Turning forces concept | ICSE Class 10 Physics Notes

This note covers translational and rotational motion, the moment of a force, clockwise and anticlockwise moments, units of moment, conditions of equilibrium, the principle of moments, its verification with a metre rule, simple calculations, and the centre of gravity of regular bodies and an irregular lamina.

How do translational and rotational motions differ?

A rigid body is an ideal body whose shape and the distances between its particles remain unchanged. A force is a push or pull. Real bodies deform under forces, but in many situations these deformations are negligible, so treating the body as rigid is useful.

In pure translational motion, every particle of a body has the same velocity at a given instant. Velocity means speed in a specified direction. A rectangular block sliding down an inclined plane without sideways movement is an example of this motion.

In rotational motion about a fixed axis, the body turns about a stationary line called its axis of rotation. Points away from this axis move in circles centred on the axis. Points on the fixed axis remain stationary.

Can one body have both kinds of motion?

A cylinder rolling down an inclined plane moves from one place to another while also turning. Its motion combines translation and rotation. Different points of the rolling cylinder do not all have the same velocity at an instant, so its motion is not pure translation.

A force changes a body's translational motion when the forces do not balance. To understand turning, however, the force's position and direction must also be considered. An opening door makes this distinction clear: its hinges constrain it to turn about the vertical hinge line.

What is the moment of a force?

Definition: The moment of a force about a point or axis measures its turning effect. It is also called torque. Its magnitude is the force multiplied by the perpendicular distance from the point or axis to the force's line of action.

The line of action is the straight line along which a force acts. The moment arm is the perpendicular distance from the turning point to this line. A pivot is a fixed support about which a body can turn.

Let M denote the magnitude of the moment, F the magnitude of the force and d the perpendicular distance from the pivot to its line of action. Then:

M = Fd

The multiplication requires the perpendicular distance, not simply the distance along the body to the point where the force is applied. When the force acts at right angles to a straight rod, the distance along that rod from the pivot is also the required perpendicular distance.

How can the same relation find a force or distance?

If the moment and its non-zero arm are known, rearrange the relation to obtain F = M/d. If the moment and a non-zero force are known, use d = M/F. Each rearrangement retains the same meaning of perpendicular distance.

For a fixed force, a larger perpendicular distance produces a larger moment. For a fixed perpendicular distance, a larger force produces a larger moment. Thus, stating only the force is insufficient to describe its turning effect about a particular pivot.

When is the moment zero?

If the force's line of action passes through the pivot, its moment arm is zero. Its moment about that pivot is therefore zero, even if the force is non-zero. A zero force also has zero moment.

Note: A force can have zero moment about one point and a non-zero moment about another. Identify the point about which the moment is required before selecting the distance.

How are clockwise and anticlockwise moments recognised?

A clockwise moment tends to turn a body in the direction of a clock's hands. An anticlockwise moment tends to turn it in the opposite direction. These descriptions refer to the turning tendency as seen from the chosen side of the body.

Imagine the effect of one force at a time while keeping the pivot fixed. For a horizontal rod viewed on the page, a downward force to the right of its pivot tends to turn it clockwise. A downward force to the left tends to turn it anticlockwise.

Reversing a force reverses its moment about the same pivot. The magnitude remains unchanged if the force magnitude and perpendicular distance remain the same. Normally, anticlockwise moments are taken as positive and clockwise moments as negative.

Why does pushing a door near its outer edge help?

A door turns about the vertical axis through its hinges. A force applied along the hinge line has no turning effect about that line. For a force of given magnitude, applying it at right angles to the door at its outer edge is most effective in producing rotation.

This explains why position matters as well as force. Moving the same perpendicular push towards the hinge line reduces its moment arm. The smaller moment arm gives a smaller turning effect.

What do a steering wheel and bicycle pedal illustrate?

A steering wheel turns about its central axis. A tangent is a straight line touching a circle at a point. A force along the tangent at the rim acts at right angles to the radius, the line from the centre to the rim. Its moment depends on the force and that perpendicular distance.

