights in its pans
- A hanging picture frame, still on its hook
Dynamic equilibrium — the body is moving with uniform velocity, in a straight line at a constant speed, and the forces on it still balance. Examples:
- A raindrop falling at its terminal velocity, where its weight is balanced by the air's viscous drag and upthrust
- A parachutist descending at a steady speed
- An aeroplane cruising at a constant height and a constant speed
- A car travelling at a constant speed on a level road, where the engine's driving force equals the total resistance
The single fact that makes dynamic equilibrium make sense. A balanced set of forces produces zero acceleration, and zero acceleration means the velocity does not change. It does not mean the velocity is zero. So a body moving steadily is just as much in equilibrium as one at rest, and this follows directly from the first law of motion.
The misconception this is set against. "No net force" does not mean "no motion". **It means no change of motion.** A car cruising at a steady km/h on a level road has balanced forces; the moment it speeds up or slows down or turns, it does not.
Worked example — a body under three forces. A body is acted on by a force of N toward the east and a force of N toward the north. What single additional force would put it in translational equilibrium?
The resultant of the two given forces, being at right angles, has magnitude
**So the third force must be N, directed exactly opposite to that resultant — that is, toward the south-west side, along the line of the resultant but reversed. Then all three forces sum to zero and the body is in translational equilibrium.
But note what this does not guarantee. If the three forces act at different points on an extended body, their moments need not cancel. Translational equilibrium alone is not complete equilibrium**, and a question asking for complete equilibrium wants both conditions stated.
How do stable, unstable and neutral equilibrium differ, and where is the centre of gravity?
Nudge the body slightly and watch its centre of gravity. If the centre of gravity rises, the equilibrium is stable; if it falls, unstable; if it stays at the same height, neutral.
First, what the centre of gravity is. The centre of gravity of a body is the point at which its whole weight appears to act, whatever its position. It is the point through which the resultant of the weights of all its particles passes.
Now the three kinds of equilibrium.
Stable equilibrium. On a slight displacement the centre of gravity rises, so the weight and the reaction create a moment that brings the body back to its original position. Examples:
- A book lying flat on a table
- A cone resting on its base
- A table lamp with a heavy, wide base
- A tumbler standing upright
Unstable equilibrium. On a slight displacement the centre of gravity falls, so the moment carries the body further away and it topples. Examples:
- A pencil balanced on its sharpened tip
- A cone resting on its apex
- A stick balanced vertically on a finger
- A tumbler balanced on its rim's edge
Neutral equilibrium. On a slight displacement the centre of gravity stays at the same height, so there is no restoring or overturning moment and the body simply stays where it is put. Examples:
- A ball resting on a horizontal table
- A cone lying on its side
- A cylinder or a pencil lying on its side
The test to apply, in one sentence. Ask whether the centre of gravity goes up, down or neither. That single question classifies every case, and it is better than trying to remember lists of examples.
Where the centre of gravity is for regular bodies, which has to be known:
- A uniform rod — at its mid-point
- A uniform circular disc or a ring — at its centre
- A triangular lamina — at its centroid, where the three medians meet
- A square, rectangular or parallelogram lamina — at the point of intersection of its diagonals
- A uniform solid or hollow cylinder — at the mid-point of its axis
- A solid or hollow sphere — at its centre
- A uniform solid cone — on its axis, at a height of from the base
And the centre of gravity may lie outside the material of the body, which is worth stating because it surprises students. A ring's centre of gravity is at its centre, where there is no metal at all; the same is true of a hollow sphere and of a horseshoe magnet.
How to find the centre of gravity of an irregular lamina — the plumb-line method.
- Make three small holes near the edge of the lamina, well spaced apart
- Suspend the lamina freely from a pin through one hole, and hang a plumb line from the same pin
- When both have come to rest, draw the line of the plumb line on the lamina
- Repeat from the second and third holes
- The three lines meet at one point, which is the centre of gravity
Why the method works. A freely suspended body comes to rest with its centre of gravity vertically below the point of suspension, because in any other position the weight would produce a moment about the pin. So the vertical through the pin passes through the centre of gravity every time, and three such verticals fix the point. Two lines would be enough in principle; the third is drawn as a check.
What the position of the centre of gravity depends on. It depends on the shape of the body and the distribution of its mass — nothing else. Change the shape and the centre of gravity moves, which is why a sheet of paper crumpled into a ball has its centre of gravity in a different place from the flat sheet.
