9th Class Physics Ch#4 PDF

9th Class Physics Chapter No. 4 4 TURNING EFFECT OF FORCES Important Short Questions and Answers Q. Define parallel forces and write name of its types. Ans. Such forces which are parallel to each o...

9th Class Physics Chapter No. 4 4 TURNING EFFECT OF FORCES Important Short Questions and Answers Q. Define parallel forces and write name of its types. Ans. Such forces which are parallel to each other are called parallel forces. There are two types of parallel forces. 1. Like parallel forces 2. Unlike parallel forces Q. Differentiate between like and unlike parallel forces. Ans. Difference between like and unlike parallel forces: Like Parallel Forces Unlike Parallel Forces Like parallel forces are the forces Unlike parallel forces are the that are parallel to each other and forces that are parallel but have have the same direction. directions opposite to each other. Example: The weight of the apples Example: Weight of a block in a bag is always in the downward hanging through a string is in direction. These forces are like downward direction but the parallel forces. tension in the string is in upward direction. These forces are unlike parallel forces. Q. What is a resultant force? Ans. A resultant force is a single force that has the same effect as the combined effect of all the forces to be added. ⃗ + B⃗ Resultant force = R⃗ = A Q. What is meant by head to tail rule? Explain with example. Ans. The graphical method of adding forces is called head to tail rule. Any number of forces can be added by head to tail rule. Example: Two vectors A ⃗ and B ⃗ are added by head to tail rule which gives resultant vector R i.e., ⃗ = A⃗ + B⃗ R Q. Define rectangular components and resolution of forces. www.rashidnotes.com Ans. Rectangular Components: The components of a vector which are mutually perpendicular to each other are called rectangular components. The rectangular components of a force F⃗ are Fx and Fy Where Fx = Fcosθ and Fy = Fsinθ For more Notes, Books & Guess Papers, visit www.ilmnext.com 1 9th Class Physics Chapter No. 4 Resolution of Forces: Splitting up of a force into two mutually perpendicular components is called the resolution of that force. Q. What is Rigid body? Ans. A body is composed of large number of small particles. If the distances between all pairs of particles of the body do not change by applying a force, then it is called a rigid body e.g., iron, stone etc. Q. Define axis of rotation. Ans. If a rigid body is rotating about a line. The particles of the body move in circles with their centers all lying on this line. This line, is called the axis of rotation. In figure. AA’ is the axis of rotation. Q. Define line of action of force and moment arm. Ans. Line of Action of Force: The line along which a force acts, is called the line of action of force. Moment Arm: The perpendicular distance between the axis of rotation and the line of action of the force is called the moment arm of the force. In figure, line AA' is the axis of rotation. Line BC is the line of action of force F ⃗ while L is the moment arm. Q. Define torque and write its unit. Ans. The turning effect of force is called torque or moment of the force. Formula: Torque = 𝜏𝜏 = F x L Unit: The SI unit of torque is newton metre (Nm). Q. What is the difference between clockwise moment and anticlockwise moment? Ans. Difference between clockwise moment and anticlockwise moment: Clockwise Moment Anticlockwise Moment The torque produced by a force The torque produced by a force that turns a body in the clockwise that turns a boy in anticlockwise direction is called the clockwise direction is called anticlockwise moment. This moment is used to moment. This moment is used to www.rashidnotes.com tighten a nut loosen a nut. Q. State the