Physics IX - Unit 4: Turning Effect of Forces PDF

Unit 4 Turning Effect of Forces STUDENT’S LEARNING OUTCOMES After studying this unit, the students will be able to:  define like and unlike parallel forces.  state...

Unit 4 Turning Effect of Forces STUDENT’S LEARNING OUTCOMES After studying this unit, the students will be able to:  define like and unlike parallel forces.  state head to tail rule of vector addition of forces/vectors.  describe how a force is resolved into its perpendicular components.  determine the magnitude and direction of a force from its perpendicular components.  define moment of force or torque as moment = force x perpendicular distance from pivot to the line of action of force.  explain the turning effect of force by relating it to everyday life.  state the principle of moments. Conceptual linkage.  define the centre of mass and centre of gravity of a body. This unit is built on Lever - Science-V  define couple as a pair of forces tending to Machines - Science-VI produce rotation. Kinematics - Physics-IX  prove that the couple has the same moments Trigonometry - Maths-IX about all points. This unit leads to:  define equilibrium and classify its types by Rotational Motion, Vectors and Equilibrium quoting examples from everyday life. - Physics-XI  state the two conditions for equilibrium of a body.  Physics IX 85 Unit 4: Turning Effect of Forces  solve problems on simple balanced systems Major Concepts when bodies are supported by one pivot only. 4.1 Forces on bodies  describe the states of equilibrium and classify 4.2 Addition of Forces them with common examples. 4.3 Resolution of Forces  explain effect of the position of the centre of mass 4.4 Moment of a Force on the stability of simple objects. 4.5 Principle of moments INVESTIGATION SKILLS 4.6 Centre of mass The students will be able to: 4.7 Couple  determine the position of centre of mass/gravity of 4.8 Equilibrium regularly and irregularly shaped objects. 4.9 Stability SCIENCE, TECHNOLOGY AND SOCIETY CONNECTION The students will be able to:  illustrate by describing a practical application of moment of force in the working of bottle opener, spanner, door/windows handles, etc.  describe the working principle of see-saw.  demonstrate the role of couple in the steering wheels and bicycle pedals.  demonstrate through a balancing toy, racing car, Force etc. that the stability of an object can be improved by lowering the centre of mass and increasing the base area of the objects. Figure 4.1: We can loose a nut with a spanner. Can the nut of the axle of a bike be loosened with hand? Normally we use a spanner as shown in figure 4.1. A spanner increases the turning effect of the force. Look at the picture on the previous page. What is the joker doing? He is trying to balance himself on a wooden plank which is placed over a cylindrical pipe. Can we do the same? A baby gradually learns to stand by balancing herself. Women and children in the villages often carry pitchers with water on their heads such as shown in figure 4.2. With a little effort we can learn to balance a stick vertically up on our finger tip. Balanced objects are said to be in equilibrium. In this unit, we will learn many interesting concepts such as torque, Figure 4.2: Children carrying equilibrium, etc. and their applications in daily life. pitchers on their heads. Physics IX 86 Unit 4: Turning Effect of Forces 4.1 LIKE AND UNLIKE PARALLEL FORCES We often come across objects on which many forces are acting. In many cases, we find all or some of the forces acting on a body in the same direction. For example, many people push a bus to start it. Why all of them push it in the same direction? All these forces are applied in the same direction so these are all parallel to each other. Such forces which are parallel to each other are called parallel forces. Figure 4.3 shows a bag with apples in it. The Weight of each apple weight of each apple weight of the bag is due to the weight of all the apples in it. Since the weight of every Figure 4.3: Like parallel forces. apple in the bag is the force of gravity acting on it vertically downwards, therefore, weights of apples are the parallel forces. All these forces are acting in the same direction. Such forces are called like parallel forces. Like parallel forces are the forces that are parallel to each other and have the same Tension direction. In figure 4.4(a), an apple is suspended by a string. The string is stretched due to weight of the apple. The forces acting on it are; weight of the apple acting vertically Weight downwards and tension in the string pulling it vertically upwards. The two forces are parallel but opposite to each (a) other. These forces are