X-Physics Textbook 2019 PDF

Chapter 1 2. Moment of Force Learning Objectives On completion of this topic, you should be able to: describe statics, and system in equilibrium. describe resultant, moment and parallelogram law of forces. explain the couple, torque of a couple and the principle of moments. A. Forces and e...

Chapter 1 2. Moment of Force Learning Objectives On completion of this topic, you should be able to: describe statics, and system in equilibrium. describe resultant, moment and parallelogram law of forces. explain the couple, torque of a couple and the principle of moments. A. Forces and equilibrium The simple descriptions of motion that we have used so far implicitly treated the motion of only one point in a body. It is a good description provided that all points in the body follow similar, parallel paths. This kind of motion is called pure translational motion. Along with translational motion, a body can also execute rotational motion (like that of a spinning wheel) and vibrational motion (like that of a shaking jelly). A body which cannot vibrate noticeably is said to be rigid; its shape and size do not change significantly when it is acted on by a system of forces. Therefore, the general kind of motion of a rigid body is a combination of translation and rotation. For example, the flight of boomerang executes translational motion due to its rotation. A couple of forces may act on a body at a same time at different or same lines of action. Figure 1.9(a) shows two forces; 8 N and 10 N, acting on a rigid body in straight line in the same direction. The total magnitude of these two forces on the body is numerically equal to the sum of the magnitudes of these individual forces and the body moves in the direction of the greater force. The net force of these forces is referred to as resultant force. The resultant force in case of Figure 1.9 (a) is 18 N and the rigid body moves in the direction of the greater force, 10 N. Resultant Force= 18 N Resultant Force= 2 N 8N 10 N 8N 10 N (a) (b) Figure 1.9 If two equal forces act in opposite direction in the same line of action, as shown in Figure 1.9 (b), then the magnitude of resultant force is equal to the difference of the magnitude of two forces, i.e., 2N and the body moves in the direction of the greater force, 10 N. m a te r ia l Reprint 2018 yr ig h t e d 8 cop  Forces and Motion When a couple of forces act on a non-rigid (elastic) body, the resultant force generally tends to change the dimension of the body and at the same time, it may execute rotational, translational and vibrational motion. For example, bouncing of a soft rubber ball undergoes all these motions. A rigid body which is currently stationary will remain stationary if the resultant force is zero for all the forces applied on it. Similarly, a moving rigid body will remain in uniform motion if the resultant force is zero for all the forces applied on it. In both the cases, the body is said to be in equilibrium. Statics is a branch of mechanics that deals with the study of forces in equilibrium. The general conditions for equilibrium are as follows: (i) The total force must be zero. (ii) The total turning effect of force about any axis must be zero. If a resultant force acts on a body then that body can be brought into equilibrium by applying an 3N 5 N = additional force that exactly balances this resultant. ltan t su Such a force is called equilibrant and is equal in Re magnitude but opposite in direction to the original 4N resultant force acting on the body. 5 N = nt Force is an influence which causes motion of any uil ibr a q rigid object. Force is a vector quantity, which has E Figure 1.10 both magnitude and direction. Hence, it follows all the properties of vectors, such as vector addition and vector subtraction. One of the vector addition technique is known as the parallelogram law of vector addition. It is a graphical representation of vectors that is used to determine the magnitude and direction of the resultant vector. This method can be easily used to determine the magnitude and direction of the resultant force. Activity 1.2 Verifying the law of parallelogram of forces Materials required: A drawing board, a sheet of plain paper, pencil, protractor, drawing pins. Procedure Step 1  Using board. the drawing pins, mount the sheet of plain paper on the drawing m a te r ia l Reprint 2018 yr ig h t e d cop 9 Chapter 1 Step 2  Consider two force vectors A of 10 N and B of 5 N. Draw arrows to represent these forces in such a way that these force vectors are tail to tail in contact