9th Class Physics · Punjab Board · Complete learning guide
Chapter 4Turning Effect of Forces
Every topic, every MCQ and every exercise question, explained step by step in simple language. Written for the Punjab Board new-syllabus Physics book, English medium.
Chapter 3 was about forces that make a body move along a line. This chapter is about forces that make a body turn, and about the conditions for a body to stay balanced.
Definition
Rigid body
A body in which the distance between any two points stays the same when forces act on it; it does not bend or stretch.
Definition
Axis of rotation
The straight line about which a rigid body turns. Every particle of the body moves in a circle whose centre lies on this line.
The body turns about the z-axis: points A and B move in circles whose centres lie on the axis.
Definition
Line of action of a force
The straight line along which a force acts, drawn through its point of application.
Definition
Like parallel forces
Parallel forces that act in the same direction.
Definition
Unlike parallel forces
Parallel forces that act in opposite directions.
F₁ and F₂ point the same way, so they are like parallel forces. F₃ points the opposite way, so it is unlike to both.
Like parallel forces
Unlike parallel forces
Weights of two children on a see-saw (both downward)
The two hands turning a steering wheel
Weights of the boxes on a shelf
The up and down pushes of a couple
Resultant = sum of the forces
Resultant = difference of the forces
Remember
A door turns about its hinges, a wheel about its axle and a see-saw about its pivot: each of these is an axis of rotation.
Topic 2
Adding forces: the head-to-tail rule
Force is a vector, so forces are added by the head-to-tail rule, not by simply adding their sizes (unless they act along one line).
Steps
Choose a scale, for example 100 N = 1 cm.
Draw the first force to scale in its direction.
Draw the next force with its tail on the head of the first, and so on.
The resultant runs from the tail of the first force to the head of the last. Measure its length and convert with the scale; measure its angle with a protractor.
Tail of each force on the head of the one before; the resultant joins the first tail to the last head.
Worked exampleAdd F₁ = 200 N at 30°, F₂ = 300 N at 45° and F₃ = 250 N at 60° to the x-axis (scale 100 N = 1 cm). (Solved Example 4.1)
Graphically: draw 2 cm at 30°, then 3 cm at 45° from its head, then 2.5 cm at 60°. Join the first tail to the last head and measure.
Check by components: ΣFx = 200 cos 30° + 300 cos 45° + 250 cos 60° = 173.2 + 212.1 + 125.0 = 510.3 N.
ΣFy = 200 sin 30° + 300 sin 45° + 250 sin 60° = 100.0 + 212.1 + 216.5 = 528.6 N.
F = √(510.3² + 528.6²) ≈ 735 N, and θ = tan⁻¹(528.6 ÷ 510.3) ≈ 46°.
Answerabout 735 N at 46° to the x-axis (the line measures about 7.35 cm)
Solved Example 4.1 drawn to scale (100 N = 1 cm): the resultant F measures 7.35 cm at 46°, so F = 735 N.
Check your book
Some solved copies read the drawing as 7.1 cm, giving 710 N at 43°. The components show the true answer is about 735 N at 46°, so a careful drawing measures 7.3–7.4 cm. In the exam, show your drawing with its scale and give the value you measure.
Topic 3
Resolution of a force: rectangular components
Definition
Resolution of a force
Splitting one force into two or more parts (components) that together have the same effect as the force.
Definition
Rectangular (perpendicular) components
The two components of a force that are at right angles to each other, usually along the x-axis and the y-axis.
F at angle θ to the x-axis splits into Fx = F cos θ along x and Fy = F sin θ along y.
Fx = F cos θ Fy = F sin θ
θ is measured from the x-axis
Trigonometry is the branch of mathematics that deals with right-angled triangles. For an angle θ: sin θ = perpendicular ÷ hypotenuse, cos θ = base ÷ hypotenuse, tan θ = perpendicular ÷ base.
θ
sin θ
cos θ
tan θ
0°
0
1
0
30°
0.5
0.866
0.577
45°
0.707
0.707
1
60°
0.866
0.5
1.732
90°
1
0
not defined
Worked exampleA force of 160 N acts on a box at 60° to the horizontal. Find its x and y components. (Solved Example 4.3)
Fx = F cos 60° = 160 × 0.5 = 80 N.
Fy = F sin 60° = 160 × 0.866 = 138.6 N.
AnswerFx = 80 N, Fy = 138.6 N
Getting the force back from its components
F = √(Fx² + Fy²) θ = tan⁻¹(Fy ÷ Fx)
Worked exampleFind the force whose components are Fx = 6 N and Fy = 8 N.
F = √(6² + 8²) = √100 = 10 N.
θ = tan⁻¹(8 ÷ 6) = tan⁻¹(1.33) ≈ 53°.
Answer10 N at about 53° to the x-axis
Exam tip
When you pull a trolley by a handle at an angle, only Fx pulls it forward; Fy tries to lift it. That is why pulling along the ground is easiest with the handle nearly level.
Topic 4
Turning effect of a force: torque
Definition
Moment arm
The perpendicular distance from the axis of rotation to the line of action of the force.
Definition
Torque (moment of a force)
The turning effect of a force, equal to the force multiplied by its moment arm. It is a vector; SI unit newton metre (N m).
τ = F × l
τ = torque, F = force, l = moment arm
The force at the end of the spanner acts at right angles, so the moment arm l is the full length from the nut.
Torque depends on
The size of the force: a bigger push turns more strongly.
The moment arm: the same force turns more strongly when it acts farther from the axis.
Direction of turning
Sign of the torque
Anticlockwise
Positive (+)
Clockwise
Negative (−)
Worked exampleA spanner 25 cm long is used to open a nut. A force of 400 N is applied at its end. Find the torque. (Solved Example 4.2)
Moment arm l = 25 cm = 0.25 m.
τ = F × l = 400 × 0.25.
Answer100 N m
Everyday example
How the moment arm is used
Door handle far from the hinges
Long moment arm, so a small push opens the door
Long spanner for a tight nut
Longer arm, less force for the same torque
Long handle on a hand pump or tap
Easier to turn
Pushing a door near the hinges
Short moment arm: hard to open
Remember
If the line of action passes through the axis, the moment arm is zero and so is the torque, however large the force.
Topic 5
Torque when the force acts at an angle
Often the force is not at right angles to the arm. Let r be the distance from the axis O to the point where F acts, and θ the angle between r and F.
F acts at P, at angle θ to OP produced. Its line of action is drawn back, and the perpendicular from O to it is the moment arm l = r sin θ.
Proof of τ = rF sin θ
Resolve F into F cos θ along r and F sin θ at right angles to r.
F cos θ acts along a line through the axis, so its torque is zero.
F sin θ acts at distance r, so its torque is r × F sin θ.
Equivalently, the moment arm is l = r sin θ, so τ = F × l = F × r sin θ.
τ = rF sin θ
Angle θ between r and F
sin θ
Torque
90° (force at right angles)
1
Largest: τ = rF
30°
0.5
Half the largest
0° (force along the arm)
0
Zero
Worked exampleA force of 20 N acts at the end of a 0.5 m spanner at 30° to the spanner. Find the torque.
τ = rF sin θ = 0.5 × 20 × sin 30°.
= 0.5 × 20 × 0.5.
Answer5 N m (half of the 10 N m a perpendicular push would give)
Exam tip
To loosen a stiff nut, always push at right angles to the spanner: that gives the most torque for your effort.
Topic 6
Principle of moments
Definition
Principle of moments
When a body is balanced, the sum of the clockwise moments about any point equals the sum of the anticlockwise moments about that point.
Σ clockwise moments = Σ anticlockwise moments
w₁ and w₂ turn the rule anticlockwise, w₃ turns it clockwise. Balanced: w₁l₁ + w₂l₂ = w₃l₃.