called unlike parallel forces. In F1 figure 4.4(b), forces F1 and F2 are also unlike parallel forces, because they are parallel and opposite to each other. But F1 and F2 are not acting along the same line and hence they are capable to rotate the body. (b) Unlike parallel forces are the forces that are parallel F2 but have directions opposite to each other. Figure 4.4: Unlike parallel forces (a) along the same line 4.2 ADDITION OF FORCES (b) can turn the object if not in Force is a vector quantity. It has both magnitude line. and direction; therefore, forces are not added by ordinary arithmetical rules. When forces are added, we get a resultant force. A resultant force is a single force that has the same effect as the combined effect of all the forces to be added. Physics IX 87 Unit 4: Turning Effect of Forces One of the methods for the addition of forces is a N A graphical method. In this method forces can be added by head to tail rule of vector addition. W E B HEAD TO TAIL RULE Figure 4.5 shows a graphical method of vector S addition. First select a suitable scale. Then draw the vectors of all the forces according to the scale; such as R B vectors A and B. Take any one of the vectors as first vector e.g., A vector A. Then draw next vector B such that its tail Figure 4.5: Adding vectors by coincides with the head of the first vector A. Similarly head to tail rule. draw the next vector for the third force (if any) with its tail coinciding with the head of the previous vector and so on. Now draw a vector R such that its tail is at the tail of vector A, the first vector, while its head is at the head of vector B, the last vector as shown in figure 4.5. Vector R It should be noted that represents the resultant force completely in magnitude head to tail rule can be and direction. used to add any number of forces. The vector EXAMPLE 4.1 representing resultant force gives the magnitude Find the resultant of three forces 12 N along and direction of the x-axis, 8 N making an angle of 45° with x-axis and 8 N resultant force. along y-axis. SOLUTION Here F = 12 N along x-axis 1 y F = 8 N along 45° with x-axis 2 F1 x F3 = 8 N along y-axis Scale: 1 cm = 2 N F3 (i) Represent the forces by vectors F1, F2 and F3 F2 according to the scale in the given direction. (ii) Arrange these forces F1, F2 and F3. The tail of D force F2 coincides with the head of force at point B as shown in figure 4.6. similarly the tail of F3 force F3 coincides with the head of force F2 at F point C. C (iii) Join point A the tail of the force F1 and point D F2 o the head of force F3. Let AD represents force F. 37 A B According to head to tail rule, force F represents F1 the resultant force. Figure 4.6: Adding forces by Head to tail rule. Physics IX 88 Unit 4: Turning Effect of Forces Some Trigonometric Ratios (iv) Measure AD and multiply it by 2 N cm1 , the scale to find the magnitude of the resultant force F. The ratios between any of its two sides of a right angled triangle are (v) Measure the angle DAB using a protractor given certain names such as sine, Which the force F makes with x-axis. This gives cosine, etc. Consider a right angled the direction of the resultant force. triangle AABC having angle 6 at A. 4.3 RESOLUTION OF FORCES B Perpendicular The process of splitting up vectors (forces) into their component forces is called resolution of forces. If a force se te nu is formed from two mutually perpendicular components po Hy then such components are called its perpendicular A  C components. Base Splitting up of a force into two mutually perpendicular components is called the resolution of that force. Consider a force F represented by line OA making an angle  with x-axis as shown in figure 4.7. Draw a perpendicular AB on x-axis from A. According to head to tail rule, OA is the resultant of vectors y represented by OB and BA. A Thus F The components OB and BA are perpendicular to Fy each other. They are called the perpendicular components of OA representing force F. Hence OB  x represents its x-component Fx and BA represents its O Fx B y-component Fy-. Therefor, equation 4.1 can be written as The magnitudes Fx and Fy of forces Fx and Fy can be found using the trigonometric ratios. In right angled triangle OBA Ratio/  0 30 45 60 90 Since 0.707 0.866 0.5 0 1 sin  0.866 0.707 Similarly 0.5 1 0 cos  0.577 1.732 Equations 4.3 and 4.4 give the perpendicular 0 1 tan  components Fx and Fy respectively. Physics IX 89 Unit 4: Turning Effect of Forces EXAMPLE 4.2 A man is pulling a trolley on a horizontal road with In a right angled triangle a force of 200 N making 30° with the road. Find the length of base is 4 cm and horizontal and vertical components of its force. its perpendicular is 3 