with each other and inclined to each other at an angle of 500. Let.1 cm = 1 N of force as shown in Figure 1.11. Step 3  Complete the parallelogram by drawing these vectors as adjacent sides of the parallelogram. Step 4  Now draw the diagonal originating from the touching tails. The diagonal represents the resultant force vector. Note the magnitude of resultant force. A B A B A and B are the force vectors to be added A and B are arranged as per Step 2 and parallelogram is completed A A +B R= Diagonal arrow represents the resultant vector Figure 1.11 B Step 5  Draw an arrow from the tails of the two vectors equal in length as the diagonal but in exactly opposite direction to it. Answer the following questions. 1. Find the resultant of vectors 7 N and 12 N inclined at an angle of 600 with each other. 2. Find the resultant of vectors 5 N and 10 N inclined at an angle of 450 to each other. 3. What are the factors on which the resultant of vectors depend? 4. What does the arrow drawn in step 5 represent? m a te r ia l Reprint 2018 yr ig h t e d 10 cop  Forces and Motion B. Couple If two forces are not acting in a straight line, then the resultant effect of the force is different. For example, when a water tap is turned to open or close, we apply two forces at the two ends of the bip cock. Here the forces are applied simultaneously, parallel and in opposite direction to each other. Thus, it results in rotational motion of bip cock. A few such examples from our day to day activities are shown in Figure 1.12. In these pictures, we use a pair of forces which are parallel to each other, equal in magnitude, and acting in the opposite directions. A pair of equal and parallel forces, acting in opposite directions is called a couple. F F F F d F F Figure 1.12 Couples In Figure 1.12, two equal and opposite forces ‘F’, which are parallel to each other act on a body. The resultant force makes a body turn. The turning effect of a force is known as the moment. If the perpendicular distance between these forces is ‘d’, then the product of the magnitude of one of the forces (F) and the perpendicular distance from the line of action of the forces give the turning effect of the force called moment of couple. The moment of couple is also called torque and is usually represented by ‘τ’ (tau). Mathematically, Moment of couple (τ) = Force (F) x Perpendicular distance (d) between forces. Example 1.1 Two forces each of 20 N is applied to a car steering wheel that has a diameter of 40 cm. If the two forces act tangentially to the steering wheel and in anti-parallel directions calculate the torque applied. Solution: Torque = F x d = 20 N x 0.4 m= 8 Nm m a te r ia l Reprint 2018 yr ig h t e d cop 11 Chapter 1 C. Principle of moments It is everyday experience that the handles on doors are Fulcrum usually located at the outer edge, so that it is easier to open or close. It is difficult to open a door if the handle is located near the hinge, that is, a much larger force is needed to open the same door. Likewise, it is easier to loosen a nut on a bolt by using a long spanner than with a short one. In all these cases, the applied force produces a turning effect Force about a pivot called moment of force. The moment of force is determined by the magnitude of force and distance of the applied force from the pivot. Similar to moment of couple, moment of a force is the product of force and the perpendicular distance of the line of action of force from Figure 1.13 Torque the pivot. Mathematically, Moment of force (τ) = Force (F) x distance from the fulcrum (d) If the force is measured in newton (N) and distance in metre(m), then the unit of moment of force is newton-metre (Nm). The moment of force brings about counter clockwise or clockwise rotation of the body. A body will not turn when it is in equilibrium even when forces act on it about a fixed pivot. This state of equilibrium is explained by the principle of moments. The principle of moments states that the total counter clockwise moment is equal to the total clockwise moment when the body is in equilibrium. In equilibrium, Total Counter Clockwise Moment = Total Clockwise Moment In case of Figure 1.14 d1 d2 Figure 1.14 F1 F2 total counter clockwise moments = total clockwise moments F1 x d1 = F2 x d2 m a te r ia l Reprint 2018 yr ig h t e d 12 cop  Forces and Motion In case of Figure 1.15 d1 d2 d3 F2 F1 Figure 1.15 F3 total counter clockwise moments = total clockwise moments F1 x d1 = (F2 x d2 ) + (F3 x d3) Activity 1.3 Verifying the principle of moments Materials required: A metre ruler, wedge, five weights of 100 g, clamp and stand, and thread. Procedure Step 1  Balance the ruler on the wedge so that it is perfectly horizontal as shown in Figure 1.16. 