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9th Class Physics · Punjab Board · Chapter 4

Chapter 4: Turning Effect of ForcesShort Questions with Answers

All 10 short questions of 9th Class Physics Chapter 4, Turning Effect of Forces, with exam-ready answers in the book's words. Punjab Board new syllabus.

Chapter 4· Turning Effect of Forces

Short Questions with Answers

Exam-ready answers. Learn the meaning, then write them in your own words.

Short Q 4.1

Define like and unlike parallel forces.

Answer
Definition
Like parallel forces

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.

F₁F₂F₃
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
xyFx = F cos θFy = F sin θFθO
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.

θθOPrFl = r sin θ
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.
AOBr₁r₂F₁F₂
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.
(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.
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.

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
v along the tangent, Fc along the radius: at right angles.
End of Chapter 4 Short QuestionsNext: Chapter 4 Long Questions →
4Chapter
Full guide: Turning Effect of Forces (notes, diagrams, all questions, self-test)
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