9th Class Physics Chapter 4 Notes PDF — Turning Effect of Force

Parhai — Free Punjab Board Notes <h2>Key Concepts</h2> <h3>Adding Forces: The Head-to-Tail Rule</h3> When several forces act on one body at the same time, we often want a single force that does exactly the same job as all of them put together. That single force is called the resultant force. To find it by drawing, we use the head-to-tail method: pick a scale, draw the first force as an arrow, then draw the second force so that its tail sits exactly on the head of the first arrow, and keep chaining arrows this way. The arrow drawn from the tail of the very first arrow to the head of the very last arrow is the resultant. Any number of forces can be combined this way — there is no limit of two or three. <h3>Splitting a Force into Components</h3> Sometimes it is easier to work with the parts of a force rather than the force itself. Breaking a force into two mutually perpendicular parts is called resolution of a vector. If a force F makes an angle θ with the x-axis, its horizontal part is Fx = F cos θ and its vertical part is Fy = F sin θ. These are called the rectangular components of the force. <h3>Building the Resultant from Components</h3> If you are given the perpendicular components instead, you can rebuild the original force. From the right-angled triangle formed by the components, the size of the force is F = √(Fx² + Fy²) and its direction angle satisfies tan θ = Fy / Fx. This is just the Pythagoras theorem applied to the component triangle. <h3>Torque: The Turning Effect of Force</h3> A force can do more than push or pull — it can also rotate a body. The turning effect of a force is called torque (also called the moment of the force). It is calculated as: τ = F × L Here F is the applied force and L is the moment arm, the perpendicular distance from the line of action of the force to the axis of rotation. The SI unit of torque is the newton-metre (N·m). Torque grows when either the force gets bigger or the moment arm gets longer. This is why a longer spanner makes it easier to tighten a nut — the same force produces a bigger turning effect. <h3>Couple: A Pair of Twisting Forces</h3> A couple is formed by two equal, opposite, parallel forces whose lines of action are different — they are not along the same line. A couple produces pure rotation with no net push in any direction. The torque due to a couple equals one of the forces multiplied by the perpendicular distance between them (the couple arm): τ = F × d. Turning a car steering wheel is a perfect everyday example: both hands apply equal and opposite forces, creating a couple that rotates the wheel. <h3>Centre of Gravity and Centre of Mass</h3> The centre of gravity of a body is the point where its entire weight seems to act vertically downward. The centre of mass is the point where an applied force moves the whole body forward without making it spin. For most everyday objects the two points sit very close together. To find the centre of gravity of an irregularly shaped flat plate, hang it from a nail through a hole near its edge so it can swing freely. Tie a plumbline (a small weight on a string) to the same nail — the string marks the vertical line. Mark that line on the plate. Hang the plate from a second hole and repeat. The two lines cross at the centre of gravity; a third hole confirms the same crossing point. <h3>Equilibrium: First and Second Conditions</h3> A body is in equilibrium when no net force and no net torque acts on it. A book resting on a table is in equilibrium, and so is a paratrooper falling at a constant terminal velocity — both have zero net force. First condition: the vector sum of all forces must be zero, written as ΣF = 0. In terms of components this means ΣFx = 0 and ΣFy = 0. Second condition: the first condition alone is not enough. Imagine two equal but opposite forces applied at different points of a body, like on a steering wheel — their vector sum is zero, yet the body clearly starts rotating. So a second condition is needed: the total torque acting on the body must also be zero, Στ = 0. <h3>Three States of Equilibrium</h3> Stable equilibrium: after a small tilt, the body returns to its original position. Its centre of gravity is at the lowest point and rises when tilted; the body comes back down by lowering its centre of gravity. Example: a book lying on a table. A body stays stable as long as the vertical line through its centre of gravity passes through its base. Unstable equilibrium: after even a tiny tilt, the body moves away