A bicycle pedal transmits a force through the crank, the arm connecting the pedal to the turning axle. The useful turning effect depends on the perpendicular distance from the axle to the pedal force's line of action. A force directed through the axle has zero moment about it.

Which units are used for force and moment?

SI means the International System of Units. The SI unit of force is the newton, symbol N. One newton gives a mass of one kilogram an acceleration of one metre per second squared. Acceleration is the rate of change of velocity.

Mass measures the quantity of matter in a body. The SI unit of mass is the kilogram, symbol kg. The SI unit of perpendicular distance is the metre, symbol m. The second, symbol s, is the SI unit of time. Thus, 1 N = 1 kg m s⁻².

The SI unit of moment of force is the newton metre, written N m. A force of 1 N with a perpendicular moment arm of 1 m produces a moment of 1 N m.

The cgs system uses centimetre, gram and second as its basic units of length, mass and time. The symbols for centimetre and gram are cm and g, respectively. Its force unit is the dyne, and its moment unit is the dyne centimetre, written dyne cm. One dyne gives one gram an acceleration of one centimetre per second squared.

The moment conversion follows by multiplying the force conversion by the length conversion: 10⁵ × 100 = 10⁷. In an SI calculation, convert a distance given in centimetres to metres before multiplying it by a force in newtons.

Note: Write a moment in N m when the force is in newtons and its perpendicular arm is in metres. A numerical answer must identify the moment unit as well as the turning direction when that direction is asked for.

What conditions are needed for a body to be in equilibrium?

Equilibrium of a rigid body requires both its forces and their turning effects to balance. An external force is one exerted on the chosen body by something outside it. The resultant is the single combined effect obtained by adding forces with their directions taken into account.

Translational equilibrium requires zero resultant external force. Let Fnet mean the resultant external force. The condition is Fnet = 0. For vertical forces alone, the total upward force must equal the total downward force.

Rotational equilibrium requires zero resultant external moment. Let Mnet mean the algebraic sum of moments about the chosen axis. The condition is Mnet = 0. Algebraic addition means including the positive or negative sign assigned to each turning direction.

Why are both conditions necessary?

Consider a light rod, where light means that its mass is negligible. Equal parallel forces acting in the same direction at its two ends have opposite moments about its midpoint. Their moments cancel, but their forces add. The forces are not in translational balance.

Now reverse one force. The equal forces cancel in their resultant, but their moments about the midpoint act in the same turning direction. The rod is in translational equilibrium but not rotational equilibrium.

A couple is a pair of equal, opposite forces acting along different parallel lines. It produces a turning effect without a resultant force. This is why zero resultant force alone does not establish full equilibrium of a body of finite size.

Static equilibrium describes a body remaining at rest with both conditions satisfied. More generally, equilibrium means that translational and rotational motion are not changing; it does not by itself require a body to be at rest.

What is the principle of moments?

Definition: For a body in rotational equilibrium, the sum of its clockwise moments about a point equals the sum of its anticlockwise moments about the same point.

Let Mcw be the total clockwise moment magnitude and Macw the total anticlockwise moment magnitude. With anticlockwise taken as positive, Mnet = Macw − Mcw. Rotational equilibrium therefore requires the two totals to be equal.

Derivation: two forces balancing a light lever

A lever is a rigid rod that can turn about a support called its fulcrum. Consider a light horizontal lever with downward forces F₁ and F₂ on opposite sides. Their perpendicular distances from the fulcrum are d₁ and d₂, respectively.

1. The force F₁ on the left produces an anticlockwise moment of magnitude F₁d₁.
2. The force F₂ on the right produces a clockwise moment of magnitude F₂d₂.
3. The upward support force acts through the fulcrum, so its moment about the fulcrum is zero.
4. For rotational equilibrium, the difference F₁d₁ − F₂d₂ must be zero.