How the centre of gravity governs stability, and this is the part with everyday consequences. A body is more stable when:
- Its centre of gravity is lower, so more tilting is needed before the weight's line of action falls outside the base
- Its base is wider, for the same reason
Which explains a set of familiar designs.
- A racing car is built low and wide, so that it can corner without overturning
- A double-decker bus has its heavier parts and its passengers' luggage low down, and loading the roof heavily makes it less stable
- A Bunsen burner and a table lamp have broad heavy bases
- A ship carries ballast low in its hull, and it becomes dangerously unstable if that ballast is removed
The condition for toppling, stated precisely. A body topples when the vertical line through its centre of gravity falls outside its base. While that line stays inside the base, the weight and the reaction produce a restoring moment and the body settles back. That is why a wide base helps: it gives the line more room before it escapes.
First, what the centre of gravity is. The centre of gravity of a body is the point at which its whole weight appears to act, whatever its position. It is the point through which the resultant of the weights of all its particles passes.
Now the three kinds of equilibrium.
Stable equilibrium. On a slight displacement the centre of gravity rises, so the weight and the reaction create a moment that brings the body back to its original position. Examples:
- A book lying flat on a table
- A cone resting on its base
- A table lamp with a heavy, wide base
- A tumbler standing upright
Unstable equilibrium. On a slight displacement the centre of gravity falls, so the moment carries the body further away and it topples. Examples:
- A pencil balanced on its sharpened tip
- A cone resting on its apex
- A stick balanced vertically on a finger
- A tumbler balanced on its rim's edge
Neutral equilibrium. On a slight displacement the centre of gravity stays at the same height, so there is no restoring or overturning moment and the body simply stays where it is put. Examples:
- A ball resting on a horizontal table
- A cone lying on its side
- A cylinder or a pencil lying on its side
The test to apply, in one sentence. Ask whether the centre of gravity goes up, down or neither. That single question classifies every case, and it is better than trying to remember lists of examples.
Where the centre of gravity is for regular bodies, which has to be known:
- A uniform rod — at its mid-point
- A uniform circular disc or a ring — at its centre
- A triangular lamina — at its centroid, where the three medians meet
- A square, rectangular or parallelogram lamina — at the point of intersection of its diagonals
- A uniform solid or hollow cylinder — at the mid-point of its axis
- A solid or hollow sphere — at its centre
- A uniform solid cone — on its axis, at a height of from the base
And the centre of gravity may lie outside the material of the body, which is worth stating because it surprises students. A ring's centre of gravity is at its centre, where there is no metal at all; the same is true of a hollow sphere and of a horseshoe magnet.
How to find the centre of gravity of an irregular lamina — the plumb-line method.
- Make three small holes near the edge of the lamina, well spaced apart
- Suspend the lamina freely from a pin through one hole, and hang a plumb line from the same pin
- When both have come to rest, draw the line of the plumb line on the lamina
- Repeat from the second and third holes
- The three lines meet at one point, which is the centre of gravity
Why the method works. A freely suspended body comes to rest with its centre of gravity vertically below the point of suspension, because in any other position the weight would produce a moment about the pin. So the vertical through the pin passes through the centre of gravity every time, and three such verticals fix the point. Two lines would be enough in principle; the third is drawn as a check.
What the position of the centre of gravity depends on. It depends on the shape of the body and the distribution of its mass — nothing else. Change the shape and the centre of gravity moves, which is why a sheet of paper crumpled into a ball has its centre of gravity in a different place from the flat sheet.
How the centre of gravity governs stability, and this is the part with everyday consequences. A body is more stable when:
- Its centre of gravity is lower, so more tilting is needed before the weight's line of action falls outside the base
- Its base is wider, for the same reason
Which explains a set of familiar designs.
- A racing car is built low and wide, so that it can corner without overturning
- A double-decker bus has its heavier parts and its passengers' luggage low down, and loading the roof heavily makes it less stable
- A Bunsen burner and a table lamp have broad heavy bases
- A ship carries ballast low in its hull, and it becomes dangerously unstable if that ballast is removed
The condition for toppling, stated precisely. A body topples when the vertical line through its centre of gravity falls outside its base. While that line stays inside the base, the weight and the reaction produce a restoring moment and the body settles back. That is why a wide base helps: it gives the line more room before it escapes.