cm. Find: SOLUTION (i) length of hypotenuse (ii) sin (iii) cos (iv) tan Since or Similarly or Thus, horizontal and vertical components of the pulling force are 173.2 N and 100 N respectively. y DETERMINATION OF A FORCE FROM ITS Q R PERPENDICULAR COMPONENTS Since a force can be resolved into two perpendicular Fy F components. Its reverse is to determine the force Fy knowing its perpendicular components. Consider Fx and Fy as the perpendicular components O  x Fx P of a force F. These perpendicular components Fx and Fy are represented by lines OP and PR respectively as Figure 4.8: Determination of a force by its perpendicular components. shown in figure 4.8. According to head to tail rule: Thus OR will completely represent the force F whose x and y-components are Fx and Fy respectively. That is The magnitude of the force F can be determined using the right angled triangle OPR Hence The direction of force F with x-axis is given by Physics IX 90 Unit 4: Turning Effect of Forces 4.4 TORQUE OR MOMENT OF A FORCE We open or close a door (Fig 4.9) by pushing or pulling it. Here push or pull turn the door about its hinge or axis of rotation. The door is opened or closed due to the turning effect of the force acting on it. RIGID BODY 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. In other words, a rigid body is the one that is not deformed by force or forces acting on it. Figure 4.9: It is easy to open and close the door by pulling or AXIS OF ROTATION pushing it at its handle. Consider a rigid body rotating about a line. The particles of the body move in circles with their centres all lying on this line. This line is called the axis of rotation of the body. Forces that produce turning effect are very common. Turning pencil in a sharpener, turning stopcock of a water tap, turning doorknob and so on are some of the examples where a force produces turning effect. QUICK QUIZ Name some more objects that work by the turning effects of forces. The turning effect of a force is called torque or moment of the force. Figure 4.10: Turning effect of Why the handle of a door is fixed near the outer forces. edge of a door? We can open or close a door more easily by applying a force at the outer edge of a door rather than near the hinge. Thus, the location where the force is applied to turn a body is very important. Physics IX 91 Unit 4: Turning Effect of Forces Let us study the factors on which torque or moment of a force depends. You might have seen that a mechanic uses a spanner as shown in figure 4.11 to loosen or tighten a nut or a bolt. A spanner having long arm helps him to do it with greater ease than the one Moment arm Moment arm Force Force Figure 4.11: It is easy to tighten a nut using a spanner of longer arm than a spanner of shorter arm. having short arm. It is because the turning effect of the force is different in the two cases. The moment produced by a force using a spanner of longer arm is greater than the torque produced by the same force but using a spanner of shorter arm. C rotation Axis of A LINE OF ACTION OF A FORCE F The line along which a force acts is called the line Moment Arm Line of action of action of the force. In figure 4.12, line BC is the line of action of force F. L of force MOMENT ARM A B The perpendicular distance between the axis of rotation and the line of action of the force is called the Figure 4.12: Factors affecting the moment of a force. moment arm of the force. It is represented by the distance L in figure 4.12. The torque or moment of a force depends upon the force F and the moment arm L of the force. Greater is a force, greater is the moment of the force. Similarly, longer is the moment arm greater is the moment of the force. Thus the moment of the force or torque is determined by the product of force F and its moment arm L. Mathematically, Torque Physics IX 92 Unit 4: Turning Effect of Forces SI unit of torque is newton-metre (Nm). A torque Mini Exercise of 1 N m is caused by a force of 1 N acting perpendicular A force of 150 N can loosen a to the moment arm 1 m long. nut when applied at the end of a spanner 10 cm long. EXAMPLE 4.3 1. What should be the length A mechanic tightens the nut of a bicycle using a of the spanner to loosen 15 cm long spanner by exerting a force of 200 N. Find the the same nut with a 60 N torque that has tightened it. force? SOLUTION 2. How much force would be sufficient to loosen it with a 6 cm long spanner? Using Thus, a torque of 30 N m is used to tighten the nut. Force 4.5 PRINCIPLE OF MOMENTS A force that turns a spanner in the clockwise direction is generally used to tighten a nut as shown in (a) figure 4.13(a). The torque or moment of the force so produced is called clockwise