F₁d₁ = F₂d₂

Let R be the upward reaction, meaning the force exerted by the support on the lever. Translational equilibrium also requires R = F₁ + F₂. The support force is absent from the moment equation because of its zero arm, not because it is absent from the body.

A horizontal bar has ends A and B and a fulcrum O between them. Downward forces F₁ and F₂ act at the ends. The upward reaction R acts at O, with distances d₁ and d₂ marked on either side.

See Fig. 6.23 in your NCERT textbook

For equal opposing moments, the larger force has the shorter perpendicular arm. A smaller force can balance a larger force if it acts with a sufficiently larger arm. The forces need not be equal, because the products of force and perpendicular distance determine their turning effects.

For several forces, add every moment in each direction. All distances must refer to the same chosen point. A body's own weight, the gravitational force on it, must also be included whenever it has a non-zero moment about that point.

How can a metre rule and two spring balances verify the principle?

A metre rule is a graduated rule one metre long. A spring balance measures force using the extension of a spring. Slotted weights are removable weights that can be placed on a hanger. Include the hanger's weight when recording the total suspended force.

Suspend the rule horizontally from two vertical spring balances at different positions. Hang slotted weights from known positions on the rule between the supports. When the apparatus is stationary, the balances exert upward forces and the rule and suspended loads exert downward forces.

What measurements and calculations are needed?

1. Determine the rule's own weight and locate its centre of gravity, the point through which its resultant weight acts. Record the support positions and the positions of all suspended loads.
2. Adjust the supports so that the loaded rule is horizontal and stationary. Record both spring-balance readings in newtons.
3. Choose one support position as the point about which to calculate moments. Measure horizontal distances from that point to each vertical force's line of action.
4. Multiply each force by its perpendicular distance. Sort the results into clockwise and anticlockwise moments, and compare the two totals.
5. Change the loads or their positions, allow the rule to settle, and repeat the comparison. Also compare the total upward force with the total downward force.

How does the rule's own weight enter?

For a rule supported at its ends, call the left end A and the right end B. Let Rₐ and Rᵦ be the respective upward balance readings and L their separation.

Let W₀ be the rule's weight and x₀ the distance of its centre of gravity from A. For two suspended loads, let W₁ and W₂ be their weights and x₁ and x₂ their respective distances from A.

RᵦL = W₀x₀ + W₁x₁ + W₂x₂

The left balance has zero moment about A. The right balance's anticlockwise moment balances the clockwise moments of the downward forces. Independently, the force balance is Rₐ + Rᵦ = W₀ + W₁ + W₂.

Draw a horizontal rule with upward force arrows at its two end supports. Mark its own weight downward at its centre of gravity and two load weights downward between the supports. Label every distance from the left support.

Use the measured centre of gravity unless the rule is known to be uniform. Its weight acts at its midpoint if it is uniform. Keep the balances vertical and read the settled apparatus carefully. Agreement within experimental uncertainty, the limits of measurement precision, verifies the moment balance.

How are simple moment calculations organised and checked?

Begin a calculation with a force diagram. Mark the support positions, each downward load and the body's own weight. Convert masses to weights when force values are needed. Let W be weight, m mass in the following formula, and g acceleration due to gravity:

W = mg

The letter m in the formula denotes mass; m written after a numerical length denotes the unit metre. A kilogram is a mass unit, whereas a newton is a force unit. Keep that distinction when entering values in a moment equation.

Derivation: balancing a bar on two supports

Let R₁ and R₂ be the upward forces at the left and right supports of a horizontal bar. Let Wb be the bar's downward weight, Wl a suspended load's weight, and l the separation of the supports.

1. Translational equilibrium requires R₁ + R₂ = Wb + Wl.
2. Take moments about the left support. Its own reaction has zero moment about that point.
3. Let a and b be the horizontal distances from the left support to the bar's centre of gravity and to the load, respectively. Assume both downward forces act between the supports.
4. The upward force at the right support balances the downward moments: R₂l = Wb a + Wl b.