10 cm x Take two different weights of your Step 2  choice and note down their values as W1 and W2. W1 W2 Figure 1.16 Step 3  Hang 1.16. the two weights on either side of the pivot as shown in Figure Step 4  Adjust the distance of the two weights from the pivot until the ruler becomes perfectly horizontal again. Step 5  Record the distance of the two weights from the pivot. Answer the following questions. 1. Calculate the moment of force in clockwise direction. 2. Calculate the moment of force in counter clockwise direction. 3. Compare the magnitude of moment of force in clockwise direction and counter clockwise direction. 4. What do you conclude from the comparison of clockwise and counter clockwise moments? m a te r ia l Reprint 2018 yr ig h t e d cop 13 Chapter 1 Questions 1. Figure 1.17 shows two forces acting on the edge of a disc making it rotate. Rotating disc 1200 N 0.20 m 1200 N Figure 1.17 a) Define torque of a couple. b) Calculate the torque produced by these forces. 2. Write down five examples of couple. http://www.splung.com/content/sid/2/page/moments http://physicsnet.co.uk/a-level-physics-as-a2/mechanics/moments/ http://www.tutorvista.com/physics/parallelogram-law-of-forces m a te r ia l Reprint 2018 yr ig h t e d 14 cop  Forces and Motion 3. Falling Objects Learning Objectives On completion of this topic, you should be able to: explain that the forces acting on falling objects change with velocity. describe terminal velocity of falling objects. A. Force on falling objects Hundreds of years ago, people thought that the mass or weight of an object was the main thing that determined how fast it would fall. This idea was put forth by the Greek philosopher Aristotle, who said that the speed of a falling object was directly proportional to how heavy it was. Although this seems reasonable at first glance, there are some big problems with this idea. The scientist Galileo Galilei (1564-1642) disproved this idea by showing that objects of the same size and shape but with different masses would hit the ground at the same time when dropped from the same height. Later, Isaac Newton (1643-1727) demonstrated not only which type of object falls fast, but also answered why do objects fall in the first place. Earth exerts a force on all objects near it which causes falling objects to speed up as they fall. He called this force as gravity. Whenever, we describe about falling object, two forces come into picture, namely; the weight of the object and the air resistance. The weight of the object acts downwards. The air resistance is the frictional force acting in the opposite direction to the 0s - 0 m/s movement of the object. 1s - 10 m/s (i) Free falling object 2s - 20 m/s A free falling object is an object that is falling under the sole influence of gravity. Any object that is being acted upon only 3s - 30 m/s by the force of gravity is said to be in a state of free fall. There are two important motion characteristics that are true of free- falling objects: Free-falling objects do not encounter air resistance. 4s - 40 m/s On Earth, all free-falling objects accelerate downwards at a rate of 9.8 m/s2.. A dot diagram of motion of free-falling objects could be used to depict an acceleration of the object. The dot diagram in Figure 1.18 depicts the acceleration of a free-falling object. 5s - 50 m/s l Figure 1.18 d m a te r ia Reprint 2018 cop yr ig h t e 15 Chapter 1 The position of the object at regular time intervals of 1 second is shown. The increasing distance travelled by the object at equal interval of time shows that the ball is speeding up as it falls downward. If an object travels downward and speeds up, then its acceleration is downward. (ii) Drag force As objects fall through air due to Earth’s gravitational Drag force force, they experience drag force or air resistance forces that act upward and oppose the force of gravity. The air drag force depends on several factors, including the speed at which the object is falling (v), the surface area of the object (A), the density of the air (d), and the drag coefficient (C), which is determined by how Weight aerodynamic the object is. Figure 1.19 Drag Force= 0.5 x d x v x A x C 2 The most important factor in determining the air drag force is the velocity of the object. As the body speeds up, the drag force becomes larger and larger. (iii) Terminal velocity Consider an object falling with increasing