Worked exampleA metre stick is pinned at end O. A force of 18 N acts at right angles at its free end (1.0 m from O). A 60 N force acts at 30° to the stick. Where must the 60 N force act so the stick does not turn? (Solved Example 4.4)
Resolve the 60 N force: the part along the stick, 60 cos 30°, passes through O and has no torque.
The part at right angles: 60 sin 30° = 30 N.
Principle of moments about O: 30 × d = 18 × 1.0.
d = 18 ÷ 30 = 0.6 m.
Answer0.6 m from the pinned end
Example 4.4: F = 60 N is resolved into Fx along the stick (it passes through O, no torque) and Fy = 30 N at right angles.
Application
How the principle of moments is used
See-saw
A heavier child sits nearer the pivot to balance a lighter one
Beam balance
Equal arms: equal weights balance, so goods weigh what the weights show
Crowbar, scissors, pliers (levers)
A small effort on a long arm overcomes a large load on a short arm
Tightrope walker with a long stick
Small shifts of the stick make large balancing moments
Remember
A balance with unequal arms cheats: if the weights go on the shorter arm, weights × short arm = goods × long arm, so the goods weigh less than the weights. With arms of 9 cm and 10 cm, 1 kg of weights balances only 0.9 kg of goods.
Topic 7
Couple
Definition
Couple
Two equal and opposite parallel forces acting at two different points of a body, along different lines of action.
Definition
Couple arm
The perpendicular distance between the lines of action of the two forces of a couple.
Torque of a couple = F × d
d = couple arm
(a) F₁ and F₂ act along one line through A, O and B: no turning. (b) They act at A and B along different lines: a couple, which turns the body about O.F₁ at A and F₂ at B are equal and opposite, and both turn the body the same way about O: τ = F₁r₁ + F₂r₂ = F × AB.
A couple has a zero resultant force (F − F = 0), so it does not move the body along a line, but it has a non-zero torque, so it turns the body. This shows that ΣF = 0 alone does not guarantee equilibrium.
Examples of couples
Turning a steering wheel with both hands.
Opening or closing a water tap.
Turning a key in a lock, or the cap of a bottle.
Winding a clock or a toy with a key.
Topic 8
Centre of mass and centre of gravity
Definition
Centre of mass
The point of a body where its whole mass can be taken to be concentrated.
Definition
Centre of gravity
The point of a body through which its whole weight appears to act, whatever the position of the body.
Remember
Where g is the same over the whole body (any everyday object on the Earth), the centre of mass and the centre of gravity are the same point. Both are often written G.
Uniform body
Centre of gravity
Rod or metre rule
Middle point
Square or rectangular sheet
Where the diagonals cross
Circular disc, sphere
Centre
Ring, hollow pipe
Centre, in the empty space (outside the material)
Triangular sheet
Where the medians cross
Cylinder
Middle of its axis
Finding the centre of gravity of an irregular lamina
Make two or three small holes near the edge of the flat sheet (lamina).
Hang it freely from one hole on a pin, so it can swing and come to rest.
Hang a plumb line from the same pin and draw its vertical line on the lamina.
Repeat from a second hole. The point where the two lines cross is the centre of gravity G.
Hang it from a third hole: that line also passes through G, which checks the result.
Fig. (a): hung from one point, the CG lies on the vertical line through it. Fig. (b): the line from a second point of suspension crosses the first at the CG.
Topic 9
Equilibrium and its conditions
Definition
Equilibrium
A body is in equilibrium if it has no acceleration: no linear acceleration and no angular (turning) acceleration.
Definition
Static equilibrium
Equilibrium of a body at rest, for example a book on a table or a picture on a wall.
Definition
Dynamic equilibrium
Equilibrium of a body moving with uniform velocity, for example a paratrooper falling at constant velocity or a car at steady speed on a straight road.
Definition
First condition of equilibrium
The vector sum of all the external forces on the body is zero: ΣF = 0, that is ΣFx = 0 and ΣFy = 0. This gives translational equilibrium.
Definition
Second condition of equilibrium
The vector sum of all the torques on the body about any point is zero: Στ = 0. This gives rotational equilibrium.
Condition
Meaning
Example of only this one holding
ΣF = 0
No linear acceleration
A couple on a steering wheel: ΣF = 0 but it turns
Στ = 0
No rotational acceleration
A box pushed by one force through its centre: no turning, but it accelerates
Remember
A body is completely in equilibrium only when both conditions hold.
Solved Example 4.5: strings at A and B, each 20 cm from the middle point O, where the weight acts.
Worked exampleA 5 N picture hangs from two vertical strings at A and B, each 20 cm from its centre of gravity O. Find the tensions. (Solved Example 4.5)
All forces are vertical: T₁ + T₂ − 5 = 0 (first condition).
Moments about O: T₁ × 0.2 = T₂ × 0.2, so T₁ = T₂ (second condition).
So 2T₁ = 5.
AnswerT₁ = T₂ = 2.5 N
Topic 10
States of equilibrium
Definition
Stable equilibrium
A body is in stable equilibrium if, after a slight tilt, it comes back to its original position. Tilting raises its centre of gravity.
Definition
Unstable equilibrium
A body is in unstable equilibrium if, after a slight tilt, it moves further away from its original position. Tilting lowers its centre of gravity.
Definition
Neutral equilibrium
A body is in neutral equilibrium if, after being disturbed, it comes to rest in its new position. Its centre of gravity stays at the same height.
A cone on its base, a ball on top of a dome, and a cylinder lying on its side.
Stable
Unstable
Neutral
After a slight tilt
Comes back
Falls further away
Stays in the new position
Centre of gravity on tilting
Rises
Falls
Stays at the same height
Examples
Cone on its base; book lying on a table; ball in a bowl
Cone on its tip; pencil on its point; ball on a dome
Cone or cylinder on its curved side; ball on a flat floor
Exam tip
One body can be in all three states: a cone is stable on its base, unstable on its tip and neutral on its side. Likewise a cylinder standing on its circular base is stable, but lying on its side it is neutral.
Topic 11
Stability and toppling
A tilted body falls back if the vertical line through its centre of gravity still falls inside its base, because its weight then turns it back. It topples if that line falls outside the base, because its weight then turns it over.
Tilted by the same angle: the line from G of the low, wide box stays inside the base; that of the tall, narrow box falls outside, so it topples.
Two ways to improve stability
Lower the centre of gravity: keep the heavy parts at the bottom.
Widen the base: spread the points of support farther apart.
Example
How stability is improved
Racing cars
Very low body, wheels set wide outside the body
Buses and trucks
Heavy loads on the floor, never on the roof
Balancing toys
Weights hang below the pivot, so G is below the point of support
Table lamps, pedestal fans
Heavy, wide base
Standing in a moving bus
Feet apart to widen the base
Remember
A body topples once the vertical line from its centre of gravity passes outside its base.
Topic 12
Centripetal force
Definition
Centripetal force
The force that keeps a body moving in a circle. It acts towards the centre of the circle, at right angles to the velocity.
Fc = mv² ÷ r
m = mass, v = speed, r = radius of the circle
The velocity is along the tangent; the centripetal force points along the radius to the centre, so the two are at right angles.
The velocity at any point of a circle is along the tangent, and the force is along the radius towards the centre. A tangent is always at right angles to the radius, so the centripetal force is always at right angles to the velocity. It changes only the direction of motion, not the speed.
Motion in a circle
What provides the centripetal force
Stone whirled on a string
Tension in the string
Car turning a corner
Friction between the tyres and the road
Moon around the Earth, satellites
Gravitational pull of the Earth
Clothes in a spin dryer
Push of the drum wall
Worked exampleA 0.5 kg stone is whirled at 4 m s⁻¹ on a string 1 m long. Find the tension in the string.
