Mechanics PDF - Past Paper

Contents 1 Introduction and basic concepts.......................... 1 1.1 Introduction................................. 1 1.2 Units..................................... 2 1.3 Scalar and vector quantities........................ 3 1.3.1 The dis...

Contents 1 Introduction and basic concepts.......................... 1 1.1 Introduction................................. 1 1.2 Units..................................... 2 1.3 Scalar and vector quantities........................ 3 1.3.1 The distinction between scalar and vector quantities..... 3 1.3.2 Addition of scalars........................ 3 1.3.3 Addition of vectors by construction............... 3 1.3.4 Components of a vector..................... 5 1.3.5 The resolution (or resolving) of vectors............. 5 1.3.6 The addition of vectors by means of components....... 6 1.3.7 The parallelogram of vectors................... 7 2 Kinematics..................................... 8 2.1 Definitions.................................. 8 2.2 Kinematic equations for uniform acceleration in one dimension..... 12 2.3 Motion under gravity............................ 14 2.4 Strategies for problem solving....................... 15 2.5 Motion in a plane.............................. 16 2.5.1 Projectile motion......................... 16 2.5.2 Uniform circular motion..................... 18 3 Dynamics...................................... 18 3.1 Newton’s first law of motion........................ 19 3.2 Newton’s second law of motion...................... 19 3.2.1 Free-body diagrams....................... 20 3.3 Newton’s third law............................. 21 3.4 Types of forces............................... 21 3.4.1 The gravitational force and weight............... 22 3.4.2 The Normal force......................... 23 3.4.3 Friction.............................. 25 3.4.4 Tension.............................. 27 3.5 Application of Newton’s laws....................... 27 3.5.1 Guidelines for solving problems involving Newton’s laws... 28 3.5.2 Equilibrium applications..................... 28 3.5.3 Non-equilibrium applications.................. 30 3.5.4 Motion on a smooth inclined plane............... 31 3.6 The centripetal force............................ 32 3.7 Satellites in circular orbits......................... 32 3.7.1 Geostationary orbits....................... 33 4 Hydrostatics..................................... 33 4.1 Density................................... 33 i  4.2 Relative density............................... 34 4.3 Pressure................................... 35 4.3.1 Summary of some laws of pressure in fluids at rest...... 35 4.3.2 Gauge pressure.......................... 36 4.4 Archimedes’ principle............................ 38 5 Work, energy and power.............................. 39 5.1 The work done by a constant force.................... 39 5.2 Energy.................................... 40 5.2.1 Kinetic energy and the work-energy theorem.......... 40 5.2.2 Potential energy......................... 41 5.3 Conservation of mechanical energy.................... 43 5.4 Power.................................... 44 6 Impulse and momentum.............................. 45 6.1 Impulse................................... 45 6.2 Momentum................................. 45 6.3 The impulse–momentum theorem..................... 46 6.4 The law of conservation of momentum.................. 46 6.5 Collisions.................................. 46 6.5.1 Classification of collisions.................... 46 7 Equilibrium of rigid bodies............................. 48 7.1 Centre of gravity.............................. 48 7.2 The moment of a force, or the torque about an axis........... 48 7.2.1 The equilibrium of a rigid body under coplanar forces..... 49 7.2.2 Stable, unstable and neutral equilibrium............ 50 8 Rotational motion.................................. 50 8.1 Angular velocity.............................. 50 8.2 Angular acceleration............................ 51 8.3 Constant angular acceleration equations of motion............ 52 8.4 Newton’s second law for rotational motion about a fixed axis...... 52 8.5 Rotational kinetic energy and moments of inertia............ 53 8.6 Work and power.............................. 54 8.7 Angular impulse and momentum..................... 