R₂ = (Wb a + Wl b)/l

After finding R₂, subtract it from the total downward force to obtain R₁. This uses both equilibrium conditions and avoids assuming equal support forces simply because there are two supports.

A homogeneous bar has the same mass per unit volume, or mass density, throughout. A uniform cross-section means that cuts perpendicular to its length have the same shape and area. Together, these conditions place its centre of gravity at its midpoint in uniform gravity.

Worked example 1. A homogeneous metal bar of uniform cross-section is 70 cm long and has mass 4.00 kg. Two knife-edge supports are 10 cm from its ends. A 6.00 kg load hangs 30 cm from the left end. Find both upward reactions, taking g = 9.8 m s⁻².

The bar's weight acts at 35 cm. Measured from the left support, the bar's weight has arm 25 cm, the load has arm 20 cm, and the right support has arm 50 cm. In metres these are 0.25 m, 0.20 m and 0.50 m.

Formula: W = mg; R₂ = (Wb a + Wl b)/l; R₁ = Wb + Wl − R₂.

Substitute: Wb = 4.00 × 9.8 = 39.20 N and Wl = 6.00 × 9.8 = 58.80 N. Therefore R₂ = (39.20 × 0.25 + 58.80 × 0.20)/0.50 = 43.12 N.

Answer: R₁ = 98.00 − 43.12 = 54.88 N and R₂ = 43.12 N. Their sum is 98.00 N, equal to the total downward weight.

The horizontal bar AB rests on supports K₁ and K₂ with upward reactions R₁ and R₂. The load acts downward at P, left of the bar's centre G. The bar's own weight acts downward at G.

See Fig. 6.26 in your NCERT textbook

Worked example 2. A metre stick initially balances on a knife edge at its centre. Two coins, each of mass 5 g, are stacked at the 12.0 cm mark. The loaded stick balances at the 45.0 cm mark. Find the stick's mass; the same acceleration due to gravity acts on the stick and coins.

The stick's weight acts at the 50.0 cm mark because the unloaded stick balances there. The coin stack has a moment arm of 45.0 − 12.0 = 33.0 cm. The stick's own weight has arm 50.0 − 45.0 = 5.0 cm.

Formula: W = mg; clockwise moment = anticlockwise moment. Let ms be the stick's mass. In kilograms and metres, ms × g × 0.050 = 0.010 × g × 0.330.

Substitute: Cancel the common non-zero g. Then ms = 0.010 × 0.330/0.050 = 0.066 kg.

Answer: The mass of the metre stick is 66 g. The smaller combined coin mass balances the stick's larger mass because the coins have the longer moment arm.

Worked example 3. A ladder of length 3 m and mass 20 kg rests against a frictionless wall, with its foot 1 m from the wall. Its weight acts at its midpoint. Find the wall and floor reactions, taking $g = 9.8\,\mathrm{m\,s^{-2}}$.

The wall reaction is horizontal. The floor supplies an upward normal reaction and friction towards the wall. By Pythagoras, the height of the ladder's top is $h = \sqrt{3^2-1^2} = 2\sqrt{2}\,\mathrm{m}$.

The weight's perpendicular arm about the foot is half the horizontal span, $a = 0.50\,\mathrm{m}$. Let $H$ be the wall reaction, $N$ the floor's normal reaction and $f$ the friction force.

Formula: $W = mg$, $N = W$, $f = H$. Taking moments about the foot gives $Hh = Wa$. The resultant floor reaction is $R = \sqrt{N^2+f^2}$.

Substitute: $W = 20\times9.8 = 196\,\mathrm{N}$. Hence $H = (196\times0.50)/(2\sqrt{2}) = 34.65\,\mathrm{N}$, so $f = 34.65\,\mathrm{N}$ and $N = 196\,\mathrm{N}$.

Then $R = \sqrt{196^2+34.65^2} \approx 199.0\,\mathrm{N}$. Its angle above the horizontal is $\alpha = \tan^{-1}(196/34.65) \approx 80^\circ$.