Fc = mv² ÷ r = 0.5 × 4² ÷ 1.
= 0.5 × 16 ÷ 1.
Answer8 N
Remember
If the string breaks, there is no centripetal force, and the stone flies off along the tangent, not outward along the radius.
The wording follows published solved exercises and every answer was re-checked by hand, so the wording or the order of options may differ slightly from your print.
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Set A: Textbook exercise 10 MCQs
The multiple-choice questions at the end of Chapter 4 in your book, in the book's option order.
Q4.1Parallel Forces and Adding Forces
A particle is simultaneously acted upon by two forces of 4 N and 3 N. The net force on the particle is:
Answer(b) between 1 N and 7 N
Step by step
Forces in the same direction add: 4 + 3 = 7 N (the largest possible resultant).
Forces in opposite directions subtract: 4 − 3 = 1 N (the smallest possible resultant).
At any other angle the resultant lies between these, for example 5 N at right angles.
The angle is not given, so the answer is "between 1 N and 7 N".
Why the others are wrong
(a) 1 N is only the special case when the forces are exactly opposite.
(c) 5 N is only the case when the forces are at right angles (3-4-5 triangle).
(d) 7 N is only the case when both forces act in the same direction.
Remember
Resultant of two forces: from (difference) to (sum).
Q4.2Resolution of Forces
A force F is making an angle of 60° with the x-axis. Its y-component is equal to:
Answer(b) F sin 60°
Step by step
Draw F at θ = 60° to the x-axis and drop a perpendicular to make a right-angled triangle.
The side along the x-axis is F cos θ; the side parallel to the y-axis (opposite θ) is F sin θ.
So Fy = F sin 60° (about 0.866 F).
Why the others are wrong
(a) F is the whole force; a component is always smaller (unless the force lies along that axis).
(c) F cos 60° is the x-component.
(d) tan is not used to find a component; tan θ = Fy ÷ Fx gives the direction.
Remember
Angle from the x-axis: x gets cos, y gets sin.
Q4.3Torque (Moment of a Force)
Moment of force is called:
Answer(d) torque
Step by step
Moment of a force = force × moment arm (τ = F × l).
This turning effect is named torque, written τ (tau), SI unit N m.
Why the others are wrong
(a) the moment arm is the perpendicular distance l, only one part of the moment.
(b) a couple is a pair of equal, opposite, parallel forces; it produces torque but is not its name.
(c) the couple arm is the distance between the two forces of a couple.
Remember
Moment of force = torque.
Q4.4Equilibrium
If F₁ and F₂ are the forces acting on a body and τ is the torque produced in it, the body will be completely in equilibrium when:
Answer(a) ΣF = 0 and Στ = 0
Step by step
First condition: ΣF = 0, so the body has no linear acceleration.
Second condition: Στ = 0, so the body has no rotational (angular) acceleration.
Only when both hold is the body completely in equilibrium.
Why the others are wrong
(b) with a resultant torque the body would start to rotate (for example a couple).
(c) with a resultant force the body would accelerate along a line.
(d) neither condition is satisfied.
Remember
Complete equilibrium: both sums zero.
Q4.5Principle of Moments
A shopkeeper sells his articles by a balance having unequal arms of the pans. If he puts the weights in the pan having the shorter arm, then the customer:
Answer(a) loses
Step by step
Let the weights W hang on the short arm a, and the goods G on the long arm b (b > a).
Principle of moments at balance: W × a = G × b.
So G = W × (a ÷ b). Since a ÷ b is less than 1, G is less than W.
The customer pays for W but receives the smaller G: he loses.
Why the others are wrong
(b) the customer would gain only if the weights were on the longer arm.
(c) that is true only for equal arms.
(d) the principle of moments tells us exactly who loses.
Remember
The longer arm needs less weight to balance; goods on the longer arm means fewer goods.
Q4.6Principle of Moments
A man walks on a tight rope. He balances himself by holding a bamboo stick horizontally. It is an application of:
Answer(c) principle of moments
Step by step
A tipping body has an unbalanced turning moment about the rope.
The long stick lets a small shift of weight act at a large distance, making a large opposite moment.
He keeps clockwise and anticlockwise moments equal: the principle of moments.
Why the others are wrong
(a) no collision or explosion is involved.
(b) the second law links force and acceleration; balancing is about turning effects.
(d) action and reaction do not explain how he stays upright.
Remember
Balancing = equal clockwise and anticlockwise moments.
Q4.7States of Equilibrium and Stability
In stable equilibrium, the centre of gravity of the body lies:
Answer(b) at the lowest position
Step by step
Tilt a body in stable equilibrium slightly: its centre of gravity rises.
Its weight then makes a moment that brings it back down.
So the resting position is the one with the centre of gravity at its lowest.
Why the others are wrong
(a) the highest position belongs to unstable equilibrium (a pencil on its tip).
(c) any position at the same height is neutral equilibrium.
(d) G may be outside some bodies (a ring), but that is not what makes them stable.
Remember
Stable = lowest G; unstable = highest G; neutral = G stays at the same height.
Q4.8Centre of Mass and Centre of Gravity
The centre of mass of a body:
Answer(d) may lie within, outside or on the surface
Step by step
Solid sphere or block: the centre of mass is inside the material.
Ring or hollow pipe: the centre of mass is at the centre, where there is no material.
So it may lie within, outside, or on the surface.
Why the others are wrong
(a) a ring's centre of mass is not in the material.
(b) a solid ball's centre of mass is inside it.
(c) it is on the surface only for some special shapes.
Remember
Think of a ring: its centre of mass is in the hole.
Q4.9States of Equilibrium and Stability
A cylinder resting on its circular base is in:
Answer(a) stable equilibrium
Step by step
Standing on its flat base, the cylinder's centre of gravity is above the middle of the base.
Tilt it slightly onto an edge: the centre of gravity rises and stays above the base.
Its weight brings it back to the upright position, so it is in stable equilibrium.
Why the others are wrong
(b) it does not fall over when tilted slightly.
(c) neutral is a cylinder lying on its curved side, which rolls and stays wherever it is put. The question says "on its circular base", which is the flat end.
(d) it is stable.
Remember
Upright on its base: stable. Lying on its side: neutral.
Q4.10Centripetal Force
Centripetal force is given by:
Answer(c) mv²/r
Step by step
Centripetal force grows with mass and with the square of the speed.
It is smaller for a larger circle.
So Fc = mv² ÷ r.
Why the others are wrong
(a) rF (with r and F perpendicular) is a torque.
(b) torque uses sin θ (rF sin θ), and it is still a torque, not a centripetal force.
(d) mv/r has the wrong units for a force; the speed must be squared.
Remember
Centripetal: m v squared over r.
Set B: More MCQs from this chapter 14 MCQs
Extra questions on the same ideas, of the kind set in board papers and class tests.
QB1Torque (Moment of a Force)
The SI unit of torque is:
Answer(c) N m
Step by step
τ = F × l.
Unit: N × m = N m (newton metre).
Why the others are wrong
(a) N m⁻¹ divides by the distance instead of multiplying.
(b) N s is the unit of impulse.
(d) kg m s⁻¹ is the unit of momentum.
Remember
Force × distance: newton metre.
QB2Torque (Moment of a Force)
A force of 20 N is applied at right angles to the end of a spanner 0.5 m long. The torque on the nut is:
Answer(b) 10 N m
Step by step
The force is perpendicular, so the moment arm is the whole length, 0.5 m.
τ = F × l = 20 × 0.5 = 10 N m.
Why the others are wrong
(a) 40 divides 20 by 0.5 instead of multiplying.