moment. On the other hand, to loosen a nut, the force is applied such that it turns the nut in the anticlockwise direction as shown in (b) figure 4.13(b). The torque or moment of the force so Force produced is called anticlockwise moment. Figure 4.13: (a) to tighten, nut is turned clockwise (b) to loosen, nut is A body initially at rest does not rotate if sum of all the turned anticlockwise. clockwise moments acting on it is balanced by the sum of all the anticlockwise moments acting on it. This is known Small Large as the principle of moments. According to the principle of distance distance moments: Large Small weight weight 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. QUICK QUIZ 1. Can a small child play with a fat child on the seesaw? Explain how? Figure 4.14: Children on 2. Two children are sitting on the see-saw, such see-saw. that they can not swing. What is the net torque in this situation? Physics IX 93 Unit 4: Turning Effect of Forces EXAMPLE 4.4 A metre rod is supported at its middle point O as shown in figure 4.15. The block of weight 10 N is suspended at point B, 40 cm from O. Find the weight of the block that balances it at point A, 25 cm from O. A O B W1 W2 Figure 4.15: Balancing a metre rod on a wedge SOLUTION Moment arm of w1= OA = 25 cm = 0.25 m Moment arm of w2 = OB = 40 cm = 0.40 m Applying principle of moments; Clockwise moments = Anticlockwise moments moment of w2 = moment of w1 or w1 x moment arm of w2 = w1 x moment arm of w1 Thus or Thus, weight of the block suspended at point A is 16 N. 4.6 CENTRE OF MASS It is observed that the centre of mass of a system moves as if its entire mass is confined at that point. A A B O force applied at such a point in the body does not produce any torque in it i.e. the body moves in the direction of net force F without rotation. Figure 4.16: Centre of mass of two unequal masses. Consider a system of two particles A and B connected by a light rigid rod as shown in figure 4.16. Physics IX 94 Unit 4: Turning Effect of Forces Let O is a point anywhere between A and B such that the F F F force F is applied at point O as shown in figure 4.17. If the O O O system moves in the direction of force F without rotation, then point O is the centre of mass of the system. Figure 4.17: A force applied at COM moves the system without rotation. Does the system still move without rotation if the force acts elsewhere on it? F F F (i) Let the force be applied near the lighter particle O O O as shown in figure 4.18. The system moves as well as rotates. Figure 4.18: The system moves as (ii) Let the force be applied near the heavier particle well as rotates when a force is applied as shown in figure 4.19. In this case, also the away from COM. system moves as well as rotates. Centre of mass of a system is such a point where an applied force causes the system to O F O F O F move without rotation. 4.6 CENTRE OF GRAVITY Figure 4.19: The system moves as well as rotates when a force is applied A body is made up of a large number of particles as away from COM. illustrated in figure 4.20. Earth attracts each of these CENTRE OF GRAVITY particles vertically downward towards its centre. The pull of the Earth acting on a particle is equal to its weight. These forces acting on the particles of a body are almost parallel. The resultant of all these parallel forces is a single force equal to the weight of the body. A point where ENTIRE WEIGHT this resultant force acts vertically towards the centre of the Earth is called the centre of gravity G of the body. A point where the whole weight of the body appears to act vertically downward is called centre of gravity of a body. It is useful to know the location of the centre of gravity of a body in problems dealing with equilibrium. Physics IX 95 Unit 4: Turning Effect of Forces CENTRE OF GRAVITY OF SOME SYMMETRICAL OBJECTS The centre of gravity of objects which have symmetrical shapes can be found from their geometry. For example, the centre of gravity of a uniform rod lies at a point where it is balanced. This balance point is its middle point G as shown in figure 4.21. G Centre of gravity w w Figure 4.21: Centre of gravity is at the middle of a uniform rod. The centre of a gravity of a uniform square or a rectangular sheet is the point of intersection of its diagonals as shown in figure 4.22 (a) and (c). The centre of gravity of a uniform circular disc is its centre as shown in figure 4.22(b). Similarly, the centre of gravity of a solid sphere or hollow sphere is the centre of the spheres as shown in figure 4.22(b). The centre of gravity of a uniform triangular sheet is the point of intersection of its medians as shown in figure 4.22 (d). The centre of gravity of a uniform