54 9 Simple harmonic motion.............................. 55 9.1 Relations in S.H.M.............................. 56 9.2 The force for S.H.M............................. 56 9.3 A simple pendulum............................. 57 10 Elasticity...................................... 57 10.1 Introduction................................. 57 10.2 Stress and strain.............................. 57 10.2.1 Measurement of Young’s modulus................ 61 10.2.2 Hooke’s law............................ 61 11 Fluid dynamics................................... 62 11.1 Steady versus non-steady flow....................... 62 11.2 Laminar versus turbulent flow....................... 62 11.3 Flow rate and an equation of continuity................. 63 11.4 Bernoulli’s equation............................. 64 11.5 Viscosity................................... 64 11.5.1 Laminar flow in tubes: Poiseuille’s law............. 64 ii  11.5.2 A spherical object moving in a fluid: Stokes’ law....... 65 11.5.3 Turbulence............................ 65 A Revision of some elementary mathematics.................... 67 A.1 Geometry.................................. 67 A.2 Trigonometry................................ 68 iii List of Examples 1 Conversion of inches and feet to SI units..................... 3 2 Conversion of km/hour to m s−1.......................... 3 3 Conversion of standard prefixes.......................... 3 4 Addition of vectors by construction........................ 4 5 Addition of vectors by adding components.................... 6 6 Velocity and speed................................. 10 7 Position-time graph................................. 10 8 Average acceleration................................ 11 9 Velocity-time graph................................. 11 10 Acceleration of a car from rest........................... 13 11 Distance covered by an accelerating spacecraft.................. 14 12 Motion under gravity................................ 14 13 Projectile motion.................................. 17 14 Free-body diagram................................. 20 15 The mass of the earth............................... 22 16 Acceleration due to gravity for a satellite in orbit................ 23 17 Weight and the normal force in an elevator.................... 23 18 Block on a horizontal surface with friction.................... 25 19 Block on an inclined plane with friction...................... 26 20 Velocity time graph for a skydiver......................... 26 21 Tension in a cord, one dimension (equilibrium case)............... 28 22 Tension in a cord, two dimensions (equilibrium case).............. 29 23 The tension in a rope (non-equilibrium)...................... 30 24 Objects connected by a rope............................ 30 25 Motion on an inclined plane............................ 31 26 Relative density of aluminium............................ 34 27 Relative density of a mixture............................ 34 28 A water barometer.................................. 37 29 Pressure due to a column of air........................... 37 30 Apparent weight of a mass suspended in a liquid................. 38 31 Work done on an object dragged over a distance................. 39 32 The work done in accelerating a car........................ 41 33 Work done in lifting an object........................... 42 34 Conservation of mechanical energy........................ 43 35 The power generated by an accelerating car................... 44 36 A bullet fired into a block of wood........................ 47 37 Moment about an axis............................... 49 38 Mass suspended from a wire............................. 60 39 Minimum diameter of a wire under stress..................... 60 iv 40 Compressing and stretching a spring........................ 62 v vi Mechanics 1 Introduction and basic concepts In this section we summarise some basic concepts and methods with which the reader should be familiar. These notes are intended as guidelines and the reader is advised to supplement their studies by reading relevant sections from the many excellent standard texts available. We recommend in particular: Fundamentals of Physics by Halliday, Resnick & Walker University Physics by Young & Freedman There is also much material freely available on the internet. Background material for the physical sciences is available (free of charge) in pdf format from the website http://www.everythingscience.co.za/grade-12 Textbooks for grades 10 and 11, as well as textbooks for mathematics for grades 10–12, are also available. 