principle of moments. Ans. A body is balanced if the sum of clockwise moments acting on the body is equal to the sum of anticlockwise moments acting on it. For more Notes, Books & Guess Papers, visit www.ilmnext.com 2 9th Class Physics Chapter No. 4 According to the principle of moments: Clockwise moments = Anticlockwise moments Q. Define center of mass. Ans. Centre of mass of a system is such a point where an applied force causes the system to move without rotation. In the given figure O is the center of mass of two unequal masses. Q. Define center of gravity. Ans. A point where the whole weight of the body appears to act vertically downward is called center of gravity of a body. Centre of gravity depends upon the geometry of object. Q. Where are the centers of gravities of uniform square and uniform triangle are located? Ans. Centers of Gravities of Uniform Square: The center of gravity of a uniform square is the point of intersection of its diagonals as shown in the figure. The center of Gravity of a Uniform Triangle: The center of gravity of a uniform triangle is the point of intersection of its medians as shown in the figure. Q. What is meant by plumbline? Ans. A plumbline consists of a small metal bob (lead or brass) supported by a string. When the bob is suspended freely by the string, then center of gravity of the bob IS exactly below its point of suspension. Plumbline is used to find the center of gravity of a body. Q. Define couple. Ans. A couple is formed by two unlike forces of the same magnitude but not along the same line. Example: When a driver turns a vehicle, he applies forces on the steering wheel www.rashidnotes.com that produce a couple. This couple turns the steering wheel. Q. State first condition of equilibrium. Ans. First Condition of Equilibrium: A body is said to satisfy first condition of equilibrium if the resultant of all the forces acting on it is zero. For more Notes, Books & Guess Papers, visit www.ilmnext.com 3 9th Class Physics Chapter No. 4 Mathematical Form: Let n number of forces F1 , F2 , F3 , … Fn are acting on a body such that: F1 + F2 + F3 + ⋯ Fn ∑F = 0 or ∑ Fx = 0 & ∑ Fy = 0 Q. Describe the second condition of equilibrium and write its mathematical form. Ans. Second Condition of Equilibrium: A body is said to satisfy second condition of equilibrium if the resultant of all the torques acting on it is zero. Mathematical Form: ∑ 𝜏𝜏 = 0 or 𝜏𝜏1 + 𝜏𝜏2 + 𝜏𝜏3 + ⋯ 𝜏𝜏𝑛𝑛 = 0 Q. Define stable equilibrium. Ans: A body is said to be in stable equilibrium if after a slight tilt it returns to its previous position. When a body is in stable equilibrium, its center of gravity is at the lowest position. Q. Define unstable equilibrium. Ans. If a body does not return to its previous position when sets free after a slightest tilt, it is said to be in unstable equilibrium. The center of gravity of the body is at its highest position in the state of unstable equilibrium. Q. What is neutral equilibrium? Ans. If a body remains in its new position when disturbed from its previous position, it is said to be in a state of neutral equilibrium. In neutral equilibrium, the center of gravity of the body remains at the same height. A ball, a sphere, a roller, an egg and a pencil lying horizontally are in neutral equilibrium. Q. In a right-angled triangle, length of base is 4 cm an its perpendicular is 3 cm. Find the length off its hypotenuse. Sol. Base = 4 cm, Perpendicular = 3 cm Hypotenuse =? We know that (Hyp)2 = (Base)2 + (Perp)2 (Hyp)2 = (4)2 + (3)2 (Hyp)2 = 16 + 9 www.rashidnotes.com (Hyp)2 = 25 Taking square root of both sides (Hyp)2 = √25 Hypotenuse = 5 cm For more Notes, Books & Guess Papers, visit www.ilmnext.com 4 9th Class Physics Chapter No. 4 Solved Exercise 1.1 Encircle the correct answer from the given choices. 