circular ring is the centre of the ring as shown in figure 4.22(e). The centre of gravity of a uniform solid or hollow cylinder is the middle point on its axis as shown in figure 4.22(f). G G G G (a) (b) (c) G G G G (d) (e) (f) Figure 4.22: Centre of gravity of some symmetrical objects. CENTRE OF GRAVITY OF AN IRREGULAR SHAPED THIN LAMINA A simple method to find the centre of gravity of a body is by the use of a plumbline. A plumbline consists of a small metal bob (lead or brass) supported by a string. When the bob is suspended freely by the string, it rests along the vertical direction due to its weight acting vertically downward as shown in figure 4.23 (a). In this state, centre of gravity of the bob is exactly below its point of suspension. Physics IX 96 Unit 4: Turning Effect of Forces EXPERIMENT A Take an irregular piece of cardboard. Make holes A, B and C as shown in figure 4.23(b) near its edge. Fix a C B nail on a wall. Support the cardboard on the nail through one of the holes (let it be A), so that the cardboard can Plumbline G Centre of swing freely about A. The cardboard will come to rest with Gravity its centre of gravity just vertically below the nail. Vertical line from A can be located using a plumbline hung from Checking line the nail. Mark the line on the cardboard behind the plumbline. Repeat it by supporting the cardboard from (a) (b) hole B. The line from B will intersect at a point G. Similarly, draw another line from the hole C. Note that this Figure 4.23: (a) Plumbline (b) Locating line also passes through G. It will be found that all the the centre of gravity of a piece of vertical lines from holes A B and C have a common point cardboard by using plumbline. G. This Common point G is the centre of gravity of the cardboard. 4.7 COUPLE When a driver turns a vehicle, he applies forces that PUSH produce a torque. This torque turns the steering wheel. These forces act on opposite sides of the steering wheel as shown in figure 4.24 and are equal in PULL magnitude but opposite in direction. These two forces form a couple. A couple is formed by two unlike parallel forces Figure 4.24: It is easy to turn a of the same magnitude but not along the same steering wheel by applying a couple. line. A double arm spanner is used to open a nut. Equal forces each of magnitude Fare applied on ends A F and B of a spanner in opposite direction as shown in figure 4.25. These forces form a couple that turns the A spanner about point O. The torques produced by both B the forces of a couple have the same direction. Thus, the O total torque produced by the couple will be F Total torque of the couple Figure 4.25: A double arm spanner. Torque of the couple Equation 4.8 gives the torque produced by a couple of forces F and F separated by distance AB. The torque of a couple is given by the product of one of Physics IX 97 Unit 4: Turning Effect of Forces the two forces and the perpendicular distance between them. 4.8 EQUILIBRIUM DO YOU KNOW? Newton's first law of motion tells us that a body continues its state of rest or of uniform motion in a straight F line if no resultant or net force acts on it. For example, a book lying on a table or a picture hanging on a wall, are at rest. The weight of the book acting downward is balanced by the upward reaction of the table. Consider a log of F wood of weight w supported by ropes as shown in figure 4.26. Here the weight w is balanced by the forces F1 A cyclist pushes the pedals of a and F2 pulling the log upward. In case of objects moving with bicycle. This forms a couple uniform velocity, the resultant force acting on them is zero. A that acts on the pedals. The pedals cause the toothed car moving with uniform velocity on a levelled road and an wheel to turn making the rear aeroplane flying in the air with uniform velocity are the wheel of the bicycle to rotate. examples of bodies in equilibrium. A body is said to be in equilibrium if no net force acts on it. A body in equilibrium thus remains at rest or moves with uniform velocity. F1 F2 Pull Pull CONDITIONS FOR EQUILIBRIUM In the above examples, we see that a body at rest or in uniform motion is in equilibrium if the resultant force Weight acting on it is zero. For a body in equilibrium, it must w satisfy certain conditions. There are two conditions for a body to be in equilibrium. Figure 4.26: The forces acting on the log are; upward forces F1, F2 and its weight w in the downward FIRST CONDITION FOR EQUILIBRIUM direction. 