1.1 Introduction Physics attempts to describe how and why our universe (including our immediate physical environment) behaves as it does. For example it explains why the sky is blue and why rainbows have colours. It also explains what keeps our moon in its orbit, and accounts for the thunder and lightning that accompany a storm. The laws of physics are remarkable for their scope, covering the behaviour both of sub- atomic particles and distant stars far greater than our sun. It is because physics is so fundamental that it is a required course for students majoring in a wide variety of other subjects. We hope you will come to see that physics is highly relevant both to you and your envi- ronment. In this course we will study the oldest branch of mechanics called classical mechanics. It is used to describe the motion of objects much bigger than atoms moving at speeds much less than the speed of light. 1 1.2 Units We shall use the SI (Système International) set of units. This has a number of base units, three of which will be important in our study of Mechanics: the metre, the kilogram and the second. Units must always be given except in the case of dimensionless quantities. Physical quantity Name of S.I. unit Symbol Length metre m Mass kilogram kg Time second s Table 1: Several commonly used base units. Units of quantities such as force and energy can be expressed as combinations of the base units and are referred to as derived units. Table 2 lists some of the derived units that will be encountered in this course. Physical quantity Name of S.I. unit Base units Symbol Force newton kg m s−2 N Energy joule kg m2 s−2 J Pressure pascal N m−2 Pa Table 2: Several commonly used derived units. The value of a quantity is often very large or very small when expressed in base or derived units. It is then convenient to express these quantities in terms of multiples of ten as given by the prefixes summarised in Table 3. Prefix Symbol Factor Tera T 1012 Giga G 109 Mega M 106 Kilo k 103 Deci d 10−1 Centi c 10−2 Milli m 10−3 Micro µ 10−6 Nano n 10−9 Pico p 10−12 Femto f 10−15 Atto a 10−18 Table 3: Standard prefixes used to denote multiples of ten. 2 Example 1: Conversion of inches and feet to SI units There are 12 inches in one foot and one inch is 2.54 cm. Calculate the height of a person who is 6 feet tall in SI units. Solution: 6 feet = 6 × 12 inches = 72 inches = 72 × 2.54 cm = 183 cm = 1.83 m. Example 2: Conversion of km/hour to m s−1 Convert 60 km h−1 to m s−1. Solution: 60 km 60 × 103 m 60 km h−1 = = = 16.7 m s−1. 1h 1 s × 60 s × 60 s Example 3: Conversion of standard prefixes Express 42 000 km in metres and 3.2 × 10−8 kg in µg. Solution: 42 000 km = 4.2 × 104 km = 4.2 × 104 × 103 m = 4.2 × 107 m. 3.2 × 10−8 kg = 3.2 × 10−8 × 103 g = 3.2 × 10−5 × 106 µg = 32 µg. 1.3 Scalar and vector quantities 1.3.1 The distinction between scalar and vector quantities Scalars are quantities that have magnitude only. Scalars are completely specified by the product of a positive or negative number and (usually) a unit. Examples of scalars are mass, temperature, distance and speed. Vectors are quantities that have magnitude and direction. The magnitude of a vector is positive. Examples of vectors are force, acceleration, displacement and velocity. Note that a vector (e.g. velocity) changes if either its magnitude or its direction changes. There are various ways of indicating that a given symbol is to represent a vector quantity. A force F could be written: F~ or F or F (i.e. heavy type or boldface). e In these notes vector quantities will be indicated using boldface symbols. 1.3.2 Addition of scalars Scalars simply add arithmetically. For example, adding several masses: 5 kg + 10 kg + 2 kg = 17 kg. 1.3.3 Addition of vectors by construction Vectors do not add arithmetically. One way of performing vector addition is with the aid of diagrams. When representing vector quantities in diagrams, arrows are used. It is understood that the length of the arrow is proportional to the magnitude of the vector and the direction of the arrow indicates the direction of the vector. 