1. Two equals but unlike parallel forces having different line of action produce (a) torque (b) a couple (c) equilibrium (d) neutral equilibrium 2. The number of forces that can be added by head to tail rule are: (a) 2 (b) 3 (c) 4 (d) any number 3. The number of perpendicular components of a force are: (a) 1 (b) 2 (c) 3 (d) 4 4. A force of 10 N is making an angle of 30° with the horizontal. Its horizontal component will be: (a) 4 N (b) 5 N (c) 7 N (d) 8.7 N 5. 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 6. 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 7. A body is in neutral equilibrium when its center of gravity: (a) is at its highest position (b) is at the lowest position (c) keeps its height if displaced (d) is situated at its bottom 8. Racing cars are made stable by: (a) increasing their speed (b) decreasing their mass (c) lowering their center of gravity (d) decreasing their width Answers: www.rashidnotes.com 1 a 2 d 3 b 4 d 5 d 6 d 7 c 8 c For more Notes, Books & Guess Papers, visit www.ilmnext.com 5 9th Class Physics Chapter No. 4 4.2 Define the following: (i) resultant vector (ii) torque (iii) center of mass (iv) center of gravity Ans. See important short questions and answers. 4.3 Differentiate the following: (i) like and unlike forces (ii) torque and couple (iii) stable and neutral equilibrium Ans. (i) Like and Unlike Parallel Forces (See important short questions and answers) (ii) Torque and Couple Torque Couple The turning effect of a force is A couple is formed by two unlike called torque or moment of the parallel forces of the same force. magnitude but not along the same line. (iii) Stable and Neutral Equilibrium: Stable Equilibrium Neutral Equilibrium A body is said to be in stable If a body remains in its new equilibrium if after a slight tilt it position when disturbed from its returns to its previous position. previous position, it is said to be in a state of neutral equilibrium. 4.4 How head to tail rule helps to find the resultant of forces? Ans. Draw the representative lines of all the force to be added in such a way that head of first force coincides with the tail of second force, head of second force coincides with the tail of third force and so on. The line obtained by joining the tail of first force with the head of last force represent resultant force. 4.5 How can a force be resolved into its perpendicular components? Ans. Resolution of Force: Splitting up of a force into two mutually perpendicular components is called the www.rashidnotes.com resolution of that force. For more Notes, Books & Guess Papers, visit www.ilmnext.com 6 9th Class Physics Chapter No. 4 Explanation: Consider a force F represented by line OA making an angle 𝜃𝜃 with x-axis. Draw a perpendicular AB on x-axis from A. According to head to tail rule, OA is the resultant of vectors represented by OB and BA. OA = AB + BA The components OB and BA are perpendicular to each other. They are called the perpendicular components of OA representing force F. Hence OB represents its x-component Fx and BA represents its y-component Fy. Therefore, above equation can be written as F = Fx + Fy ------------- (1) The magnitudes of forces Fx and Fy can be found using the trigonometric ratios. In right angled triangle OBA Fx OB Since = = cosθ F OA Fx = cosθ ------------- (2) Fy AB Similarly, = = sinθ F OA Fy = Fsinθ ------------- (3) Equations 1 and 2 give the perpendicular components Fx and Fy respectively. 