Answer: The wall reaction is 34.6 N away from the wall. The floor reaction is approximately 199 N, directed upwards towards the wall at about $80^\circ$ above the horizontal.

Worked example 4. A car of mass 1800 kg has its axles 1.8 m apart. Its centre of gravity is 1.05 m behind the front axle. Find the upward ground force on each front and back wheel, using $g = 9.8\,\mathrm{m\,s^{-2}}$.

Let $A$ and $B$ be the total reactions at the front and back axles. Each axle's reaction is shared equally by its two wheels. The weight acts between the axles, with a perpendicular arm of 1.05 m about the front axle.

Formula: $W = mg$, $A+B = W$. Taking moments about the front axle gives $B\times1.8 = W\times1.05$. Each wheel carries half its axle's reaction.

Substitute: $W = 1800\times9.8 = 17640\,\mathrm{N}$. Therefore $B = 17640\times1.05/1.8 = 10290\,\mathrm{N}$, and $A = 17640-10290 = 7350\,\mathrm{N}$.

The force on each front wheel is $7350/2 = 3675\,\mathrm{N}$; the force on each back wheel is $10290/2 = 5145\,\mathrm{N}$.

Answer: Each front wheel has an upward reaction of 3675 N and each back wheel 5145 N. Checking the total gives $2\times3675+2\times5145 = 17640\,\mathrm{N}$, equal to the car's weight.

Worked example 5. A non-uniform horizontal bar of length 2 m and weight $W$ hangs at rest from two strings of negligible weight. The left and right strings make angles of $36.9^\circ$ and $53.1^\circ$ with the vertical. Find its centre of gravity's distance from the left end.

Let $T_L$ and $T_R$ be the string tensions. A tension's vertical component is its magnitude multiplied by the cosine of its angle with the vertical; its horizontal component uses the sine.

Formula: Horizontal force balance gives $T_L\sin36.9^\circ = T_R\sin53.1^\circ$. Vertical balance gives $T_L\cos36.9^\circ+T_R\cos53.1^\circ = W$.

Substitute: Use $\sin36.9^\circ \approx 0.6$, $\cos36.9^\circ \approx 0.8$, $\sin53.1^\circ \approx 0.8$ and $\cos53.1^\circ \approx 0.6$. Horizontal balance gives $0.6T_L = 0.8T_R$, so $T_L = 4T_R/3$.

Vertical balance becomes $0.8(4T_R/3)+0.6T_R = W$, hence $T_R = 0.6W$. Its vertical component is therefore $0.6T_R = 0.36W$.

Take moments about the left end. The left tension and the right tension's horizontal component have zero moment. If the centre of gravity is at distance $d$, then $Wd = (0.36W)(2\,\mathrm{m})$.

Answer: Cancelling $W$ gives $d = 0.72\,\mathrm{m}$, so the centre of gravity is approximately 0.72 m from the left end. The bar is non-uniform, so its centre of gravity need not be at its midpoint.

What is the centre of gravity of a regular body?

The centre of gravity of a body is the point through which its resultant weight acts. The total moment of all its gravitational forces about this point is zero. This lets the distributed weight of a body be represented by one downward force in a moment diagram.

A body's centre of mass is the point representing the mass-weighted average position of its particles, so heavier portions contribute more to its location. In uniform gravity, where gravitational acceleration is the same throughout the body, the centre of gravity coincides with the centre of mass.

For a small body near the Earth's surface, the variation of gravity across it can be neglected. The locations found from symmetry for a uniform mass distribution can then be used as locations of the centre of gravity.

How do shape and uniformity help?

A uniform body has an even distribution of mass appropriate to its shape. A lamina is a thin, flat plate. For regular uniform bodies in uniform gravity, symmetry identifies the following positions.