3 Example 4: Addition of vectors by construction Suppose a car travels from point A in an easterly direction for 10 km to point B, and then travels another 5 km in a direction 60° north of east to point C. Determine the total distance travelled and the displacement of the car. Solution: The total distance travelled is 15 km, which is the sum of the distances travelled from A to B and from B to C. C m 5k = s2 60° A s1 = 10 km B To find the displacement, we first draw AB 10 units long representing the displacement in the easterly direction. From the end point of AB, we draw BC 5 units long at 60° north of east to represent the displacement in the north easterly direction. The total displacement (or resultant) is then represented by AC. The magnitude of the displacement is the length of AC whilst the direction is indicated by the arrow. Measurement shows that the length of AC is 13.3 km and the direction is 19.1° north of east. Any number of vectors may be added together in a similar way. Consider the vectors shown in Figure 1a. To add these vectors, draw a scaled representation of any of the vectors (here the vector pointing west was chosen — the starting point is labelled ‘P’ in Figure 1b). Now take any of the other vectors and construct it from the endpoint of the previous vector. Continue in this way until all the vectors have been constructed. The resultant is then the vector drawn from the starting point to the end point of the last vector drawn (in this case the vector pointing south — the point labelled ‘Q’ in Figure 1b). 1 2 2 1 1 Q 1 Resultant 1 1 P (a) (b) Figure 1: Adding several vectors by construction. Note that, if in representing a number of vectors in a scale diagram the ‘finishing’ point coincides with the ‘starting’ point, then the resultant is zero. In such a case, if the vectors 4 are forces acting on a body, then the body will be in equilibrium (see also Section 3.5.2). 1.3.4 Components of a vector Finding the resultant of a number of vectors by construction is neither accurate nor conve- nient. In order to manipulate vectors algebraically, we need to introduce the concept of the components of a vector. An arbitrary vector A is shown in Figure 2. The x and y axes are drawn so that the origin coincides with the starting point of the vector A. (In this course we will only consider vectors in two dimensions.) It is evident from Figure 2 that the vector A may be obtained by adding the vectors Ax and Ay. Ax and Ay are known as the vector components of A. The lengths of the components Ax and Ay are written Ax and Ay and are called the scalar components of A. The scalar components Ax and Ay are positive if they point along the positive x and y axes respectively, and negative if they point along the negative x and y axes. +y A Ay θ +x Ax Figure 2: A vector Aand its vector components Ax and Ay. 1.3.5 The resolution (or resolving) of vectors If the magnitude and direction of a vector are known, we can find the components of the vector. This process is known as resolving a vector into its components. If the magnitude of the vector A in Figure 2 is A, and the direction of A is θ then Ax = A cos θ , (1) and Ay = A sin θ. (2) If the components Ax and Ay of a vector A are known, it is possible to find the magnitude and direction of A. The magnitude may be found using the theorem of Pythagoras: q A = A2x + A2y. (3) The direction is found using the tangent, thus   −1 Ay θ = tan. (4) Ax 5  The values of Ax and Ay depend on the orientation of the axes (and indeed on the chosen coordinate system). The choice of orientation is purely a matter of convenience. In problems where it is necessary to resolve a vector into its components, a suitable choice of orientation can lead to substantial simplification of the solution. An important example is an object (weight W) at rest on an inclined plane (see Section 3.5.4) 1.3.6 The addition of vectors by means of components y By B R A Ay β α x Ax Bx Figure 3: Addition of vectors using the components of the vectors. Consider two vectors A and B (see Figure 3), we wish to find the resultant R. It should be obvious from figure 3 that the scalar components of R are given by Rx = Ax + Bx and Ry = Ay + By , where Ax , Ay , and Bx , By are the scalar components of A and B respectively. The method of adding (or subtracting) vectors by adding (or subtracting) the components in this way may be extended to any number of vectors. We illustrate this by means of an example. Example 5: Addition of vectors by adding components Find the resultant of the three vectors in the drawing below. The vectors A and B are perpendicular to each other, and the magnitudes of A, B and C are 10, 20 and 15 units respectively. 