(c) 20.5 adds the numbers.
(d) 0.025 divides 0.5 by 20.
Remember
Torque = force × moment arm (in metres).
QB3Torque (Moment of a Force)
A force produces no torque about an axis when:
Answer(a) its line of action passes through the axis
Step by step
τ = F × l, where l is the perpendicular distance from the axis to the line of action.
If the line of action passes through the axis, l = 0.
So τ = 0: pushing a door straight at its hinges does not turn it.
Why the others are wrong
(b) a larger force gives a larger torque, unless the moment arm is zero.
(c) a long moment arm gives a large torque.
(d) a perpendicular force gives the largest torque.
Remember
Push at the hinge: nothing turns.
QB4Resolution of Forces
A force of 100 N acts at 30° to the horizontal. Its horizontal component is:
Answer(d) 86.6 N
Step by step
Fx = F cos θ.
cos 30° = 0.866.
Fx = 100 × 0.866 = 86.6 N.
Why the others are wrong
(a) 50 N is F sin 30°, the vertical component.
(b) 57.7 N is F tan 30°, which is not a component.
(c) 100 N is the whole force.
Remember
cos 30° ≈ 0.866, sin 30° = 0.5.
QB5Resolution of Forces
The rectangular components of a force are Fx = 6 N and Fy = 8 N. The magnitude of the force is:
Answer(d) 10 N
Step by step
The components are at right angles, so use Pythagoras.
F = √(6² + 8²) = √(36 + 64) = √100.
F = 10 N.
Why the others are wrong
(a) 14 N adds the components as if they were in the same direction.
(b) 2 N subtracts them.
(c) 48 multiplies them.
Remember
6-8-10 is a doubled 3-4-5 triangle.
QB6Principle of Moments
A 20 N weight hangs 30 cm from the pivot of a balanced metre rule. What weight hung 20 cm from the pivot on the other side keeps it balanced?
Answer(c) 30 N
Step by step
Clockwise moment = anticlockwise moment.
20 × 30 = w × 20.
w = 600 ÷ 20 = 30 N. (The units of distance cancel, so cm can be used on both sides.)
Why the others are wrong
(a) 13.3 N puts the distances the wrong way round (20 × 20 ÷ 30).
(b) 600 is the moment in N cm, not the weight.
(d) 40 N does not balance: 40 × 20 = 800, not 600.
Remember
Closer to the pivot needs a bigger weight.
QB7Couple
A couple acting on a body produces:
Answer(b) rotational motion only
Step by step
ΣF = F − F = 0: the body does not move as a whole.
The forces are on different lines, so Στ = F × d ≠ 0.
So the body only rotates.
Why the others are wrong
(a) the resultant force of a couple is zero, so there is no translation.
(c) the zero resultant force rules out translation.
(d) the torque of the couple turns the body.
Remember
A couple turns but does not shift.
QB8Centre of Mass and Centre of Gravity
The centre of gravity of a uniform ring lies:
Answer(d) at its centre, where there is no material
Step by step
For a uniform symmetrical body, the centre of gravity is at its geometric centre.
The centre of a ring is in the hole.
So the centre of gravity lies outside the material.
Why the others are wrong
(a) no single point on the ring is the centre of the ring.
(b) the ring is symmetrical, so G is not at the top.
(c) the ring is symmetrical, so G is not at the bottom.
Remember
Ring, bangle, tyre: G is in the hole.
QB9Equilibrium
A car moving with uniform velocity on a straight road is in:
Answer(a) dynamic equilibrium
Step by step
Equilibrium means no acceleration.
Uniform velocity in a straight line: acceleration = 0, so ΣF = 0.
It is moving, so it is dynamic (not static) equilibrium.
Why the others are wrong
(b) static equilibrium is for a body at rest.
(c) unstable equilibrium is about what happens after a small tilt, not about moving at constant velocity.
(d) zero acceleration means it is in equilibrium.
Remember
At rest: static. Uniform velocity: dynamic.
QB10States of Equilibrium and Stability
A ball lying on a flat, horizontal floor is in:
Answer(c) neutral equilibrium
Step by step
Push the ball slightly so it rolls.
Its centre of gravity neither rises nor falls.
It stays wherever it stops: neutral equilibrium.
Why the others are wrong
(a) it does not come back to where it was.
(b) it does not move further away on its own.
(d) the forces on it balance, so it is in equilibrium.
Remember
Neutral: G at the same height, stays in its new place.
QB11States of Equilibrium and Stability
Racing cars are built low and wide mainly to:
Answer(d) lower their centre of gravity and widen their base
Step by step
A body topples when the vertical line from its centre of gravity falls outside its base.
A low G and a wide base keep that line inside the base even at a large tilt.
So the car stays stable on fast turns.
Why the others are wrong
(a) the design is for stability, not looks.
(b) racing cars are kept as light as possible.
(c) grip comes from tyres; being low and wide is about not toppling.
Remember
Stability: low G, wide base.
QB12Centripetal Force
A body moves in a circle of fixed radius. If its speed is doubled, the centripetal force needed becomes:
Answer(c) four times
Step by step
Fc = mv² ÷ r with m and r fixed, so Fc ∝ v².
New force = m(2v)² ÷ r = 4 × mv² ÷ r.
So the force becomes four times as large.
Why the others are wrong
(a) the force increases, it does not decrease.
(b) doubling would be true if Fc ∝ v; the speed is squared.
(d) Fc depends on speed.
Remember
Speed squared: double the speed, four times the force.
QB13States of Equilibrium and Stability
A body is tilted slightly. Its centre of gravity rises and it comes back to its original position. The body was in:
Answer(a) stable equilibrium
Step by step
Tilting raises G.
The weight then makes a moment that turns the body back.
It returns to its original position: stable equilibrium.
Why the others are wrong
(b) an unstable body moves further away when tilted, and its G falls.
(c) a neutral body stays in its new position.
(d) dynamic equilibrium is about moving with uniform velocity.
Remember
Comes back = stable.
QB14Torque (Moment of a Force)
Door handles are fixed far from the hinges so that:
Answer(b) a small force gives a large torque
Step by step
τ = F × l.
The farther the handle from the hinge, the larger l.
For the same torque, a larger l needs a smaller F.
Why the others are wrong
(a) the position of the handle does not change the door's weight.
(c) the hinges carry the same weight wherever the handle is.
(d) the handle is there to turn the door, not to keep it still.
Parallel forces that act in the same direction, for example the weights of two children sitting on a see-saw.
Definition
Unlike parallel forces
Parallel forces that act in opposite directions, for example the two forces of the hands on a steering wheel.
F₁ and F₂ are like parallel forces; F₃ is unlike to them.
Short Q 4.2
What are rectangular components of a vector and their values?
Answer
The two components of a vector that are at right angles to each other, usually along the x-axis and y-axis, are its rectangular components. For a force F at angle θ to the x-axis:
Fx = F cos θ Fy = F sin θ
Rectangular components of a force.
Short Q 4.3
What is the line of action of a force?
Answer
The line of action of a force is the straight line along which the force acts, drawn through its point of application. The moment arm of a force is measured from the axis to this line.
Short Q 4.4
Define moment of a force. Prove that τ = rF sin θ, where θ is the angle between r and F.
Answer
Definition
Moment of a force (torque)
The product of the force and its moment arm: τ = F × l. SI unit N m.
Proof: Let F act at a point at distance r from the axis O, at angle θ to r. The moment arm is the perpendicular distance from O to the line of action of F, which is l = r sin θ.
τ = F × l = F × (r sin θ) = rF sin θ
Equally, only the part F sin θ of the force is at right angles to r; the part F cos θ passes through O and has no torque.
Moment arm l = r sin θ.