Exam tip
Which habits protect the marks in a moments numerical?
Draw the rule or the beam, mark the pivot, and label every force with its distance from the pivot before writing any equation. The diagram is worth a mark and it prevents every sign error.
- Measure distances from the pivot, not from the end of the rule — a weight at the cm mark with a pivot at cm is cm away, not cm
- **Ask whether the pivot is at the cm mark.** If it is not, the uniform rule's own weight acts at cm and has a moment that must be included
- Label each moment anticlockwise or clockwise as you compute it
- Take moments about the point where an unknown force acts, so that force drops out of the equation
- Use the force condition as well when there are two unknown reactions
- Keep the units consistent: newtons with metres, or gram-force with centimetres, and convert before multiplying
- Check the sense of the answer: a lighter weight must sit further from the pivot than a heavier one
- Check a beam answer by taking moments about the other support
- State the units of a moment as N m, dyne cm, kgf m or gf cm — a bare number earns nothing
- For an equilibrium question, state both conditions unless only one is asked for
The misconception to name. Equilibrium does not mean at rest. A body moving with uniform velocity is in dynamic equilibrium, because balanced forces mean zero acceleration rather than zero speed. A raindrop at terminal velocity and a cruising aeroplane are both in equilibrium, and writing that an object in equilibrium must be stationary is a marked error.
A second trap. Forgetting the weight of the metre rule when the pivot is off centre. **In the worked example with the pivot at the cm mark, the rule's gf acting at the cm mark supplied a moment of gf cm** — the whole of the answer. Omitting it would have left an equation with no solution at all, since nothing else would be balancing the suspended weight.
- Measure distances from the pivot, not from the end of the rule — a weight at the cm mark with a pivot at cm is cm away, not cm
- **Ask whether the pivot is at the cm mark.** If it is not, the uniform rule's own weight acts at cm and has a moment that must be included
- Label each moment anticlockwise or clockwise as you compute it
- Take moments about the point where an unknown force acts, so that force drops out of the equation
- Use the force condition as well when there are two unknown reactions
- Keep the units consistent: newtons with metres, or gram-force with centimetres, and convert before multiplying
- Check the sense of the answer: a lighter weight must sit further from the pivot than a heavier one
- Check a beam answer by taking moments about the other support
- State the units of a moment as N m, dyne cm, kgf m or gf cm — a bare number earns nothing
- For an equilibrium question, state both conditions unless only one is asked for
The misconception to name. Equilibrium does not mean at rest. A body moving with uniform velocity is in dynamic equilibrium, because balanced forces mean zero acceleration rather than zero speed. A raindrop at terminal velocity and a cruising aeroplane are both in equilibrium, and writing that an object in equilibrium must be stationary is a marked error.
A second trap. Forgetting the weight of the metre rule when the pivot is off centre. **In the worked example with the pivot at the cm mark, the rule's gf acting at the cm mark supplied a moment of gf cm** — the whole of the answer. Omitting it would have left an equation with no solution at all, since nothing else would be balancing the suspended weight.
Did you know
Why does a wide heavy base make something so much harder to tip over?
Stand a tall empty bottle on a table and give it a small push near the top. It rocks and comes back. Push a little harder and it goes over. Somewhere in between there is a critical tilt, and what happens at that tilt is worth understanding exactly.
As the bottle tips, its weight continues to act vertically downward through its centre of gravity. The table pushes up at the edge the bottle is pivoting on. While the centre of gravity is still over the base, those two forces form a moment that rotates the bottle back upright. The instant the vertical through the centre of gravity crosses outside the base, the same two forces produce a moment the other way and the bottle falls.
So the whole question of stability is: how far can it tilt before that line escapes the base? And two things extend the margin:
- A wider base means the line has further to travel before leaving it
- A lower centre of gravity means a larger tilt is needed to move the line the same horizontal distance
Fill the bottle with water and it becomes noticeably harder to tip, because the centre of gravity has dropped toward the bottom. Empty it and put the water on top as a heavy cap and it becomes easier — same total weight, different distribution, different stability.
Which is why the design examples in this chapter all look the same once you see the principle. A racing car, a Bunsen burner, a table lamp, a ship with ballast, a double-decker bus with its passengers below rather than above: every one of them is an attempt to put the mass low and spread the base wide.