6  A 240° B C Solution: Using Equation (1) and (2) we first find the x and y components of each vector. We place the origin of the coordinate system at the intersection of the vectors, with the positive x axis along the vector B. Thus Ax = 0 and Ay = 10, Bx = 20 and By = 0, and Cx = 15 cos 240° = −7.5 and Cy = 15 sin 240° = −13. To find the resultant R we first calculate the scalar components of R: Rx = Ax + Bx + Cx = 0 + 20 − 7.5 = 12.5 and Ry = Ay + By + Cy = 10 + 0 − 13 = −3. Using Equation (3) the magnitude of R is therefore q p R = Rx2 + Ry2 = 12.52 + (−3)2 = 12.85 The direction of R is obtained from Equation (4):   −1 −3 θ = tan = −13.5°. 12.5 1.3.7 The parallelogram of vectors B B R θ φ A A (a) (b) Figure 4: Parallelogram of vectors. 7 Consider two vectors A and B at an angle θ to each other as in Figure 4a. If we construct lines parallel and equal to A and B, we obtain a parallelogram with the resultant R as the diagonal (Figure 4b). The magnitudes of A and B and R are related by R2 = A2 + B 2 − 2AB cos φ, (5) where φ = 180° − θ. 2 Kinematics Kinematics is the study of the motion of bodies without reference to any forces. Forces will be considered in Section 3. To describe the motion of an object, we need to specify where the object is with respect to some reference point. A frame of reference combines a reference point with a set of directions. The Cartesian coordinate system is a frame of reference, it consists of an origin and a set of mutually perpendicular directions, familiar to us as the x, y and z axes. 2.1 Definitions Displacement is the change in position of an object in a specified time interval. Displacement is a vector quantity that has both magnitude and direction. In the diagram (1) alongside, an object is displaced from A to B. The ∆s displacement is independent of the choice of path, A B thus the paths indicated by (1) and (2) correspond to the same displacement even though the distance (2) covered in each case is clearly different. Note that the distance an object moves is always positive, whereas the displacement ∆s can be either positive or negative. We now consider a particle which starts from the origin O and moves in the xy plane along the dotted path shown in the figure below. y A ∆s B s1 s2 x O At time t1 the particle it is at A and at t2 it is at B. In going from A to B, the particle’s displacement is ∆s = s2 − s1. Let ∆t = t2 − t1. This is the time taken to go from A to B. The speed of an object is a measure of how fast it is moving. Average speed is the total distance travelled divided by the time taken. 8  distance AB along the dotted line average speed =. (6) ∆t The SI unit of speed is the metre per second (m s−1 ). Uniform speed is achieved when equal distances are covered in equal times. When the distance travelled is plotted against the time taken, uniform speed produces a straight line graph, with the slope of the graph equal to the speed (see Figure 5). distance distance time time (a) (b) Figure 5: Graphs of distance vs time for (a) uniform speed and (b) non-uniform speed. In order to describe the motion of an object, we need to know the direction in which it is moving as well as how fast it is moving. This information is given by the vector quantity velocity. The average velocity in a given direction is the total displacement in that direction divided by the total time taken. ∆s ∆v = , (7) ∆t where ∆t = tf − ti is the total time. Now suppose we let ∆t become very ‘small’. Then ∆s is small and we obtain the instan- taneous velocity. The instantaneous velocity is the velocity at any instant during the motion of an object. ∆s v = lim (8) ∆T →0 ∆t The acceleration of an object is a measure of how fast the velocity of an object is changing. The average acceleration of an object in a given direction is the change in velocity divided by the total time taken. Acceleration like velocity is a vector quantity: ∆v ∆a = , (9) ∆t where ∆v = vf − vi and ∆t = tf − ti. 9 The instantaneous acceleration is the