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 are acting on a body such that The symbol is a Greek letter called sigma used for summation. Equation 4.9 is called the first condition for Figure 4.27: A wall hanging is equilibrium. in equilibrium Physics IX 98 Unit 4: Turning Effect of Forces The first condition for equilibrium can also be stated in terms of x and y-components of the forces acting on the body as: and A book lying on a table or a picture hanging on a wall, are at rest and thus satisfy first condition for equilibrium. A paratrooper coming down with terminal velocity (constant velocity) also satisfies first condition for Figure 4.28: A paratrooper equilibrium and is thus in equilibrium. coming down with terminal velocity is in equilibrium. EXAMPLE 4.5 A block of weight 10 N is hanging through a cord as shown in figure 4.29. Find the tension in the cord. T SOLUTION Weight of the block w = 10 N 10 N Tension in the cord T = ? Applying first condition for equilibrium w There is no force acting along x-axis. Figure 4.29 F2 F1 (a) Thus, the tension in the cord is 10 N. F1 SECOND CONDITION FOR EQUILIBRIUM F2 First condition for equilibrium does not ensure (b) that a body is in equilibrium. This is clear from the following example. Consider a body pulled by the Figure 4.30: (a) Two equal forces F1 and F2 as shown in figure 4.30(a). The two and opposite forces acting along the same lines (b) Two forces are equal but opposite to each other. Both are equal and opposite forces acting along the same line, hence their resultant will be acting along different lines zero. According to the first condition, the body will be in Physics IX 99 Unit 4: Turning Effect of Forces equilibrium. Now shift the location of the forces as shown in figure 4.30(b). In this situation, the body is not in equilibrium although the first condition for equilibrium is still satisfied. It is because the body has the tendency to rotate. This situation demands another condition for equilibrium in addition to the first condition for equilibrium. This is called second condition for equilibrium. According to this, a body satisfies second condition for equilibrium when the resultant torque acting on it is zero. Mathematically, QUICK QUIZ 1. A ladder leaning at a wall as shown in figure 4.31 is in equilibrium. How? 2. The weight of the ladder in figure 4.31 produces an anticlockwise torque. The wall pushes the ladder at its top end thus produces a clockwise torque. Does the ladder satisfy second condition for equilibrium? 3. Does the speed of a ceiling fan go on increasing all the time? 4. Does the fan satisfy second condition for equilibrium when rotating with uniform speed? EXAMPLE 4.6 A uniform rod of length 1.5 m is placed over a wedge at 0.5 m from its one end. A force of 100 N is applied at one of its ends near the wedge to keep it horizontal. Find the weight of the rod and the reaction of the wedge. R A O G B 0.5 m 0.25 m 0.75 m F w SOLUTION A rod balanced over a wedge Physics IX 100 Unit 4: Turning Effect of Forces Applying second condition for equilibrium, taking torques about O. or Applying first condition for equilibrium Thus, weight of the rod is 200 N and reaction of the wedge is 300 N. STATES OF EQUILIBRIUM There are three states of equilibrium; stable equilibrium, unstable equilibrium and neutral equilibrium. A body may be in one of these three states of equilibrium. STABLE EQUILIBRIUM G G (a) (b) Figure 4.33: Stable equilibrium (a) A book is lying on a table (b) The book returns to its previous position when let free after a slight tilt. Consider a book lying on the table. Tilt the book slightly about its one edge by lifting it from the opposite side as shown in figure 4.33. It returns to its previous Physics IX 101 Unit 4: Turning Effect of Forces position when sets free. Such a state of the body is called stable equilibrium. Thus 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 centre of gravity is at the lowest position. When it is tilted, its centre of gravity rises. It returns to its stable state by lowering its centre of gravity. A body remains in stable equilibrium as long as the centre of gravity acts through the base of the body. Consider a block as shown in figure 4.34. When the block is tilted, its centre of gravity G rises. If the vertical line through G passes through its base in the tilted position as Can you do this without falling? shown in figure 4.34 (b), the block returns to its previous position. If the vertical line through G gets out of its base as shown in figure 4.34(c), the block does not return to its previous position. It topples over its base and moves to new G G G (a) (b) (c) Figure 4.34 (a) Block in stable equilibrium (b) Slightly tilted block is returning to its previous position, (c) A more tilted block topples over its base and does not return to its previous position. Figure 4.35: A double decker bus stable equilibrium position. That is why a