4.6 When a body is said to be in equilibrium? Ans. Ans. A body is said to be in equilibrium if it is at rest or moving with uniform velocity. In equilibrium, the sum of all the forces and torques acting on body is zero i.e., ∑F = 0 and ∑ τ = 0 4.7 Explain the first condition for equilibrium. Ans. A body is said to satisfy first condition for equilibrium if the resultant of all the forces acting on it is zero. Let n number of forces F1 , F2 , F3 , … Fn acting on a body such that F1 + F2 + F3 + − − − + Fn = 0 ∑F = 0 This equation is called the first condition for equilibrium. www.rashidnotes.com The first condition for equilibrium can also be stated in terms of x and y- components of the forces acting on the body as: F1x + F2x + F3x + − − − + Fnx = 0 and F1y + F2y + F3y + − − − + Fny = 0 For more Notes, Books & Guess Papers, visit www.ilmnext.com 7 9th Class Physics Chapter No. 4 or ∑ Fx = 0 and ∑ Fy = 0 A book lying on a table or a picture hanging on a wall, are at rest and thus satisfy first condition for equilibrium. 4.8 Why there is a need of second condition for equilibrium if a body satisfies first condition for equilibrium? Ans. When two equal and opposite forces act on a body along the same line, it will be in equilibrium and no linear acceleration is produced in it. But when two equal and opposite forces act on a body not along the same line, the body is not in equilibrium because rotational acceleration is produced in the body although first condition is still satisfied. Hence in this case for the body to be in equilibrium second condition is needed 4.9 What is second condition for equilibrium? Ans. See important short questions and answers. 4.10 Give an example of a moving body which is in equilibrium. Ans. A paratrooper coming down with terminal velocity is in equilibrium. This type of equilibrium is known as dynamic equilibrium. 4.11 Think of a body which is at rest but not in equilibrium. Ans. A ball thrown upward becomes at rest at the top. At this state it is not in equilibrium although it is at rest. 4.12 Why a body cannot be in equilibrium due to single force acting on it? 4.13 Why the height of vehicles is kept as low as possible? Ans. To make bodies stable, their center of mass must be kept as low as possible. Hence, the height of vehicles especially racing cars is kept as low as possible so that their center of mass become as low as possible and they become stable. 4.14 Explain what is meant by stable, unstable and neutral equilibrium. Give one example in each case. Ans. See important short questions and answers. www.rashidnotes.com For more Notes, Books & Guess Papers, visit www.ilmnext.com 8 9th Class Physics Chapter No. 4 Numerical Problems 4.2 Find the perpendicular components of a force of 50 N making an angle of30° with x axis. (Imp) Sol. F = 50 N, θ = 30°, Fx =? and Fy =? Fx =? Fx =? Fx = Fcosθ Fy = Fsinθ Fx = 50 × cos30 Fy = 50 × sin30 Fx = 50 × 0.866 Fy = 50 × 0.5 Fx = 43.3 N Fy = 25 N 4.3 Find the magnitude and direction of a force, if its x-component is12 N and y- component is 5 N. Sol. Fx = 12 N , Fy = 5 N, F =? and θ =? We know that We know that Fy F = (Fx )2 + (F𝑦𝑦 )2 θ = tan−1 Fx 5 F = (12)2 + (5)2 θ= tan−1 12 −1 ( F = √144 + 25 θ = tan 0.416) F = √169 θ = 22.6° F = 13 N 4.4 A force of 100 N is applied perpendicularly on a spanner at a distance of 10 cm from a nut. Find the torque produced by the force. 10 Sol. F = 100 N, F = 10 cm = = 0.1 m, Torque = 𝜏𝜏 =? 100 We know that τ=F×L τ = 100 × 0.1 = 10 Nm 4.5 A force is acting on a body making an angle of 30° with the horizontal. The horizontal component of the force is 20 N. Find the force. Sol. θ = 30°, Fx = 20 N, F =? We know that Fx = Fcosθ www.rashidnotes.com 20 = Fcos30 20 = F × 0.866 20 F= = 23.1 N 0.866 For more Notes, Books & Guess Papers, visit www.ilmnext.com 9 9th Class Physics Chapter No. 4 4.6 The steering of a car has a radius 16 cm. Find the torque produced by a couple of 50 N. (Imp) Sol. F = 50 N 32 AB = 16 + 16 = 32 cm = = 0.32 m 100 Torque = τ =? We know that τ = F × AB τ = 50 × 0.32 τ = 16 Nm 4.7 A picture frame is hanging by two vertical strings. The