acceleration at any instant during the motion of an object. The instantaneous acceleration is defined in a similar way to the instantaneous velocity. ∆v a = lim (10) ∆T →0 ∆t Uniform acceleration is obtained when equal changes of velocity occur in equal time intervals. The graph of velocity versus time for uniform acceleration is a straight line (see Figure 6). The slope of the graph gives the magnitude of the acceleration. It can be shown that the area under the graph of velocity versus time is equal to the total displacement. (See Example 9 and also Section 2.2.) velocity velocity vf vi time time tf (a) (b) Figure 6: Graphs of velocity vs time in one dimension for (a) uniform acceleration and (b) non–uniform acceleration. In future, when we speak about ‘the velocity’ or ‘the acceleration’ we will mean the in- stantaneous quantities. Example 6: Velocity and speed A jogger goes for her usual afternoon run. She leaves from her house and jogs a round trip of 10 km in 1 hour. What is her average speed and her average velocity after 1 hour? Solution: We determine the average speed from Equation (6). Hence distance 10 km average speed = = = 10 km h−1. time 1h Since the jogger starts and finishes at the same point, the total displacement is zero. Hence the average velocity is also zero. Example 7: Position-time graph A cyclist rides at constant velocity and then stops for lunch. After lunch she rides back to the place from where she started, at a different constant velocity. Find the average velocity in each of the regions A, B and C indicated on the graph below. 10 Solution: B 20 position (km) A C 10 0 0 1 2 3 4 time (h) The displacement is the change in position. We may choose the coordinate system so that the position is along the x axis. In segment A of the journey, the initial position si = 0 x̂ km and the final position is sf = 20 x̂ km. The displacement ∆s = 20−0 = 20 x̂ km. In segment B, si = sf so the displacement is zero, and in segment C, ∆s = sf −si = 0−20 = −20 x̂ km. The average velocities may be found using Equation (7). Thus ∆s 20 km Segment A ∆v = = = 20 x̂ km h−1 ∆t 1h ∆s 0 km Segment B ∆v = = = 0 x̂ km h−1 ∆t 1h ∆s −20 km Segment C ∆v = = = −10 x̂ km h−1 ∆t 2h Example 8: Average acceleration An aeroplane coming in to land is travelling at a speed of 100 m s−1. What is the average acceleration of the aeroplane if it comes to rest in a time of 10 s? Solution: Assume the aeroplane is travelling in the positive-x direction. The average acceleration is given by Equation (9). The initial and final velocities are vi = 100 x̂ m s−1 and vf = 0 x̂ m s−1 corresponding to times ti = 0 s and tf = 10 s. Thus ∆v 0 − 100 ∆a = = = −10 x̂ m s−2. ∆t 10 − 0 Note that the acceleration is negative. This indicates that the acceleration is in the oppo- site direction to the velocity and that the aeroplane is slowing down. An object that slows down is said to decelerate or experience a retardation. Example 9: Velocity-time graph A car accelerates uniformly from rest, drives at constant velocity for a short while before decelerating, and coming to rest at a traffic light. Find the acceleration in each of the segments A, B and C indicated on the graph below. Find also the total displacement during segments A, B and C. 11 Solution: B velocity (m s−1 ) 10 A C 5 0 0 2 4 6 8 time (s) We again choose the direction of motion along the x axis. The acceleration is given by the slope of the graph and the displacement by the area under the graph. In segment A the initial velocity vi = 0 x̂ m s−1 and the final velocity is vf = 10 x̂ m s−1. The change in velocity in segment A is therefore 10 m s−1. In segment B the slope of the graph is zero so the change in velocity is also zero, i.e. there is no acceleration. In the final segment, the change in velocity ∆v = vf − vi = 0 − 10 = −10 x̂ m s−1. The acceleration is in the opposite direction to the velocity, indicating that the car slows down. The average acceleration in each segment may be found using Equation (9). Thus ∆v 10 Segment A ∆a = = = 2.5 x̂ m s−2 ∆t 4 ∆v 0 Segment B ∆a = = = 0 x̂ m s−2 ∆t 2 ∆v −10 Segment C ∆a = = = −5 x̂ m s−2 ∆t 2 The displacement equals the area under the graph. We sum the area of triangle A, the rectangle B and the triangle C to obtain 1 s= 2 × 4 × 10 + 2 × 10 + 12 × 2 × 10 = 50 x̂ m. 2.2 Kinematic equations for uniform