Short Q 4.5
With the help of a diagram, show that the resultant force is zero but the resultant torque is not zero.
Answer
Apply two equal and opposite parallel forces F at two different points of a body, so that their lines of action are a distance d apart (a couple).
Resultant force: F + (−F) = 0.
Resultant torque: both forces turn the body the same way, so τ = F × d, which is not zero.
F₁ down at A and F₂ up at B: ΣF = 0, but both turn the body the same way about O, so Στ = F₁r₁ + F₂r₂ ≠ 0.
Short Q 4.6
Identify the state of equilibrium in each case in the figure given below (a cone on its base, a ball on top of a hemisphere, a cylinder lying on its side).
Answer
(a) Cone on its base — stable: a slight tilt raises its centre of gravity, so it falls back.
(b) Ball on top of the hemisphere — unstable: a slight push lowers its centre of gravity, so it rolls off.
(c) Cylinder lying on its side — neutral: it rolls, its centre of gravity stays at the same height, and it rests in its new position.
Stable, unstable and neutral equilibrium.
Short Q 4.7
Give an example of a body which is moving yet in equilibrium.
Answer
A paratrooper falling with uniform (terminal) velocity. His weight downward is balanced by the air resistance upward, so the net force is zero and he has no acceleration. He is in dynamic equilibrium. A car moving at a steady speed on a straight level road is another example.
Short Q 4.8
Define centre of mass and centre of gravity of a body.
Answer
Definition
Centre of mass
The point of a body where its whole mass can be taken to be concentrated.
Definition
Centre of gravity
The point through which the whole weight of the body appears to act. Near the Earth's surface, where g is uniform, the two points coincide.
Short Q 4.9
What are two basic principles of stability in physics which are applied in designing balancing toys and racing cars?
Answer
Lowering the centre of gravity: racing cars are built very low, and balancing toys carry weights below the pivot.
Widening the base: racing cars have their wheels set far apart, outside the body.
Both keep the vertical line through the centre of gravity inside the base even when the body tilts, so it does not topple.
Short Q 4.10
How can you prove that the centripetal force always acts perpendicular to the velocity?
Answer
For a body moving in a circle, the velocity at any point is along the tangent to the circle at that point. The centripetal force acts along the radius, towards the centre. A tangent to a circle is always perpendicular to the radius at that point, so the centripetal force is always perpendicular to the velocity.
v along the tangent, Fc along the radius: at right angles.
These test whether you really understand. Give a reason every time.
CRQ 4.1
A car travels at the same speed around two curves with different radii. For which radius does the car experience more centripetal force? Prove your answer.
Answer
Answer
The car needs more centripetal force on the curve with the smaller radius (the sharper curve).
Proof
Fc = mv² ÷ r
The mass m and speed v are the same on both curves, so Fc ∝ 1/r: the smaller r, the larger Fc.
Worked exampleExample: a 1000 kg car at 10 m s⁻¹
On a curve of radius 50 m: Fc = 1000 × 10² ÷ 50 = 2000 N.
On a curve of radius 25 m: Fc = 1000 × 10² ÷ 25 = 4000 N.
Answerhalf the radius, twice the force
Remember
This force comes from friction between the tyres and the road, which is why sharp bends must be taken slowly.
CRQ 4.2
A ripe mango does not normally fall from the tree. But when the branch of the tree is shaken, the mango falls down easily. Can you tell the reason?
Answer
Hanging still, the mango is in equilibrium: its weight is balanced by the pull of its weak stalk.
When the branch is shaken, the branch moves suddenly but the mango, because of its inertia, tends to stay where it was. The stalk must then pull hard on the mango to make it move with the branch, and it swings the mango away from its resting position. The ripe stalk is weak, so this extra force breaks it and the mango falls.
Remember
Shaking disturbs the mango's equilibrium: the forces on it no longer balance, and the weak stalk cannot supply the extra force needed.
CRQ 4.3
Discuss the concepts of stability and centre of gravity in relation to objects toppling over. Give an example where an object's centre of gravity affects its stability, and explain how altering its base of support can influence stability.
Answer
Stability and the centre of gravity
When a body is tilted, its weight acts down through its centre of gravity G. If the vertical line through G still falls inside the base, the weight turns the body back; if it falls outside the base, the weight turns it over and it topples. The lower G is, the farther the body can tilt before that happens.
The line from G inside the base: the body falls back. Outside: it topples.
Example
A double-decker bus with passengers on the upper deck, or a truck with heavy goods on its roof, has a high centre of gravity and can overturn on a sharp turn. With the load on the floor, G is low and the bus is stable.
Effect of the base
A wider base means the body must tilt farther before the line from G passes outside it, so it is more stable. That is why a table lamp has a heavy wide base and why we stand with our feet apart in a moving bus.
CRQ 4.4
Why can an accelerated body not be considered in equilibrium?
Answer
A body is in equilibrium only if it has no acceleration. If a body accelerates, then by F = ma there must be a resultant force on it, so ΣF ≠ 0 and the first condition of equilibrium is not satisfied. Hence an accelerated body cannot be in equilibrium.
Remember
Even a body moving in a circle at constant speed is accelerating (its direction changes), so it is not in equilibrium.
CRQ 4.5
Two boxes of the same weight but different heights are lying on the floor of a truck. If the truck makes a sudden stop, which box is more likely to tumble over? Why?
Answer
The taller box is more likely to tumble over.
When the truck stops suddenly, both boxes tend to keep moving forward (inertia), and friction at their bases tilts them forward. The taller box has a higher centre of gravity, so even a small tilt carries the vertical line from its G outside the front edge of its base, and it topples. The shorter box, with its low G, can tilt more and still fall back.
Tall, narrow box: the line from G leaves the base after only a small tilt.
Set A: Textbook comprehensive questions 4 questions
The long questions at the end of the chapter in your book.
Long Q 4.1
Explain the principle of moments with an example.
Answer
Statement
Definition
Principle of moments
When a body is balanced, the sum of the clockwise moments about any point equals the sum of the anticlockwise moments about that point.
Experiment
Balance a metre rule on a wedge at its centre of gravity so that it is horizontal. Hang two weights w₁ and w₂ on one side at distances l₁ and l₂ from the wedge, and move a third weight w₃ on the other side until the rule balances again at distance l₃.
Diagram
Metre rule balanced on a wedge with weights on both sides.
Explanation
w₁ and w₂ try to turn the rule anticlockwise; their moments are w₁ × l₁ and w₂ × l₂.
w₃ tries to turn it clockwise; its moment is w₃ × l₃.
When the rule is balanced: w₁l₁ + w₂l₂ = w₃l₃.
Σ anticlockwise moments = Σ clockwise moments
Example
Worked examplew₁ = 2 N at 20 cm and w₂ = 1 N at 40 cm on the left; find w₃ at 30 cm on the right
Anticlockwise: 2 × 20 + 1 × 40 = 80 N cm.
Clockwise: w₃ × 30.
w₃ = 80 ÷ 30 ≈ 2.67 N.
Answerw₃ ≈ 2.67 N
Applications
See-saws and beam balances.
Levers such as a crowbar, scissors and a wheelbarrow.
A tightrope walker's balancing stick.
Conclusion
The principle of moments is the second condition of equilibrium (Στ = 0) written for a body that can turn about a pivot.
Long Q 4.2
Describe how you could determine the centre of gravity of an irregular-shaped lamina experimentally.
Answer
Principle
A body hanging freely from a point comes to rest with its centre of gravity directly below the point of suspension, because only then does its weight have no turning effect. So G lies on the vertical line through the point of suspension.