And the failures follow the same rule. A bus with heavy luggage strapped to the roof has a raised centre of gravity and tips more easily on a curve; a cyclist carrying a load on their head is less stable than one with the same load in a pannier. Nothing about the weight changed — only its height.
One observation about self-righting objects. A toy that always rocks back upright has a very heavy, rounded base, so that its centre of gravity is extremely low and rises steeply whenever it is tilted. That is stable equilibrium taken to its extreme, and the toy cannot be balanced on its head for the same reason a cone cannot be balanced on its apex — the centre of gravity would be at its highest, which is unstable.
A last connection back to moments. Everything in this section is the principle of moments again, with the pivot at the tipping edge of the base rather than at the middle of a metre rule. The restoring moment is the weight times the horizontal distance from the tipping edge to the centre of gravity — and when that distance reaches zero, the body is on the point of toppling. One formula, two very different-looking topics.
As the bottle tips, its weight continues to act vertically downward through its centre of gravity. The table pushes up at the edge the bottle is pivoting on. While the centre of gravity is still over the base, those two forces form a moment that rotates the bottle back upright. The instant the vertical through the centre of gravity crosses outside the base, the same two forces produce a moment the other way and the bottle falls.
So the whole question of stability is: how far can it tilt before that line escapes the base? And two things extend the margin:
- A wider base means the line has further to travel before leaving it
- A lower centre of gravity means a larger tilt is needed to move the line the same horizontal distance
Fill the bottle with water and it becomes noticeably harder to tip, because the centre of gravity has dropped toward the bottom. Empty it and put the water on top as a heavy cap and it becomes easier — same total weight, different distribution, different stability.
Which is why the design examples in this chapter all look the same once you see the principle. A racing car, a Bunsen burner, a table lamp, a ship with ballast, a double-decker bus with its passengers below rather than above: every one of them is an attempt to put the mass low and spread the base wide.
And the failures follow the same rule. A bus with heavy luggage strapped to the roof has a raised centre of gravity and tips more easily on a curve; a cyclist carrying a load on their head is less stable than one with the same load in a pannier. Nothing about the weight changed — only its height.
One observation about self-righting objects. A toy that always rocks back upright has a very heavy, rounded base, so that its centre of gravity is extremely low and rises steeply whenever it is tilted. That is stable equilibrium taken to its extreme, and the toy cannot be balanced on its head for the same reason a cone cannot be balanced on its apex — the centre of gravity would be at its highest, which is unstable.
A last connection back to moments. Everything in this section is the principle of moments again, with the pivot at the tipping edge of the base rather than at the middle of a metre rule. The restoring moment is the weight times the horizontal distance from the tipping edge to the centre of gravity — and when that distance reaches zero, the body is on the point of toppling. One formula, two very different-looking topics.
Exam relevance
How do moments and equilibrium prepare you for JEE and NEET?
This is foundation work for Class 11 Systems of Particles and Rotational Motion, and for the Laws of Motion chapter, both of which are examined in JEE Main, JEE Advanced and NEET Physics.
Where the moment of a force leads. Class 11 renames it torque and writes it as a vector cross product, so its magnitude becomes — which is the same "force times perpendicular distance" with the perpendicular distance written as . **The requirement that the distance be perpendicular, which you learn here as a word in the definition, is exactly what the encodes**, and the case giving zero torque is the door-pushed-toward-its-hinges case.
Where the principle of moments leads. It becomes the second condition of equilibrium for a rigid body, , used alongside . The technique of taking moments about the point where an unknown force acts, so that it drops out, is used in every Class 11 problem on ladders, beams, hinged rods and balanced planks — and JEE Main sets those regularly as numericals.
Where the centre of gravity leads. Class 11 introduces the centre of mass, computed as , and in a uniform gravitational field it coincides with the centre of gravity. **The results you learn here — mid-point of a rod, centroid of a triangular lamina, for a solid cone — are the answers that formula produces, and Class 11 derives several of them by integration.
Where the toppling condition leads. It reappears as the condition for a body on an incline to topple rather than slide, a standard JEE Advanced comparison, and the answer turns on whether the vertical through the centre of mass leaves the base before friction gives way.
Where the equilibrium types lead. Stable and unstable equilibrium become statements about potential energy — stable at a minimum, unstable at a maximum, neutral where it is constant. The centre of gravity rising or falling is precisely the potential energy increasing or decreasing, and that connection is made explicit in Class 11 and used again in oscillations.