A median of a triangle joins a vertex to the midpoint of the opposite side. The medians meet at the centroid. The centre of gravity of a uniform triangular lamina is at this intersection, not at a point chosen by visual estimation.

The centre of gravity need not lie in the material of a body. The centre of a uniform ring lies in its empty central region. Shape alone is insufficient if the mass distribution is not uniform, so the uniformity condition belongs with each regular-body statement.

Worked example 6. Three particles of masses 100 g, 150 g and 200 g occupy the vertices of an equilateral triangle of side 0.5 m. Find the centre of mass, placing the first two particles at $(0,0)$ and $(0.5,0)$, with the third above them.

The third vertex lies vertically above the midpoint of the base. Its height is $h = \sqrt{0.5^2-0.25^2} = 0.25\sqrt{3}\,\mathrm{m}$, so its coordinates in metres are $(0.25,0.25\sqrt{3})$.

Formula: Each centre of mass coordinate is a mass-weighted average: $X = (m_1x_1+m_2x_2+m_3x_3)/(m_1+m_2+m_3)$ and $Y = (m_1y_1+m_2y_2+m_3y_3)/(m_1+m_2+m_3)$.

Substitute: The total mass is $100+150+200 = 450\,\mathrm{g}$. Using grams consistently in numerator and denominator cancels the mass unit, leaving each coordinate in metres.

Thus $X = [100(0)+150(0.5)+200(0.25)]/450 = 125/450 = 5/18\,\mathrm{m}$.

Similarly, $Y = [100(0)+150(0)+200(0.25\sqrt{3})]/450 = \sqrt{3}/9\,\mathrm{m}$.

Answer: The centre of mass is at $X \approx 0.278\,\mathrm{m}$ and $Y \approx 0.192\,\mathrm{m}$, or 0.278 m along the base and 0.192 m above it. Unequal particle masses mean this is not the triangle's geometrical centre.

How is the centre of gravity of an irregular lamina found?

An irregular lamina lacks the simple outline used to locate a centre by symmetry. Its centre of gravity can be found by suspension. When it hangs freely at rest, the vertical line through its suspension point passes through its centre of gravity.

The reason is moment balance. The supporting force passes through the suspension point and has no moment about it. If the line of action of the lamina's weight missed that point, the weight would produce a turning effect and the lamina would not remain in equilibrium.

What is the suspension procedure?

1. Suspend the lamina freely from a point near its edge and let it come to rest.
2. Use a plumb line, a thread carrying a weight that hangs vertically, to identify the vertical through the suspension point. Mark this line on the lamina.
3. Suspend the lamina from a different point and mark the new vertical in the same way.
4. Locate the intersection of the marked lines. Repeat with another suspension point to check the position of the centre of gravity.

An irregular outline has suspension points A, B and C. The vertical through A is marked AA₁. Dashed lines associated with the other suspension positions meet this line at G, the centre of gravity.

See Fig. 6.25 in your NCERT textbook

A related method is to balance a small cardboard lamina horizontally on a narrow pencil tip. The successful support point is its centre of gravity. The upward support balances the weight, and the gravitational turning effects about that point balance.

The lamina's outline can be irregular without changing the equilibrium principle. The suspension method uses the direction of weight and the absence of a resultant moment, so it does not require the lamina to have a geometrical centre identified from its outline.

Glossary

• Rigid body — An ideal body whose shape and distances between its constituent particles remain unchanged.
• Translation — Motion in which all particles of a body have the same velocity at each instant.
• Axis of rotation — The line about which a body turns during rotational motion.
• Moment of a force — The turning effect measured by force multiplied by its perpendicular moment arm.
• Line of action — The straight line along which a particular force acts on a body.
• Moment arm — The perpendicular distance from a pivot to a force's line of action.
• Fulcrum — The support about which a lever can turn under applied forces.
• Reaction — The force exerted by a support on the body it supports.
• Translational equilibrium — The condition in which the resultant external force acting on a body is zero.
• Rotational equilibrium — The condition in which the resultant external moment acting on a body is zero.
• Couple — Two equal and opposite forces acting along different parallel lines, producing a turning effect.
• Centre of gravity — The point through which the resultant weight of a body acts.
• Lamina — A thin flat plate whose thickness is small compared with its other dimensions.
• Plumb line — A thread carrying a weight that indicates the vertical when hanging at rest.