Apparatus
Irregular cardboard sheet (lamina) with three small holes near its edge
Stand with a pin or nail
Plumb line (thread with a small weight)
Pencil and ruler
Method
Hang the lamina on the pin through hole A so it can swing freely, and let it come to rest.
Hang the plumb line from the same pin in front of the lamina.
Mark two points along the plumb line and join them with a straight line.
Hang the lamina from hole B and draw the plumb line in the same way.
The point where the two lines cross is the centre of gravity G.
Diagram
The vertical lines from two points of suspension cross at the centre of gravity.
Check and precautions
Hang the lamina from a third hole: the third line should pass through the same point G.
The lamina and the plumb line must hang freely, not touching the stand.
Make the holes small and near the edge, far apart from each other.
Result
The centre of gravity is the point where the vertical lines from different points of suspension intersect.
Long Q 4.3
State and explain the conditions of equilibrium. Show that if two equal and opposite forces act at two different points of a body, the resultant force is zero but the resultant torque is not zero.
Answer
First condition of equilibrium
Definition
Statement
The vector sum of all the external forces acting on a body is zero.
ΣF = 0, that is ΣFx = 0 and ΣFy = 0
It ensures translational equilibrium: the body has no linear acceleration, so it stays at rest or moves with uniform velocity.
Second condition of equilibrium
Definition
Statement
The vector sum of all the torques acting on a body about any point is zero.
Στ = 0
It ensures rotational equilibrium: the body has no angular acceleration, so it does not start to turn (or keeps turning steadily).
Complete equilibrium
A body is completely in equilibrium only if both conditions hold. A book lying on a table is an example.
Equal and opposite forces at two different points
(a) F₁ and F₂ on one line through A, O, B: no turning. (b) On different lines at A and B: a couple.
Case (a): F₁ and F₂ are equal and opposite and act along the same line. ΣF = 0 and neither has a moment about the point where they act, so the body stays in equilibrium.
Case (b): F₁ and F₂ are equal and opposite but act along different lines, a distance d apart. Still ΣF = F − F = 0, so the first condition holds. But both forces turn the body the same way, so Στ = F × d ≠ 0 and the body rotates.
Conclusion
The first condition alone is not enough: a couple satisfies ΣF = 0 but not Στ = 0, so a body acted on by a couple is not in equilibrium.
Long Q 4.4
How can the stability of an object be improved? Give a few examples to support your answer.
Answer
Meaning of stability
A body is stable if it returns to its position after being tilted. It topples when the vertical line through its centre of gravity falls outside its base.
Lowering the centre of gravity, so the body must tilt much farther before the line from G leaves the base.
Widening the base, so the line from G has farther to go before it passes outside the base.
Examples
Example
What makes it stable
Low armchair
More stable than a tall stool: low centre of gravity
Loading a bus or truck
Heavy goods on the floor keep G low; luggage on the roof raises G and the bus can overturn on a sharp turn
Racing cars
Very low body and wheels wide apart, outside the main body
Balancing toys
G kept below the pivot; when tilted, G rises, so the toy swings back
Table lamps, pedestal fans
Heavy, wide base
Conclusion
To make anything stable: keep its centre of gravity low and its base wide.
Set B: More important long questions 6 questions
Long questions often set in board papers from this chapter, answered in the same full style.
Imp. LQ 1
What is torque? Explain the factors on which it depends, with examples. Write its formula and unit.
Answer
Definition
Definition
Torque (moment of a force)
The turning effect of a force, equal to the product of the force and its moment arm.
τ = F × l
SI unit: newton metre (N m)
Definition
Moment arm
The perpendicular distance from the axis of rotation to the line of action of the force.
Diagram
Force F at the end of a spanner, moment arm l.
Factors on which torque depends
Magnitude of the force: a larger force gives a larger torque. A tight nut needs a harder push.
Moment arm: the farther the line of action is from the axis, the larger the torque. A longer spanner opens a nut more easily; a door is easiest to open at the handle, far from the hinges.
Angle of the force: at an angle θ to the arm, τ = rF sin θ. It is largest at 90° and zero when the force points along the arm or through the axis.
Sign of torque
Turning
Sign
Anticlockwise
Positive
Clockwise
Negative
Example
Worked exampleA 300 N force is applied at right angles to a door handle 1.2 m from the hinges
τ = F × l = 300 × 1.2.
Answer360 N m
Conclusion
Pushing a door at its hinges does not open it, because the moment arm is zero.
Imp. LQ 2
What is meant by resolution of a force? Find the rectangular components of a force, and explain how a force can be found from its components.
Answer
Definition
Definition
Resolution of a force
Splitting a force into parts (components) that together have the same effect as the single force.
Definition
Rectangular components
The two components at right angles to each other, along the x-axis and y-axis.
Diagram
F at angle θ, with its components Fx and Fy.
Finding the components
Draw F as OA at angle θ with the x-axis. From A drop a perpendicular AB onto the x-axis. OB is the x-component and BA is the y-component. In the right-angled triangle OBA:
cos θ = OB ÷ OA = Fx ÷ F, so Fx = F cos θ.
sin θ = BA ÷ OA = Fy ÷ F, so Fy = F sin θ.
Finding a force from its components
By Pythagoras, OA² = OB² + BA²:
F = √(Fx² + Fy²) θ = tan⁻¹(Fy ÷ Fx)
Example
Worked exampleA force of 200 N acts at 30° to the horizontal
Fx = 200 cos 30° = 200 × 0.866 = 173.2 N.
Fy = 200 sin 30° = 200 × 0.5 = 100 N.
Check: √(173.2² + 100²) = √40 000 = 200 N.
AnswerFx = 173.2 N, Fy = 100 N
Imp. LQ 3
What is a couple? Explain with examples, and show that the torque of a couple equals F × d.
Answer
Definition
Definition
Couple
Two equal and opposite parallel forces acting at different points of a body along different lines of action.
Definition
Couple arm
The perpendicular distance d between the lines of action of the two forces.
Diagram
(b) shows a couple: two equal and opposite forces a distance d apart.
Torque of a couple
Take moments about the point midway between the two forces. Each force is d/2 from it.
Torque of each force = F × d/2, and both turn the body the same way.
Total torque = F × d/2 + F × d/2 = F × d.
τ = F × d
Effect of a couple
Since the two forces are equal and opposite, the resultant force is zero and the body does not move along a line. But the torque F × d is not zero, so the body rotates.
Examples
Turning a steering wheel with both hands.
Opening a tap or a bottle cap.
Turning a key in a lock.
Winding a clock.
Conclusion
A couple produces rotation only, never translation.
Imp. LQ 4
What are the states of equilibrium? Explain each with examples in terms of the centre of gravity.
Answer
Introduction
A body in equilibrium may react to a slight push in three ways. What happens depends on whether the push raises, lowers or does not change the height of its centre of gravity G.
Diagram
Stable, unstable and neutral equilibrium.
Stable equilibrium
Definition
Definition
After a slight tilt the body comes back to its original position.
Tilting raises G, and the weight turns the body back. Examples: a cone on its base, a book lying on a table, a ball in a bowl, a cylinder standing on its base.
Unstable equilibrium
Definition
Definition
After a slight tilt the body moves further away from its original position.
Tilting lowers G, and the weight turns the body farther over. Examples: a cone balanced on its tip, a pencil on its point, a ball on top of a dome.
Neutral equilibrium
Definition
Definition
After being disturbed the body comes to rest in its new position.
G stays at the same height, so the weight has no turning effect. Examples: a ball on a flat floor, a cylinder or cone lying on its curved side.
State
G on tilting
Result
Stable
Rises
Returns
Unstable
Falls
Topples
Neutral
Same height
Stays in the new position
Imp. LQ 5
Define centripetal force. Explain it with examples and state the factors on which it depends.