vehicle is made heavy being under test for stability. at its bottom to keep its centre of gravity as low as possible. A lower centre of gravity keeps it stable. Moreover, the base of a vehicle is made wide so that the vertical line passing through its Topple over centre of gravity should not get out of its base during a turn. G UNSTABLE EQUILIBRIUM G Take a pencil and try to keep it in the vertical position on its tip as shown in figure 4.36. Whenever you leave it, the pencil topples over about its tip and falls (a) (b) down. This is called the unstable equilibrium. In Figure 4.36: Unstable equilibrium (a) pencil just balanced at its tip with unstable equilibrium, a body may be made to stay only centre of gravity G at the highest position, (b) Pencil topples over caused by the torque of its weight acting at G. Physics IX 102 Unit 4: Turning Effect of Forces for a moment. Thus a body is unable to keep itself in the state of unstable equilibrium. Thus If a body does not return to its previous position when sets free after a slightest tilt is said to be in unstable equilibrium. The centre of gravity of the body is at its highest Vehicles are made heavy at the position in the state of unstable equilibrium. As the body bottom. This lowers their centre of topples over about its base (tip), its centre of gravity moves gravity and helps to increase their towards its lower position and does not return to its stability. previous position. NEUTRAL EQUILIBRIUM Take a ball and place it on a horizontal surface as shown in figure 4.37. Roll the ball over the surface and leave it after displacing from its previous position. It (a) (b) remains in its new position and does not return to its Figure 4.37: Neutral equilibrium previous position. This is called neutral equilibrium. (a) a ball is placed on a horizontal surface (b) the ball remains in its new If a body remains in its new position when displaced position. disturbed from its previous position, it is said to be in a state of neutral equilibrium. In neutral equilibrium, all the new states in which a body is moved, are the stable states and the body, remains in its new state. In neutral equilibrium, the centre of gravity of the body remains at the same height, irrespective to its new position. There are various objects which have neutral equilibrium such as a ball, a sphere, a roller, a pencil lying horizontally, an egg lying horizontally on a flat surface etc. 4.9 STABILITY AND POSITION OF CENTRE OF MASS Needle As we have learnt that position of centre of mass of Cork an object plays an important role in their stability. To make Fork them stable, their centre of mass must be kept as low as possible. It is due to this reason, racing cars are made heavy at the bottom and their height is kept to be minimum. Can Circus artists such as tight rope walkers use long poles to lower their centre of mass. In this way they are prevented from topple over. Here are few examples in which lowering Figure 4.38: A needle is made of centre of mass make the objects stable. These to balance at its tip.  Physics IX 103 Unit 4: Turning Effect of Forces objects return to their stable states when disturbed. In each case centre of mass is vertically below their point of support. This makes their equilibrium stable. Figure 4.38 shows a sewing needle fixed in a Card cork. The cork is balanced on the tip of the needle by hanging forks. The forks lower the centre of mass of the Spar system. Figure 4.39(a) shows a perched parrot which is made heavy at its tail. Figure 4.39(b) shows a toy that keeps itself upright when tilted. It has a heavy semi- spherical base. When it is tilted, its centre of mass rises. It Figure 4.39: (a) A perchd parrot returns to its upright position at which itscentre of mass is (b) A self righting toy at the lowest. SUMMARY  Parallel forces have their lines of action on a body in equilibrium is equal to the parallel to each other. sum of anticlockwise moments acting  If the direction of parallel forces is the on it. same, they are called like parallel  Centre of mass of a body is such a point forces. If two parallel forces are in where a net force causes it to move opposite direction to each other, then without rotation. they are called unlike parallel forces.  The centre of gravity of a body is a point  The sum of two or more forces is called where the whole weight of a body acts the resultant force. vertically downward.  A graphical method used to find the  A couple is formed by two parallel resultant of two or more forces is called forces of the same magnitude but head to tail rule. acting in opposite direction along  Splitting up a force into two different lines of action. components perpendicular to each  A body is in equilibrium if net force other is called resolution of that force. acting on it is zero. A body in These components are equilibrium either remains at rest or moves with a uniform velocity.  