Common errors and misconceptions

• Misconception: Force alone determines the turning effect. Correct: The perpendicular distance from the chosen pivot to the line of action also determines the moment.
• Misconception: Any distance from the pivot can be used. Correct: The moment arm must be perpendicular to the force's line of action.
• Misconception: A zero resultant force proves complete equilibrium. Correct: The resultant moment must also be zero; a couple has zero resultant force but a turning effect.
• Misconception: A force through the pivot must be absent. Correct: It can be non-zero, although its moment about that pivot is zero.
• Misconception: The two supports of a bar must carry equal forces. Correct: Their reactions depend on the positions and magnitudes of all the downward forces.
• Misconception: A rule's own weight can be left out. Correct: Include its weight in force balance and include its moment whenever its line of action misses the chosen pivot.
• Misconception: Centre of gravity means the geometrical centre of every outline. Correct: Mass distribution matters; simple symmetry results require the stated uniformity conditions.
• Misconception: Kilograms and newtons can be substituted interchangeably. Correct: Kilograms measure mass; convert mass to weight using the given gravitational acceleration when calculating force in newtons.

Exam-style questions with model answers

Q1. Define the moment of a force about a pivot and name its SI unit. [2 marks]

1. The moment is the force's turning effect about the pivot, measured by force multiplied by the perpendicular distance to its line of action.
2. Its SI unit is the newton metre, written N m.

Q2. A non-zero force acts along a line passing through a door's hinge axis. Explain its turning effect about that axis. Then explain why applying the same force at right angles to the door at its outer edge is more effective. [3 marks]

1. The moment arm is the perpendicular distance from the axis to the force's line of action. A line through the hinge axis has zero moment arm.
2. The moment is force multiplied by this arm, so the first force has zero moment about the hinge axis despite being non-zero.
3. The perpendicular force at the outer edge has a larger moment arm and therefore produces a larger turning effect for the same force magnitude.

Q3. State both conditions for complete equilibrium of a rigid body. Explain why two equal opposite forces along different parallel lines do not satisfy both conditions. [3 marks]

1. The resultant external force must be zero for translational equilibrium. This means that the forces balance when their directions are taken into account.
2. The resultant external moment must also be zero for rotational equilibrium, with all moments calculated about the same chosen point.
3. The given forces cancel as forces but form a couple: their turning effects add instead of cancelling. Consequently, they do not provide rotational equilibrium.

Q4. A light horizontal lever has downward force F₁ on its left at perpendicular distance d₁ from its fulcrum, and downward force F₂ on its right at distance d₂. An upward support reaction R acts at the fulcrum. Derive the moment balance and state the force balance for equilibrium. [4 marks]

1. The left force produces an anticlockwise moment F₁d₁, found by multiplying the force by its perpendicular distance from the fulcrum.
2. The right force produces a clockwise moment F₂d₂. The reaction R has zero moment because its line of action passes through the fulcrum.
3. Rotational equilibrium therefore requires F₁d₁ − F₂d₂ = 0, or F₁d₁ = F₂d₂.
4. Translational equilibrium requires the upward force to equal the total downward force: R = F₁ + F₂. The lever's own weight is negligible because it is light.