Answer
Definition
Definition
Centripetal force
The force that keeps a body moving in a circular path. It is always directed towards the centre of the circle.
Fc = mv² ÷ r
Diagram
Velocity along the tangent; centripetal force towards the centre.
Explanation
By Newton's first law a moving body goes straight unless a force acts. To move in a circle its direction must change all the time, so a force must keep pulling it towards the centre. This force is at right angles to the velocity, so it changes the direction but not the speed.
Factors
Factor
Effect on Fc
Mass m
Fc ∝ m: double the mass, double the force
Speed v
Fc ∝ v²: double the speed, four times the force
Radius r
Fc ∝ 1/r: half the radius, twice the force
Examples
Stone on a string: the tension in the string pulls the stone towards the hand.
Car on a curve: friction between tyres and road provides it.
Moon and satellites: the Earth's gravity provides it.
If the force stops
If the string breaks, the stone flies off along the tangent at that point, moving in a straight line.
Worked exampleA 0.5 kg stone at 4 m s⁻¹ on a 1 m string
Fc = 0.5 × 4² ÷ 1.
Answer8 N
Imp. LQ 6
Explain the addition of forces by the head-to-tail rule, with the help of an example.
Answer
Introduction
Forces are vectors: their directions matter, so they cannot simply be added as numbers. The head-to-tail rule adds them by drawing.
Method
Choose a suitable scale, for example 100 N = 1 cm.
Draw the first force to scale in its direction from a point O.
Draw the second force with its tail at the head of the first, then the third with its tail at the head of the second.
Join O (the first tail) to the head of the last force. This line is the resultant.
Measure its length (convert with the scale) and the angle it makes with the x-axis.
Diagram
The resultant runs from the first tail to the last head.
Example (Solved Example 4.1)
Worked exampleF₁ = 200 N at 30°, F₂ = 300 N at 45°, F₃ = 250 N at 60°
Draw 2 cm at 30°, then 3 cm at 45°, then 2.5 cm at 60°, head to tail.
Join the first tail to the last head: it measures about 7.35 cm at 46° to the x-axis.
Resultant = 7.35 × 100 ≈ 735 N.
Answerabout 735 N at 46°
(a) The three forces from O. (b) Added head to tail to scale: F = 7.35 cm = 735 N at 46°.
Check your book
The components confirm this: ΣFx = 510.3 N, ΣFy = 528.6 N, so F = √(510.3² + 528.6²) ≈ 735 N and θ ≈ 46°. Answers read off a rough drawing, such as 710 N at 43°, are a little out.
Conclusion
The order in which the forces are drawn does not change the resultant.
A force of 200 N is acting on a cart at an angle of 30° with the horizontal direction. Find the x and y components of the force.
Given Data
F = 200 N
θ = 30°
To Find
x-component Fx
y-component Fy
Formulas
Fx = F cos θ
Fy = F sin θ
Calculation
Fx = F cos θ = 200 × cos 30° = 200 × 0.866 = 173.2 N.
Fy = F sin θ = 200 × sin 30° = 200 × 0.5 = 100 N.
Result
Fx = 173.2 N, Fy = 100 N
Numerical 4.2
A force of 300 N is applied perpendicularly at the knob of a door to open it. If the knob is 1.2 m away from the hinge, what is the torque applied? Is it positive or negative torque?
The knob is 1.2 m from the hinge line, the axis of rotation.
Given Data
F = 300 N (perpendicular to the door)
moment arm l = 1.2 m
To Find
Torque τ, and its sign
Formula
τ = F × l
Calculation
The force is perpendicular, so the moment arm is the full 1.2 m.
τ = F × l = 300 × 1.2 = 360 N m.
In the book's figure the door turns anticlockwise, and anticlockwise torque is taken as positive.
Result
τ = 360 N m, positive (anticlockwise)
Numerical 4.3
Two weights hang from a metre rule that is balanced at its centre of gravity (C.G). An unknown weight w hangs 40 cm to the left of the C.G, and a 4 N weight hangs 30 cm to the right. Find w.
The rule balances on the wedge at its C.G.
Given Data
4 N weight at 30 cm = 0.3 m from the C.G (clockwise)
w at 40 cm = 0.4 m from the C.G (anticlockwise)
To Find
Unknown weight w
Formula
clockwise moments = anticlockwise moments
Calculation
By the principle of moments: 4 × 0.3 = w × 0.4.
1.2 = 0.4 w.
w = 1.2 ÷ 0.4 = 3 N.
Result
w = 3 N
Numerical 4.4
A see-saw is balanced with two children sitting near either end. Child A has a mass of 30 kg and sits 2 m from the pivot; child B has a mass of 40 kg and sits 1.5 m from the pivot. Calculate the moment on each side and decide whether the see-saw is in equilibrium. (g = 10 m s⁻²)
Given Data
m₁ = 30 kg, l₁ = 2 m
m₂ = 40 kg, l₂ = 1.5 m
g = 10 m s⁻²
To Find
Moment of each child
Is the see-saw in equilibrium?
Formulas
W = mg
τ = W × l
Calculation
Weights: W₁ = 30 × 10 = 300 N and W₂ = 40 × 10 = 400 N.
Moment of A: τ₁ = 300 × 2 = 600 N m (one way).
Moment of B: τ₂ = 400 × 1.5 = 600 N m (the other way).
The two moments are equal and opposite, so Στ = 0.
Result
Each moment is 600 N m, so the see-saw is in equilibrium
Note: The question's "weighs 30 kg" means a mass of 30 kg; weight is a force in newtons.
Numerical 4.5
A crowbar is used to lift a box. A downward force of 250 N is applied at the end of the bar, 30 cm from the pivot. The box rests on the other end, 5 cm from the pivot. How much weight does that end bear? The crowbar itself has negligible weight.
Effort 30 cm from the pivot, load 5 cm from it.
Given Data
F = 250 N
l₁ = 30 cm = 0.3 m
l₂ = 5 cm = 0.05 m
To Find
Load (weight) W on the short end
Formula
F × l₁ = W × l₂
Calculation
By the principle of moments about the pivot: 250 × 0.3 = W × 0.05.
75 = 0.05 W.
W = 75 ÷ 0.05 = 1500 N.
Result
W = 1500 N
Note: A 250 N push lifts 1500 N: the long arm is six times the short arm.
Numerical 4.6
A 30 cm long spanner is used to open the nut of a car. If the torque required is 150 N m, how much force should be applied at the end of the spanner?
F acts at right angles at the end of the spanner, so the moment arm l is the full 30 cm.
Given Data
l = 30 cm = 0.3 m
τ = 150 N m
To Find
Force F
Formula
τ = F × l, so F = τ ÷ l
Calculation
F = τ ÷ l = 150 ÷ 0.3.
F = 500 N.
Result
F = 500 N (applied at right angles to the spanner)
Numerical 4.7
A 5 N ball hanging from a rope is pulled to the right by a horizontal force F. The rope makes an angle of 60° with the ceiling. Determine the force F and the tension T in the rope.
Three forces on the ball: T along the rope, F to the right, weight 5 N down.
Given Data
W = 5 N
rope at θ = 60° to the ceiling (horizontal)
To Find
Horizontal force F
Tension T
Formulas
ΣFy = 0: T sin θ = W
ΣFx = 0: F = T cos θ
Calculation
Resolve T: horizontal part T cos 60° (to the left), vertical part T sin 60° (up).
Vertical balance: T sin 60° = 5, so T = 5 ÷ 0.866 = 5.77 N.
Horizontal balance: F = T cos 60° = 5.77 × 0.5 = 2.89 N.
Result
T ≈ 5.77 N, F ≈ 2.89 N
Note: Some solved copies round T to 5.8 N first and get F = 2.9 N; the exact value is F = 5 ÷ tan 60° = 2.89 N.