velocity under the influence of gravity or a constant driving force. The object experiences drag force due to air resistance. It will ultimately reach a maximum velocity where the drag force is equal to the driving force. This final, constant velocity of motion is called terminal velocity. Initially, ball falls As speed As speed Speed increases to a point slowly, therefore increases, air increases air when air resistance equals air resistance is resistance resistance gravitational force. Ball is in negligible. Ball increases. Ball increases further. dynamic equilibrium when accelerates at accelerates at Ball accelerates ball moves with terminal 10m/s2. (each 10m/s2. (each downward. (not velocity) second speed second speed nearly as fast) FDrag= 10 N increases by 10 increases by 10 m/s) FDrag= 6 N m/s) FDrag= 3 N Net Force = -10 N Net Force = -7 N Net Force = -4 N Net Force = 0 N Fg= 10 N Fg= 10 N Fg= 10 N Fg= 10 N Figure 1.20 m a te r ia l Reprint 2018 yr ig h t e d 16 cop  Forces and Motion Objects falling through a fluid eventually reach terminal velocity, when the resultant force acting on them is zero and they are moving at a steady speed. For example, when a feather and a coin of roughly same surface area is dropped from a height, they experience the same air resistance. During the fall, the air resistance increases and soon balances the weight of the feather. So the feather now falls at its terminal velocity. On the other hand, the coin continues to move downward with increasing velocity, because it is relatively heavier. The coin travels faster since a larger air resistance is needed to balance its weight. The coin hits the ground before it reaches its terminal velocity. (iv) Stages of falling objects We can identify three stages of a falling object before it hits the surface of the Earth: 1. At the start, the object accelerates downwards due to its weight. At this point the air resistance is very minimal or absent. Therefore, a resultant force acts downward that makes the object to accelerate downward. 2. With acceleration, the velocity of the falling object increases. The air resistance on the object increases but its weight remains same. At this stage, the weight of the object is still greater than the air resistance and the fall continues. 3. The air resistance continues to increase, and eventually it balances the object’s weight. At this stage the resultant force is zero and the object reaches a constant velocity called terminal velocity. The sky diver accelerates as fall begins As the sky diver speeds up, the air resistance increases At terminal velocity, the air resistance and weight are equal so that speed is constant. Speed The parachute opens and the air resistance increases and slows the fall of sky diver. The sky diver continues to slow down until the new air resistance and weight are equal again ( new terminal velocity). Time Figure 1.21 Speed-time graph for a skydiver’s descent m a te r ia l Reprint 2018 yr ig h t e d cop 17 Chapter 1 Summary  The body with larger mass will have more gravity and larger gravitational field.  A point on a body through which the whole weight of a body appears to act is known as centre of the gravity of the body. It is usually represented as C.G. The point can be either within or outside the body.  Lower the C.G. of the body, more stable is the body and vice versa.  The surface area in contact with another surface which supports a body is the area of base of support.  Increasing the base of a body increases the stability of a body.  The study of forces in equilibrium is called statics.  A body is said to be in stable equilibrium when a body has an ability to regain its original position even when displaced by an external force.  A body is in state of unstable equilibrium when the line of gravity of its weight falls outside the B.S. on slight displacement and it loses its ability to regain its original position.  If a body neither takes a new position nor regains its original position when displaced slightly by an external force, then the body is said to be in neutral equilibrium.  A pair of equal and parallel forces, acting in opposite direction is called a couple.  A body is said to be in equilibrium if both the sum of forces acting on it and the sum of moments of forces is also zero.  The equilibrant of any number of forces is the single force required to produce equilibrium, and is equal in magnitude but opposite in direction to the resultant force.  