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9th Class Physics · Punjab Board · Chapter 4

Chapter 4: Turning Effect of ForcesLong Questions with Answers

10 long (comprehensive) questions of 9th Class Physics Chapter 4, Turning Effect of Forces, answered in full with headings, diagrams and tables, including important board-paper questions beyond the book. Punjab Board new syllabus.

Chapter 4· Turning Effect of Forces

Long Questions with Answers

Long answers, explained in full with headings. Learn the structure: definition, explanation, diagram or table, example.

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

CGl₁l₂l₃w₁w₂w₃w₁l₁ + w₂l₂ = w₃l₃
Metre rule balanced on a wedge with weights on both sides.

Explanation

  1. w₁ and w₂ try to turn the rule anticlockwise; their moments are w₁ × l₁ and w₂ × l₂.
  2. w₃ tries to turn it clockwise; its moment is w₃ × l₃.
  3. 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
  1. Anticlockwise: 2 × 20 + 1 × 40 = 80 N cm.
  2. Clockwise: w₃ × 30.
  3. 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

  1. Hang the lamina on the pin through hole A so it can swing freely, and let it come to rest.
  2. Hang the plumb line from the same pin in front of the lamina.
  3. Mark two points along the plumb line and join them with a straight line.
  4. Hang the lamina from hole B and draw the plumb line in the same way.
  5. The point where the two lines cross is the centre of gravity G.

Diagram

Fig. (a)CGpoints ofsuspensionFig. (b)
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

AOBF₂F₁(a)BOAF₂F₁(b)
(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.

(a) Low and wideGline from G falls inside the basefalls back(b) Tall and narrowGline from G falls outside the basetopples over
Low, wide box falls back; tall, narrow box topples.

Two ways to improve stability

  1. Lowering the centre of gravity, so the body must tilt much farther before the line from G leaves the base.
  2. Widening the base, so the line from G has farther to go before it passes outside the base.

Examples

ExampleWhat makes it stable
Low armchairMore stable than a tall stool: low centre of gravity
Loading a bus or truckHeavy goods on the floor keep G low; luggage on the roof raises G and the bus can overturn on a sharp turn
Racing carsVery low body and wheels wide apart, outside the main body
Balancing toysG kept below the pivot; when tilted, G rises, so the toy swings back
Table lamps, pedestal fansHeavy, 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

Fnut (axis)moment arm lτ = F × l
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

TurningSign
AnticlockwisePositive
ClockwiseNegative

Example

Worked exampleA 300 N force is applied at right angles to a door handle 1.2 m from the hinges
  1. 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

xyFx = F cos θFy = F sin θFθO
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
  1. Fx = 200 cos 30° = 200 × 0.866 = 173.2 N.
  2. Fy = 200 sin 30° = 200 × 0.5 = 100 N.
  3. 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

AOBF₂F₁(a)BOAF₂F₁(b)
(b) shows a couple: two equal and opposite forces a distance d apart.

Torque of a couple

  1. Take moments about the point midway between the two forces. Each force is d/2 from it.
  2. Torque of each force = F × d/2, and both turn the body the same way.
  3. 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

(a) StableGtilt it: G rises,so it falls back(b) UnstableGpush it: G falls,so it rolls off(c) NeutralGroll it: G stays atthe same height
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.

StateG on tiltingResult
StableRisesReturns
UnstableFallsTopples
NeutralSame heightStays 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

centreFcrvFc = mv² ÷ r• Fc points to the centre• v is along the tangent• Fc is at right angles to v, soonly the direction changes• string cut: the body flies offalong the tangent
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

FactorEffect on Fc
Mass mFc ∝ m: double the mass, double the force
Speed vFc ∝ v²: double the speed, four times the force
Radius rFc ∝ 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
  1. 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

  1. Choose a suitable scale, for example 100 N = 1 cm.
  2. Draw the first force to scale in its direction from a point O.
  3. 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.
  4. Join O (the first tail) to the head of the last force. This line is the resultant.
  5. Measure its length (convert with the scale) and the angle it makes with the x-axis.

Diagram

ABR = A + Btail of B on head of A;R runs from the first tailto the last head
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°
  1. Draw 2 cm at 30°, then 3 cm at 45°, then 2.5 cm at 60°, head to tail.
  2. Join the first tail to the last head: it measures about 7.35 cm at 46° to the x-axis.
  3. Resultant = 7.35 × 100 ≈ 735 N.

Answerabout 735 N at 46°

xyOF₁F₂F₃30°45°60°(a) the three forcesO30°45°60°F = 7.35 cm = 735 N at 46°F₁ (2.0 cm)F₂ (3.0 cm)F₃ (2.5 cm)(b) head to tail, 100 N = 1 cm
(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.

End of Chapter 4 Long QuestionsNext: Chapter 4 Numericals →
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Full guide: Turning Effect of Forces (notes, diagrams, all questions, self-test)
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