Numerical 4.8
A 200 N signboard hangs from two vertical steel wires fixed 2 m on either side of its centre of gravity. Find the tension in each wire.
The wires are 2 m on either side of the board's centre of gravity.
Given Data
W = 200 N
each wire 2 m from the centre of gravity
wires vertical
To Find
Tensions T₁ and T₂
Formulas
ΣF = 0: T₁ + T₂ = W
Στ = 0 about G: T₁ × 2 = T₂ × 2
Calculation
Moments about G: T₁ × 2 = T₂ × 2, so T₁ = T₂.
Forces: T₁ + T₂ − 200 = 0, so 2T₁ = 200.
T₁ = T₂ = 100 N.
Result
T₁ = T₂ = 100 N
Numerical 4.9
A girl of mass 30 kg sits 1.6 m from the axis of a see-saw. Another girl of mass 40 kg wants to sit on the other side so that the see-saw stays in equilibrium. How far from the axis should she sit? (g = 10 m s⁻²)
Given Data
m₁ = 30 kg, l₁ = 1.6 m
m₂ = 40 kg
g = 10 m s⁻²
To Find
Distance l₂ of the second girl
Formula
m₁g × l₁ = m₂g × l₂
Calculation
By the principle of moments: 30 × 10 × 1.6 = 40 × 10 × l₂.
480 = 400 l₂.
l₂ = 480 ÷ 400 = 1.2 m.
Result
l₂ = 1.2 m from the axis
Note: g cancels from both sides, so 30 × 1.6 = 40 × l₂ gives the same answer.
Numerical 4.10
A 150 N block hangs from a knot. One string from the knot goes to the ceiling at 60° to it; a second string goes horizontally from the knot to a wall. Find the tension in each string.
T₁ at 60° to the ceiling, resolved into T₁ sin 60° (up) and T₁ cos 60° (left); T₂ horizontal; 150 N hanging from the knot.
Given Data
W = 150 N
T₁ at θ = 60° to the ceiling
T₂ horizontal
To Find
Tension T₁
Tension T₂
Formulas
ΣFy = 0: T₁ sin θ = W
ΣFx = 0: T₂ = T₁ cos θ
Calculation
Vertical balance at the knot: T₁ sin 60° = 150, so T₁ = 150 ÷ 0.866 = 173.2 N.
Horizontal balance: T₂ = T₁ cos 60° = 173.2 × 0.5 = 86.6 N.
A body in which the distance between any two points stays the same when forces act on it; it does not bend or stretch.
Axis of rotation
The straight line about which a rigid body turns. Every particle of the body moves in a circle whose centre lies on this line.
Line of action of a force
The straight line along which a force acts, drawn through its point of application.
Like parallel forces
Parallel forces that act in the same direction.
Unlike parallel forces
Parallel forces that act in opposite directions.
Resolution of a force
Splitting one force into two or more parts (components) that together have the same effect as the force.
Rectangular (perpendicular) components
The two components of a force that are at right angles to each other, usually along the x-axis and the y-axis.
Moment arm
The perpendicular distance from the axis of rotation to the line of action of the force.
Torque (moment of a force)
The turning effect of a force, equal to the force multiplied by its moment arm. It is a vector; SI unit newton metre (N m).
Principle of moments
When a body is balanced, the sum of the clockwise moments about any point equals the sum of the anticlockwise moments about that point.
Couple
Two equal and opposite parallel forces acting at two different points of a body, along different lines of action.
Couple arm
The perpendicular distance between the lines of action of the two forces of a couple.
Centre of mass
The point of a body where its whole mass can be taken to be concentrated.
Centre of gravity
The point of a body through which its whole weight appears to act, whatever the position of the body.
Equilibrium
A body is in equilibrium if it has no acceleration: no linear acceleration and no angular (turning) acceleration.
Static equilibrium
Equilibrium of a body at rest, for example a book on a table or a picture on a wall.
Dynamic equilibrium
Equilibrium of a body moving with uniform velocity, for example a paratrooper falling at constant velocity or a car at steady speed on a straight road.
First condition of equilibrium
The vector sum of all the external forces on the body is zero: ΣF = 0, that is ΣFx = 0 and ΣFy = 0. This gives translational equilibrium.
Second condition of equilibrium
The vector sum of all the torques on the body about any point is zero: Στ = 0. This gives rotational equilibrium.
Stable equilibrium
A body is in stable equilibrium if, after a slight tilt, it comes back to its original position. Tilting raises its centre of gravity.
Unstable equilibrium
A body is in unstable equilibrium if, after a slight tilt, it moves further away from its original position. Tilting lowers its centre of gravity.
Neutral equilibrium
A body is in neutral equilibrium if, after being disturbed, it comes to rest in its new position. Its centre of gravity stays at the same height.
Centripetal force
The force that keeps a body moving in a circle. It acts towards the centre of the circle, at right angles to the velocity.
Common mistakes to avoid
Using the length of the arm instead of the perpendicular distance to the line of action as the moment arm.
Forgetting to change cm to m before calculating a torque in N m.
Swapping sin and cos: with θ from the x-axis, Fx = F cos θ and Fy = F sin θ.
Thinking ΣF = 0 means equilibrium: a couple has ΣF = 0 but still turns the body.
Writing the mass in kg as a force in a moments question: multiply by g, or cancel g on both sides.
Calling a cylinder on its flat base neutral: it is stable; only on its curved side is it neutral.
Saying a stone flies outward when its string breaks: it flies off along the tangent.
Saying centripetal force speeds a body up: it is at right angles to the velocity and changes only its direction.
Extra practice tap a question to see its answer
Find the torque when 50 N acts at right angles 0.4 m from a hinge.Show answer
50 × 0.4 = 20 N m
What force at the end of a 0.25 m spanner gives a torque of 50 N m?Show answer
50 ÷ 0.25 = 200 N
A 10 N force acts at 30° to a 2 m rod at its end. Find the torque about the other end.Show answer
2 × 10 × sin 30° = 10 N m
Find the components of a 50 N force at 60° to the x-axis.Show answer
Fx = 50 × 0.5 = 25 N; Fy = 50 × 0.866 = 43.3 N
A force has components 5 N and 12 N. Find its magnitude.Show answer
√(25 + 144) = √169 = 13 N
A 6 N weight hangs 20 cm left of a pivot. Where must a 4 N weight hang to balance it?Show answer
6 × 20 = 4 × d, so d = 30 cm on the right
A 500 N man sits 1 m from a see-saw pivot. What weight 2.5 m away on the other side balances him?Show answer
500 × 1 ÷ 2.5 = 200 N
Two forces of 10 N form a couple with a couple arm of 0.3 m. Find the torque.Show answer
10 × 0.3 = 3 N m
An 80 N picture hangs symmetrically from two vertical strings. Find the tension in each.Show answer
80 ÷ 2 = 40 N
A 2 kg ball moves at 3 m s⁻¹ in a circle of radius 1.5 m. Find the centripetal force.Show answer
2 × 9 ÷ 1.5 = 12 N
The radius of a circular path is halved at the same speed. What happens to the centripetal force?Show answer
It doubles (Fc ∝ 1/r)
Name the state of equilibrium of a pencil balanced on its point.Show answer
Unstable: a slight tilt lowers G and it falls over
Where is the centre of gravity of a uniform rectangular sheet?Show answer
Where its diagonals cross
Give the two ways of making a body more stable.Show answer
Lower its centre of gravity and widen its base
Final self-test
Final self-test: all 24 MCQs, no answers
Give yourself 25 minutes. Pick one option for each question, then submit to see your score and which questions to revise. Any question you get wrong tells you which topic to read again.