and does not return. Its centre of gravity is at the highest point and sinks further as the body topples. Example: a pencil balanced vertically on its tip. Neutral equilibrium: the body simply stays wherever you leave it. Its centre of gravity keeps the same height in every position. Examples: a ball, a sphere, a roller, or a pencil lying horizontally. <h3>Like and Unlike Parallel Forces</h3> Like parallel forces run parallel to each other and point in the same direction. Unlike parallel forces are also parallel but point in opposite directions. A couple is the special case of two unlike parallel forces that are equal in size but act along different lines. <h2>Questions and Answers</h2> Q1. What is a resultant force? How is it found with the head-to-tail rule? A resultant force is a single force that produces exactly the same effect as all the individual forces acting together. In the head-to-tail rule, the arrows representing all forces are drawn to a suitable scale with the tail of each arrow placed at the head of the previous one. The arrow joining the tail of the first to the head of the last is the resultant. Q2. What is meant by resolution of a vector? Splitting a vector (such as a force) into two mutually perpendicular components is called resolution. For a force F at angle θ to the x-axis, the components are Fx = F cos θ and Fy = F sin θ. Q3. How do you get the resultant force when its rectangular components are known? Use the right-angled triangle of the components: F = √(Fx² + Fy²) and tan θ = Fy / Fx, where θ is the angle the resultant makes with the x-axis. Q4. Define torque and name the factors it depends on. The turning effect of a force is called torque, τ = F × L, measured in N·m. It depends on (i) the size of the applied force F and (ii) the moment arm L (perpendicular distance from the axis of rotation). Q5. What is a couple, and what is the torque produced by it? A couple is a pair of equal, opposite, parallel forces acting along different lines. Its torque equals one of the forces multiplied by the perpendicular distance between them (the couple arm): τ = F × couple arm. Q6. Define centre of mass and centre of gravity. The centre of mass is the point where an applied force moves the whole body forward without rotating it. The centre of gravity is the point where the entire weight of the body appears to act vertically downward. Q7. When is a body said to be in equilibrium? State both conditions. A body is in equilibrium when it is at rest or moving with uniform velocity, meaning no net force and no net torque act on it. First condition: ΣF = 0. Second condition: Στ = 0. Q8. Why is the second condition for equilibrium needed even when the first is satisfied? Two equal and opposite forces acting along different lines add up to zero, so the first condition is satisfied — but they form a couple and rotate the body. Only the second condition (zero total torque) rules out this rotation. Q9. Give an example of a moving body that is in equilibrium. A paratrooper descending with constant (terminal) velocity: the air resistance exactly balances his weight, so the net force is zero and he is in equilibrium while moving. Q10. Can a body be at rest but not in equilibrium? Explain. Yes. A person sitting in a bus that is accelerating may be at rest relative to the other passengers, yet a net force acts on him (he is being accelerated), so he is not in equilibrium. Q11. Why can a body not be in equilibrium under a single force? A single force always gives a non-zero net force (ΣF ≠ 0), so the first condition fails. Also, a single force acting off-centre always produces a torque about any other point, so the second condition fails too. Q12. Why is the height of vehicles kept as low as possible? Lowering the body of a vehicle lowers its centre of gravity. A lower centre of gravity means the vehicle can tilt further before the vertical line through its centre of gravity falls outside its base — so it is more stable and less likely to topple. Q13. Explain stable, unstable and neutral equilibrium with one example each. Stable: a body returns to its original position after a slight disturbance — a book lying on a table. Unstable: the body moves away and does not return — a pencil standing vertically. Neutral: the body stays in its new position — a ball lying on the ground, whose centre of gravity stays at the same height. Q14. How does the head-to-tail rule help find the resultant of forces? By chaining the force arrows tail-to-head to a chosen scale, the single closing arrow from the