tensions in the strings are 3.8 N and 4.4 N. Find the weight of the picture frame. Sol. Picture is at equilibrium; therefore, the weight of the picture is equal to the total tension in the strings i.e., w = T1 + T2 − − − −(1) T1 = 3.8 N T2 = 4.4 N By putting the value of T1 and T2 in equation 1, we get w = 3.8 + 4.4 = 8.2 N 4.8 Two blocks of masses 5 kg and 3 kg are suspended by the two strings. Find the tension in each string. Sol. For block A m = 3 kg At equilibrium w = T = mg T = 3 × 10 = 30 N For block A m = 3 + 5 = 8 kg At equilibrium w = T = mg T = 8 × 10 = 80 N 4.9 A nut has been tightened by a force of 200 N using 10 cm long spanner. What www.rashidnotes.com length of a spanner is required to loosen the same nut with 150 N force? (imp) Sol. L1 = 10 cm, F1 = 200 N, F2 = 150 N L2 =? According to the principle of moments: For more Notes, Books & Guess Papers, visit www.ilmnext.com 10 9th Class Physics Chapter No. 4 L1 F1 = L2 × F2 L1 ×F1 or L2 = F2 10×200 2000 L2 = = = 13.33 cm 150 150 4.10 A block of mass 10 kg is suspended at a distance of 20 cm from the center of a uniform bar1 m long. What force is required to balance it at its center of gravity by applying the force at the other end of the bar? Sol. m1 = 10 kg F1 = m1 g F1 = 10 × 10 = 100 N L1 = 20 cm L2 = 50 cm According to the principle of moments: L1 F1 = L2 × F2 F1 ×L1 or F2 = L2 100×20 2000 L2 = = = 40 N 50 50 Example: A mechanic tightens the nut of a bicycle using a 15 cm long spanner by exerting a force of 200 N. Find the torque that has tightened it. Sol. F = 200 N 15 L = 15 cm = = 0.15 m 100 𝜏𝜏 =? We know that τ=F×L τ = 200 × 0.15 = 30 Nm www.rashidnotes.com For more Notes, Books & Guess Papers, visit www.ilmnext.com 11 9th Class Physics Chapter No. 4 Extra MCQs 1. The forces that are parallel to each other and have the same direction are called: (a) like parallel forces (b) unlike parallel forces (c) resultant forces (d) net parallel forces 2. The number of forces that can be added by head to tail rule are: (OR) The number of vectors that can be added by head to tail rule are: (a) 2 (b) 3 (c) 4 (d) any number 3. The number of perpendicular components of a force is: OR The numbers of perpendicular components of a vector are: (a) 1 (b) 2 (c) 3 (d) 4 4. A force of 10 N is making an angle of 30 0 with the horizontal. Its horizontal component will be: (a) 4 (b) 5 (c) 7 (d) 8.7 N 5. If a force of 1-0 N is making an angle of 60 0 with x-axis, then its nonzonal component will be: (a) 4 (b) 5 (c) 7 (d) 8.7 N 6. If 𝐅𝐅𝐲𝐲 = 4 N and 𝐅𝐅𝐱𝐱 = 3 N, what is magnitude of resultant force? (a) 7 N (b) 5 N (c) 12 N (d) 10 N 𝐅𝐅𝐱𝐱 7. Complete the equation: =−−−− 𝐅𝐅𝐲𝐲 (a) sinθ (b) cosθ (c) tanθ (d) cosecθ 8. The direction of force F with x-axis is given by: Fx Fy (a) θ = tan−1 (b) θ = tan−1 Fy Fx Fx Fy www.rashidnotes.com (c) θ = tan (d) θ = tan Fy Fx 9. The direction of resultant vector can be found by: Fx Fy (a) θ = tan (b) θ = tan Fy Fx For more Notes, Books & Guess Papers, visit www.ilmnext.com 12 9th Class Physics Chapter No. 4 Fx Fy (c) θ = tan−1 (d) θ = tan−1 Fy Fx 10. 𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬 = − − − − − base perpendicular (a) (b) perpendicular base perpendicular base (c) (d) hypotenuse hypotenuse 11. 𝐜𝐜𝐜𝐜𝐜𝐜𝐜𝐜 = − − − − − base perpendicular (a) (b) hypotenuse hypotenuse perpendicular hypotenuse (c) (d) base base 12. 𝐜𝐜𝐜𝐜𝐜𝐜𝐜𝐜𝐜𝐜° = − − − − − (a) 0 (b) 0.5 (c) 0.707 (d) 1 13. 𝐭𝐭𝐭𝐭𝐭𝐭𝐭𝐭𝐭𝐭° = − − − − − (a) 0.5 (b) 1.732 (c) 0.577 (d) 1 14. 𝐜𝐜𝐜𝐜𝐜𝐜𝐜𝐜𝐜𝐜° = − − − − − (a) 0.5 (b) 1.732 (c) 0.866 (d) 0.577 15. 𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬° = − − − − − (a) 0 (b) 1 (c) 10 (d) 0.5 16. 