Q5. A homogeneous bar of uniform cross-section, length 70 cm and mass 4.00 kg, rests horizontally on supports 10 cm from each end. A 6.00 kg load hangs 30 cm from the left end. With g = 9.8 m s⁻², calculate the upward reaction at each support. [5 marks]

1. The bar's weight acts at its midpoint, 35 cm from the left end. The weights are 4.00 × 9.8 = 39.20 N for the bar and 6.00 × 9.8 = 58.80 N for the load.
2. Let R₁ and R₂ be the left and right reactions. Vertical force balance gives R₁ + R₂ = 39.20 + 58.80 = 98.00 N.
3. About the left support, the perpendicular distances are 0.25 m for the bar's weight, 0.20 m for the load, and 0.50 m for the right reaction.
4. The moment balance is R₂ × 0.50 = 39.20 × 0.25 + 58.80 × 0.20. Hence R₂ = 43.12 N upward.
5. Substitute into the force balance: R₁ = 98.00 − 43.12 = 54.88 N upward. The two upward reactions together balance the total downward weight.

Q6. Describe how to verify the principle of moments using a horizontal metre rule suspended by two vertical spring balances, with slotted weights hanging between the supports. Include the rule's own weight and a force-balance check. [5 marks]

1. Determine the rule's weight and centre of gravity. Record the support positions, the total weights of the suspended loads including their hangers, and their positions along the rule.
2. Suspend the rule horizontally from the two balances. Once the loaded apparatus is stationary, record both upward force readings in newtons.
3. Choose one support as the moment point. Measure every perpendicular distance from it to the other support, each load and the rule's centre of gravity.
4. Calculate force multiplied by perpendicular distance for each force. Compare the clockwise and anticlockwise totals; they should agree within experimental uncertainty.
5. Check that the two upward readings add to the rule's weight plus the suspended weights. Repeat with changed loads or positions to check the moment balance again.

Q7. Explain how suspending an irregular cardboard lamina from different points locates its centre of gravity. Include why each marked vertical must contain that point. [4 marks]

1. Suspend the lamina freely near its edge and let it come to rest. Use a plumb line to mark the vertical through the suspension point.
2. The support force has zero moment about the suspension point. For equilibrium, the weight's line of action must also pass through that point, so the centre of gravity lies on the marked vertical.
3. Suspend the lamina from a different point and mark the new vertical. The centre of gravity must lie on this line as well.
4. The intersection of the marked lines gives the centre of gravity. A further suspension and vertical line provide a check on the position.

Key takeaways

• A force's turning effect depends on its magnitude and its perpendicular distance from the chosen pivot.
• The moment arm reaches the force's line of action, so a distance measured along the body is not automatically correct.
• Clockwise and anticlockwise moments have opposite signs when moments are added about the same point.
• Complete equilibrium requires both zero resultant external force and zero resultant external moment on the body.
• The principle of moments equates total clockwise and anticlockwise moments for a body in rotational equilibrium.
• A support force has zero moment about its own point of application but still enters the force balance.
• Include a rule's own weight and locate its centre of gravity before calculating its contribution to the moment balance.
• The centre of gravity coincides with the centre of mass in uniform gravity; symmetry results also require the specified mass distribution.
• Intersecting vertical lines from different suspension positions locates the centre of gravity of an irregular lamina.

Test yourself

What distance is used when finding a moment?

Use the perpendicular distance from the chosen pivot to the force's line of action.

Can a non-zero force have zero moment?

Yes. Its moment about a pivot is zero if its line of action passes through that pivot.

What changes when a force is reversed at the same position?

Its turning direction reverses. Its moment magnitude stays unchanged if the force magnitude and perpendicular distance stay unchanged.

What is the cgs unit of moment?

It is the dyne centimetre; one newton metre equals 10⁷ dyne centimetres.

Why is zero resultant force insufficient for complete equilibrium?

The forces could form a couple with a non-zero moment. Rotational equilibrium must also be checked.

Why can the support reaction disappear from a moment equation?

If moments are taken about that support, its reaction has zero perpendicular distance and therefore zero moment.

Where is the centre of gravity of a uniform ring in uniform gravity?

It is at the ring's geometrical centre, which lies in the empty region enclosed by the ring.

What does the intersection of suspension verticals identify?

It identifies the centre of gravity, since that point lies on the vertical through each suspension point in equilibrium.