Where dynamic equilibrium leads. It is the working content of the first law, and it underlies terminal velocity in fluid mechanics, which NEET asks about directly for a sphere falling through a viscous liquid.
Question types to expect. At this level: moment calculations, metre-rule balancing, identifying types of equilibrium, and centre-of-gravity locations. In competitive papers: torque numericals, ladder and beam equilibrium, centre-of-mass calculations, toppling versus sliding, and terminal-velocity problems.
The single trap that costs marks. Using the distance along the force's own direction instead of the perpendicular distance. A force whose line of action passes through the pivot has zero moment however large it is**, and the Class 11 version of the error is dropping the from the torque formula.
A second trap. Believing that zero net force means the body is at rest. It means zero acceleration, so uniform velocity qualifies — and in NEET the terminal-velocity question is built on exactly that reading. A body can be in equilibrium and moving fast.
Board versus competitive emphasis. The ICSE paper marks the definition, the units, the labelled diagram, the two sides of the moment equation and the named type of equilibrium; a competitive paper marks a torque, a reaction force or a critical angle. The transferable habit is choosing the pivot to make an unknown disappear — it is the single most efficient move in rigid-body statics, at this level and at every level after it.
Where the moment of a force leads. Class 11 renames it torque and writes it as a vector cross product, so its magnitude becomes — which is the same "force times perpendicular distance" with the perpendicular distance written as . **The requirement that the distance be perpendicular, which you learn here as a word in the definition, is exactly what the encodes**, and the case giving zero torque is the door-pushed-toward-its-hinges case.
Where the principle of moments leads. It becomes the second condition of equilibrium for a rigid body, , used alongside . The technique of taking moments about the point where an unknown force acts, so that it drops out, is used in every Class 11 problem on ladders, beams, hinged rods and balanced planks — and JEE Main sets those regularly as numericals.
Where the centre of gravity leads. Class 11 introduces the centre of mass, computed as , and in a uniform gravitational field it coincides with the centre of gravity. **The results you learn here — mid-point of a rod, centroid of a triangular lamina, for a solid cone — are the answers that formula produces, and Class 11 derives several of them by integration.
Where the toppling condition leads. It reappears as the condition for a body on an incline to topple rather than slide, a standard JEE Advanced comparison, and the answer turns on whether the vertical through the centre of mass leaves the base before friction gives way.
Where the equilibrium types lead. Stable and unstable equilibrium become statements about potential energy — stable at a minimum, unstable at a maximum, neutral where it is constant. The centre of gravity rising or falling is precisely the potential energy increasing or decreasing, and that connection is made explicit in Class 11 and used again in oscillations.
Where dynamic equilibrium leads. It is the working content of the first law, and it underlies terminal velocity in fluid mechanics, which NEET asks about directly for a sphere falling through a viscous liquid.
Question types to expect. At this level: moment calculations, metre-rule balancing, identifying types of equilibrium, and centre-of-gravity locations. In competitive papers: torque numericals, ladder and beam equilibrium, centre-of-mass calculations, toppling versus sliding, and terminal-velocity problems.
The single trap that costs marks. Using the distance along the force's own direction instead of the perpendicular distance. A force whose line of action passes through the pivot has zero moment however large it is**, and the Class 11 version of the error is dropping the from the torque formula.
A second trap. Believing that zero net force means the body is at rest. It means zero acceleration, so uniform velocity qualifies — and in NEET the terminal-velocity question is built on exactly that reading. A body can be in equilibrium and moving fast.
Board versus competitive emphasis. The ICSE paper marks the definition, the units, the labelled diagram, the two sides of the moment equation and the named type of equilibrium; a competitive paper marks a torque, a reaction force or a critical angle. The transferable habit is choosing the pivot to make an unknown disappear — it is the single most efficient move in rigid-body statics, at this level and at every level after it.
Key takeaways
What must you be able to do from this part?
One product, one balance condition and one point that carries all the weight.