𝐜𝐜𝐜𝐜𝐜𝐜𝐜𝐜𝐜𝐜° = − − − − − (a) 1 (b) 0.866 (c) 0.707 (d) 0 17. In a right-angled triangle length of base is 4 cm and perpendicular is 3 cm, length of diagonal (hypotenuse) will be: (a) 2 cm (b) 5 cm (c) 4 cm (d) 6 cm 18. In a right-angled triangle length of base is 4 cm and perpendicular is 3 cm, then its 𝐜𝐜𝐜𝐜𝐜𝐜𝐜𝐜 is equal to: (a) 0.8 (b) 0.75 www.rashidnotes.com (c) 1 (d) 0.6 19. In a right-angle triangle, base is 4 cm and perpendicular is 3 cm. Its 𝐭𝐭𝐭𝐭𝐭𝐭𝐭𝐭 is equal to: (a) 0.75 (b) 0.8 For more Notes, Books & Guess Papers, visit www.ilmnext.com 13 9th Class Physics Chapter No. 4 (c) 0 89 (d) 0.6 20. Turning effect of a force is called: (a) torque (b) moment (c) couple (d) torque and momentum 21. The formula of torque is: (a) L = F x E (b) E = F x T (c) τ = L x F (d) E = F x L 22. S.I. unit of torque is: (a) Nm (b) Ns (c) Nm-1 (d) Ns-1 23. Torque depends on: (a) force and mass (b) mass and velocity (c) force and moment arm (d) force and velocity 24. Number of factors on which torque depends: (a) 2 (b) 3 (c) 4 (d) 5 25. If force is 200 N and length of spanner is 0.15 m then torque will be: (a) 30 Nm (b) 15 Nm (c) 20 Nm (d) 10 Nm 26. The moment arm is represented by a sign: (a) T (b) L (c) F (d) N 27. Such a point where an applied force causes the system to move without rotation: (a) center of gravity (b) center of mass (e) center of weight (d) none of these 28. The center of gravity of uniform is the point of intersection of its diagonals. (a) triangle sheet (b) solid cylinder (c) circular ring (d) square sheet 29. The center of gravity of a triangle is at: (a) center (b) point of intersection of medians www.rashidnotes.com (c) center of axis (d) point of intersection of diagonals 30. The center of gravity of a sphere is: (a) center of sphere (b) outside the sphere (c) radius of sphere (d) none of these For more Notes, Books & Guess Papers, visit www.ilmnext.com 14 9th Class Physics Chapter No. 4 31. The center of gravity of an irregular shaped body can be found with the help of: (a) screw gauge (b) plumbline (c) metre rule (d) wedge 32. Two equals but unlike parallel forces having different line of action produce: (OR) Two unlike parallel forces of the same magnitude but not along the same line forms: (a) a torque (b) couple (c) equilibrium (d) neutral equilibrium 33. 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 34. The states of equilibrium are: (a) 2 (b) 3 (c) 4 (d) 5 35. The symbol of sigma is: (a) α (b) Σ (c) µ (d) θ 36. First condition of equilibrium is: ‘OR’ Mathematically first condition of equilibrium is represented as: (a) ΣF = 0 (b) Στ = 0 (c) ΣF = 0, Στ = 0 (d) All of these 37. According to second condition of equilibrium must be zero: (a) angular acceleration (b) linear acceleration (c) rotational force (d) sum of torques 38. A body is in equilibrium when its: (a) acceleration is uniform (b) speed is uniform (c) speed and acceleration are uniform www.rashidnotes.com (d) acceleration is zero 39. The net torque acting on a rotating body with uniform speed is: (a) 1 (b) 2 (c) 5 (d) 0 For more Notes, Books & Guess Papers, visit www.ilmnext.com 15 9th Class Physics Chapter No. 4 40. A body is in neutral equilibrium when its center of gravity: (a) is at its highest position (b) is at lowest position (c) keeps its height if displaced (d) is situated at its bottom 41. Racing cars are made stable by: (a) increasing their speed (b) decreasing their mass (c) lowering their center of gravity (d) decreasing their width 42. When center of gravity is at the highest position, body will be in: (a) neutral equilibrium (b) stable equilibrium (c) unstable equilibrium (d) none of these Answers 1 a 2 d 3 b 4 d 5 b 6 b 7 c 8 b 9 d 10 c 11 a 12 c 13 d 14 a 15 b 16 d 17 b 18 a 19 a 20 a 21 c 22 a 23 c 24 a 25 a 26 b 27 b 28 d 29 b 30 a 31 b 32 b 33 d 34 b 35 b 36 a 37 d 38 d 39 d 40 c 41 c 42 c www.rashidnotes.com For more Notes, Books & Guess Papers, visit www.ilmnext.com 16