If two forces simultaneously acting at a point, can be represented in magnitude and direction by two adjacent sides of a parallelogram, then the resultant will be represented by diagonal of the parallelogram passing through the intersection point of these two sides.  The constant maximum velocity reached by a body falling through the atmosphere under the attraction of gravity is known as its terminal velocity. m a te r ia l Reprint 2018 yr ig h t e d 18 cop  Forces and Motion Exercises I. Fill in the blanks. 1. A cone on its base is in.................equilibrium. 2. Stability of bodies depends upon …………………and …………………… 3. A body is said to be in equilibrium if the resultant forces acting on the body is………… 4. If we suspend lamina at different positions, its center of gravity will still lie along the line of................. 5. To form a couple, forces should be................. in magnitude. II. Match the following. Column A Column B 1. Torque A. Line of centre of gravity 2. Centre of gravity B. Maximum constant velocity 3. Plumb line C. Not in motion 4. Terminal velocity D. Point of action of weight 5. Statics E. Turning effect of force F. Study of forces III. Multiple Choice Questions. 1. Two forces of equal magnitude in opposite direction and acting along the different line of action give rise to A torque. B couple. C motion. D rotation. 2. A force of 2 N applied by Tshomo brings about 16 N m moment of force. At what distance from the pivot is Tshomo applying the effort? A 32 m. B 8 m. C 14 m. D 18 m. m a te r ia l Reprint 2018 yr ig h t e d cop 19 Chapter 1 3. The centre of gravity is usually located where A no mass is concentrated. B average mass is concentrated. C less mass is concentrated. D more mass is concentrated. 4. A body in motion is said to be in equilibrium when it is A moving with uniform velocity. B at rest. C accelerated by a force. D moving in an indefinite path. 5. Pair of forces that cause steering wheel of a car to rotate is called A couple. B friction. C normal force. D weight. 6. When the net force acting on a droplet of water becomes zero, then it falls with A final velocity. B initial velocity. C terminal velocity. D zero velocity. 7. Dorji is turning a wheel in clockwise direction with a force of 10 N, while Pema applies a force of 15 N in opposite direction. What additional force must be added to produce equilibrium? A 5 N acting in the same direction as the 10N force B 5 N acting in the same direction as the 15 N force C 10 N acting in the same direction as the 10 N force D 25 N acting in the same direction as the 15 N force m a te r ia l Reprint 2018 yr ig h t e d 20 cop  Forces and Motion 8. The greater the force, the larger will be the A axis of rotation. B torque. C mass. D centre of mass. 9. Karma applies a force of 4 N on a handle that is about 1.5 m away from door hinge. The moment of force will be A 5.50 Nm. B 2.66 Nm. C 6.00 Nm. D 2.50 Nm. IV. State ‘True’ or ‘False’. Rewrite the false statements correctly. 1. The body at terminal velocity will have less weight than the drag force. 2. The line of gravity is always within the base of support of the body in stable equilibrium. 3. The centre of gravity of objects will be always on the body itself. 4. The force that accelerates a falling body is the force of friction. 5. A couple always turns the body in a single direction. V. Answer the following questions. 1. Explain the term ‘equilibrant’. 2. The Figure 1.22 shows a body A of weight 400N placed on the left of a seesaw, 2.5 m away from the pivot. A body B of weight 300 N is placed on the right hand side 3.5m away from the pivot. A B 400 N 300 N 2.5 m 3.5 m Figure 1.22 a) Who will turn the seesaw clockwise? b) What is the clockwise moment? m a te r ia l Reprint 2018 yr ig h t e d cop 21 Chapter 1 c) What is the counter clockwise moment? d) Where must a body C of weight 10 N be placed to balance the see saw? 3. A uniform metre ruler, acted upon by three forces is balanced on a pivot as shown in Figure 1.23. 20 N 30 cm 70 cm 14 N Figure 1.23 6N a) Which one of the forces has no moment about the pivot? Explain your answer. b) Calculate the moments of the other two forces about the pivot. c) Will the ruler turn? Explain. 4. The velocity of free-falling ball at different time intervals on unknown planet is shown in Figure 1.24. 0s - 0 m/s a) What is the acceleration due to gravity of the 1s - 3.75m/s planet? b) What is the velocity at positions A and B? 2s - 7.50 m/s c) What happens to the magnitude of velocity with passage of time? Why? 3s - 11.25 m/s d) What factor(s) can influence the falling object on the Earth? A 4s B 5s m a te r ia l Figure 1.24 22 copyrig Reprint 2018 hted