- Moment of a force force perpendicular distance of its line of action from the pivot
- SI unit N m; CGS unit dyne cm; gravitational units kgf m and gf cm
- ** N dyne and m cm, so N m dyne cm**; and kgf N
- Moment is a vector, anticlockwise taken positive and clockwise negative
- The moment is zero if the force is zero, or if its line of action passes through the pivot
- ** N on a cm spanner gives N m**; a N m moment at cm needs N; the same N at cm gives only N m
- **A tangential N on a wheel of radius m gives N m
- Principle of moments: in rotational equilibrium, the sum of the anticlockwise moments equals the sum of the clockwise moments
- Measure distances from the pivot.** A gf weight at the cm mark with the pivot at cm gives gf cm, balanced by gf at the cm mark
- **The uniform rule's own weight acts at the cm mark** and matters whenever the pivot is elsewhere — pivoted at cm, a gf rule needs gf at the cm mark
- ** gf at cm and gf at cm** give against gf cm, so a gf weight at the cm mark restores balance
- **A kgf child m out balances a kgf child at m — the lighter one sits further
- Take moments about a point where an unknown force acts** to remove it: a m beam of kgf with a kgf load m from the left gives kgf and kgf
- Complete equilibrium needs both: resultant force zero and resultant moment zero
- Static equilibrium is at rest — a book on a table; dynamic equilibrium is uniform velocity — a raindrop at terminal velocity, a cruising aeroplane
- Zero net force means zero acceleration, not zero speed
- The centre of gravity is the point where the whole weight appears to act, and it may lie outside the material — as in a ring
- Stable equilibrium raises the centre of gravity on displacement; unstable lowers it; neutral leaves it unchanged
- Positions: rod at its mid-point, disc and sphere at the centre, triangular lamina at the centroid, parallelogram at the intersection of the diagonals, cylinder at the mid-point of its axis, solid cone at from the base
- Irregular lamina: suspend from three holes, draw the plumb lines, and they meet at the centre of gravity
- Stability improves with a lower centre of gravity and a wider base, and a body topples when the vertical through its centre of gravity leaves its base
The best self-test needs a ruler and two coins. Balance the ruler on a pencil, put one coin on one side and two stacked coins on the other, and predict where the pair must sit before you slide them there — then move the pencil off the middle and see whether your prediction still works without accounting for the ruler's own weight.
- Moment of a force force perpendicular distance of its line of action from the pivot
- SI unit N m; CGS unit dyne cm; gravitational units kgf m and gf cm
- ** N dyne and m cm, so N m dyne cm**; and kgf N
- Moment is a vector, anticlockwise taken positive and clockwise negative
- The moment is zero if the force is zero, or if its line of action passes through the pivot
- ** N on a cm spanner gives N m**; a N m moment at cm needs N; the same N at cm gives only N m
- **A tangential N on a wheel of radius m gives N m
- Principle of moments: in rotational equilibrium, the sum of the anticlockwise moments equals the sum of the clockwise moments
- Measure distances from the pivot.** A gf weight at the cm mark with the pivot at cm gives gf cm, balanced by gf at the cm mark
- **The uniform rule's own weight acts at the cm mark** and matters whenever the pivot is elsewhere — pivoted at cm, a gf rule needs gf at the cm mark
- ** gf at cm and gf at cm** give against gf cm, so a gf weight at the cm mark restores balance
- **A kgf child m out balances a kgf child at m — the lighter one sits further
- Take moments about a point where an unknown force acts** to remove it: a m beam of kgf with a kgf load m from the left gives kgf and kgf
- Complete equilibrium needs both: resultant force zero and resultant moment zero
- Static equilibrium is at rest — a book on a table; dynamic equilibrium is uniform velocity — a raindrop at terminal velocity, a cruising aeroplane
- Zero net force means zero acceleration, not zero speed
- The centre of gravity is the point where the whole weight appears to act, and it may lie outside the material — as in a ring
- Stable equilibrium raises the centre of gravity on displacement; unstable lowers it; neutral leaves it unchanged
- Positions: rod at its mid-point, disc and sphere at the centre, triangular lamina at the centroid, parallelogram at the intersection of the diagonals, cylinder at the mid-point of its axis, solid cone at from the base
- Irregular lamina: suspend from three holes, draw the plumb lines, and they meet at the centre of gravity
- Stability improves with a lower centre of gravity and a wider base, and a body topples when the vertical through its centre of gravity leaves its base
The best self-test needs a ruler and two coins. Balance the ruler on a pencil, put one coin on one side and two stacked coins on the other, and predict where the pair must sit before you slide them there — then move the pencil off the middle and see whether your prediction still works without accounting for the ruler's own weight.