A body is said to satisfy second  A force can be determined from its condition for equilibrium if the resultant perpendicular components as torque acting on it is zero.  A body is said to be in stable equilibrium if after a slight tilt it returns to its  Torque or moment of a force is the previous position. turning effect of the force. Torque of a  If a body does not return to its previous force is equal to the product of force position when sets free after slightly tilt and moment arm of the force. is said to be in unstable equilibrium.  According to the principle of moments, the sum of clockwise moments acting  Physics IX 104 Unit 4: Turning Effect of Forces  A body that remains in its new position position is said to be in a state of when disturbed from its previous neutral equilibrium. 4.1 Encircle the correct answers from (d) acceleration is zero the given choices: vii. A body is in neutral equilibrium i. Two equal but unlike parallel when its centre of gravity: forces having different line of (a) is at its highest position action produce (b) is at the lowest position (c) keeps its height if displaced (a) a torque (d) is situated at its bottom (b) a couple (c) equilibrium viii. Racing cars are made stable by: (d) neutral equilibrium (a) increasing their speed (b) decreasing their mass ii. The number of forces that can (c) lowering their centre of be added by head to tail rule gravity are: (d) decreasing their width (a) 2 (b) 3 4.2 Define the following: (c) 4 (d) any number (i) resultant vector iii. The number of perpendicular (ii) torque components of a force are: (iii) centre of mass (a) 1 (b) 2 (iv) centre of gravity (c) 3 (d) 4 4.3 Differentiate the following: iv. A force of 10 N is making an (i) like and unlike forces angle of 30° with the horizontal. (ii) torque and couple Its horizontal component will be : (iii) stable and neutral (a) 4 N (b) 5 N equilibrium (c) 7 N (d) 8.7 N 4.4 How head to tail rule helps to find v. A couple is formed by the resultant of forces? (a) two forces perpendicular to 4.5 How can a force be resolved into each other its perpendicular components? (b) two like parallel forces (c) two equal and opposite forces 4.6 When a body is said to be in in the same line equilibrium? (d) two equal and opposite forces 4.7 Explain the first condition for not in the same line equilibrium. vi. A body is in equilibrium when its: 4.8 Why there is a need of second (a) acceleration is uniform condition for equilibrium if a body (b) speed is uniform satisfies first condition for (c) speed and acceleration are equilibrium? uniform Physics IX 105 Unit 4: Turning Effect of Forces 4.9 What is second condition for 4.12 Why a body cannot be in equilibrium? equilibrium due to single force acting on it? 4.10 Give an example of a moving body 4.13 Why the height of vehicles is kept which is in equilibrium. as low as possible? 4.11 Think of a body which is at rest but 4.14 Explain what is meant by stable, unstable and neutral equilibrium. not in equilibrium. Give one example in each case. 4.1 Find the resultant of the following 4.7 A picture frame is hanging by two forces: vertical strings. The tensions in (i) 10 N along x-axis the strings are 3.8 N and 4.4 N. (ii) 6 N along y-axis and Find the weight of the picture (iii) 4 N along negative x-axis. frame. (8.2 N) (8.5 N making 45° with x-axis) 4.8 Tw o b l o c k s o f 4.2 Find the perpendicular components masses 5 kg and 3 kg are suspended of a force of 50 N making an angle of by the two strings 30° with x axis. (43.3 N, 25 N) as shown. Find the tension in each 4.3 Find the magnitude and direction string. (80 N, 30 N) of a force, if its x-component is 5 kg 4.9 A nut has been 12 N and y- component is 5 N. tightened by a force (13 N making 22.6° with x-axis) of 200 N using 10 cm 4.4 A force of 100 N is applied long spanner. What length of a spanner perpendicularly on a spanner at a is required to loosen 3 kg distance of 10 cm from a nut. Find the same nut with the torque produced by the 150 N force? (13.3 cm) force. (10 Nm) 4.5 A force is acting on a body making 4.10 A block of mass 10 kg is an angle of 30° with the horizontal. suspended at a distance of 20 cm The horizontal component of from the centre of a uniform bar the force is 20 N. Find the 1 m long. What force is required to force. (23.1 N) balance it at its centre of gravity 4.6 The steering of a car has a radius by applying the force at the other 16 cm. Find the torque produced end of the bar? (40 N) by a couple of 50 N. (16 Nm)