start of the chain to its end automatically gives both the size (from its length) and the direction of the resultant. <h2>Multiple Choice Questions</h2> <ol> <li>Two equal but unlike parallel forces with different lines of action produce: (a) a torque (b) a couple (c) equilibrium (d) neutral equilibrium</li> <li>The number of forces that can be added by the head-to-tail rule is: (a) 2 (b) 3 (c) 4 (d) any number</li> <li>The number of perpendicular components of a force is: (a) 1 (b) 2 (c) 3 (d) 4</li> <li>A force of 10 N makes 30° with the horizontal. Its horizontal component is: (a) 4 N (b) 5 N (c) 7 N (d) 8.7 N</li> <li>A couple is formed by: (a) two forces perpendicular to each other (b) two like parallel forces (c) two equal and opposite forces in the same line (d) two equal and opposite forces not in the same line</li> <li>A body is in equilibrium when its: (a) acceleration is uniform (b) speed is uniform (c) speed and acceleration are uniform (d) acceleration is zero</li> <li>A body is in neutral equilibrium when its centre of gravity: (a) is at its highest position (b) is at its lowest position (c) keeps its height if displaced (d) is situated at its bottom</li> <li>Racing cars are made stable by: (a) increasing their speed (b) decreasing their mass (c) lowering their centre of gravity (d) decreasing their width</li> </ol> <h2>Key Formulas</h2> <table border="1"> <tr><th>Quantity</th><th>Formula</th><th>Unit</th></tr> <tr><td>Rectangular components</td><td>Fx = F cos θ, Fy = F sin θ</td><td>N</td></tr> <tr><td>Resultant from components</td><td>F = √(Fx² + Fy²), tan θ = Fy/Fx</td><td>N</td></tr> <tr><td>Torque (moment of force)</td><td>τ = F × L</td><td>N·m</td></tr> <tr><td>Torque due to a couple</td><td>τ = F × couple arm</td><td>N·m</td></tr> <tr><td>First condition of equilibrium</td><td>ΣF = 0 (so ΣFx = 0, ΣFy = 0)</td><td>—</td></tr> <tr><td>Second condition of equilibrium</td><td>Στ = 0</td><td>—</td></tr> <tr><td>Weight</td><td>w = mg (take g = 10 m/s²)</td><td>N</td></tr> </table> <h2>Solved Numericals</h2> N1. Find the resultant of: (i) 10 N along the x-axis, (ii) 6 N along the y-axis, (iii) 4 N along the negative x-axis. Solution: Take x and y components. Fx = 10 − 4 = 6 N; Fy = 6 N. F = √(6² + 6²) = √72 ≈ 8.5 N; θ = tan⁻¹(6/6) = 45° with the x-axis. Resultant ≈ 8.5 N at 45° to the x-axis. N2. Find the perpendicular components of a 50 N force making 30° with the x-axis. Solution: Fx = F cos θ = 50 × cos 30° = 50 × 0.866 = 43.3 N. Fy = F sin θ = 50 × sin 30° = 50 × 0.5 = 25 N. Fx = 43.3 N, Fy = 25 N. N3. A force has x-component 12 N and y-component 5 N. Find its magnitude and direction. Solution: F = √(12² + 5²) = √169 = 13 N. θ = tan⁻¹(5/12) ≈ 22.6° with the x-axis. F = 13 N at 22.6° to the x-axis. N4. A 100 N force is applied perpendicularly on a spanner 10 cm from a nut. Find the torque. Solution: L = 10 cm = 0.1 m. τ = F × L = 100 × 0.1 = 10 N·m. Torque = 10 N·m. N5. A force acting at 30° to the horizontal has horizontal component 20 N. Find the force. Solution: Fx = F cos θ → 20 = F × cos 30° = F × 0.866. F = 20 / 0.866 ≈ 23.1 N. Force ≈ 23.1 N. N6. A car steering wheel has radius 16 cm. Find the torque produced by a 50 N couple. Solution: Couple arm = diameter = 2 × 16 cm = 32 cm = 0.32 m. τ = F × couple arm = 50 × 0.32 = 16 N·m. Torque = 16 N·m. N7. A picture frame hangs from two vertical strings with tensions 3.8 N and 4.4 N. Find its weight. Solution: For vertical equilibrium, upward tensions balance the weight: w = T₁ + T₂ = 3.8 + 4.4 = 8.2 N. Weight = 8.2 N. N8. Two blocks of masses 5 kg and 3 kg hang one below the other from strings. Find the tension in each string. (g = 10 m/s²) Solution: Weight of 5 kg block = 50 N; weight of 3 kg block = 30 N. Upper string supports both: T₁ = 50 + 30 = 80 N. Lower string supports only the 3 kg block: T₂ = 30 N. T₁ = 80 N, T₂ = 30 N. N9. A nut is tightened with 200 N on a 16 cm spanner. What spanner length is needed to loosen it with 150 N? Solution: Torque applied = F × L = 200 × 0.16 = 32 N·m. The same torque must be produced by 150 N: L = τ / F = 32 / 150 ≈ 0.213 m = 21.3 cm. Spanner length ≈ 21.3 cm. N10. A 20 kg block hangs 20 cm from the centre of a 1 m uniform bar. What force applied at the other end (50 cm from centre) balances it about the centre? Solution: Weight = 20 × 10 = 200 N. Taking moments about the centre: 0.2 × 200 = F × 0.5 → 40 = 0.5 F → F = 80 N. Force = 80 N. More free notes at Parhai.

9th Class Physics Chapter 4 Notes - Turning Effect of Force

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