acceleration in one dimension In this section we derive the kinematic equations for an object travelling in a straight line with uniform acceleration. For motion in a straight line we drop the vector notation. For convenience, we assume that the object is initially located at the origin. Thus si = 0 at t = 0 and after a time t, the object is located at sf. To simplify the notation we write the elapsed time t = ∆t, the displacement s = ∆s = sf − si , the initial velocity u = vi and the final velocity (after a time t), v = vf. For uniform acceleration, the average acceleration is the same as the instantaneous accel- eration (∆a = a). From Equation (9) we obtain ∆v v−u =. a= t t Rearranging this with v the subject, we find v = u + at. (11) 12  Variable Symbol Unit time t s displacement s m initial velocity u m s−1 final velocity v m s−1 acceleration a m s−2 Table 4: Kinematic variables in one dimension. The graph of Equation (11) is a straight line graph with slope a and intercept u (see Figure 6a). From Equation (7) with ∆s = s, the average velocity is s v̄ =. (12) t If the acceleration is constant, the velocity increases or decreases at a constant rate, the average velocity is therefore midway between the initial and final velocities. That is v̄ = 21 (u + v). (13) Combining Equations (12) and (13) we find that s = 21 (u + v)t. (14) Equation (14) represents the area under the graph of velocity versus time. (See Figure 6a and also Example 9.) Substituting Equation (11) in Equation (14) we obtain s = ut + 12 at2. (15) Finally if we arrange Equation (11) with t as the subject and substitute this in Equation (14),   v−u 1 2
1 s = 2 (v + u) = v − u2 , a 2a and hence v 2 = u2 + 2as. (16) Example 10: Acceleration of a car from rest A car accelerates uniformly from rest to 36 km h−1 in 4 s. Calculate the magnitude of the acceleration and the distance covered during this 4 s interval. Solution: We have the following information: u = 0 v = 36 km h−1 = 10 m s−1 t = 4s a=? s=? We can use Equation (11) to find the acceleration. Thus v = u + at gives v−u 10 − 0 a= = = 2.5 m s−2. t 4 13  The displacement may be found using Equation (15). s = ut + 21 at2 = 0 + 12 × 2.5 × 42 = 20 m. Example 11: Distance covered by an accelerating spacecraft A spacecraft is travelling with a velocity of 3000 m s−1 when it fires its retrorockets and begins to slow down with an acceleration whose magnitude is 10 m s−2. Determine the velocity of the spacecraft when its displacement is 200 km relative to the point at which the retrorockets were fired. Solution: Since the spacecraft slows down, the acceleration is in the opposite direction to the velocity. If we take the direction of motion of the spacecraft as the positive direction, then the acceleration is negative. Data: u = 3000 m s−1 a = −10 m s−2 s = 200 000 m v=? We can use Equation (16) to find v. Hence v 2 = u2 + 2as = 30002 + 2 × (−10) × 200 000 = 5 000 000 m2 s−2. The velocity is therefore ±2236 m s−1. Both answers are acceptable: the negative value indicates a velocity in the opposite direction to the direction in which the spacecraft was initially moving. In other words, the retrorockets may have been fired long enough to slow the spacecraft to a halt and accelerate it in the opposite direction. 2.3 Motion under gravity The acceleration g due to gravity, close to the earth’s surface, is constant for all bodies at a given place in the absence of air resistance. The magnitude of the acceleration due to gravity on earth is approximately g ≃ 9.8 m s−2. For motion under gravity the equations of motions can be applied with the magnitude of the acceleration given by the value for g above. It must be remembered though that an object moving upwards suffers a retardation since the acceleration is in the opposite direction to the velocity. Example 12: Motion under gravity Calculate the time taken for a ball thrown vertically upwards, with an initial speed of 19.6 m s−1 , to return to its starting point, neglecting air resistance. Solution: Since the ball returns to its starting position, the displacement is zero. If we take the upward direction as positive, we must use a = −g = −9.8 m s−2. We can then use Equa- tion (15) with the following values: s=0 u = 19.6 m s−1 a = −9.8 m s−2 t=? 14  Thus s = ut + 12 at2 gives 0 = ut + 21 at2 , or −2u −2 × 19.6 t= = = 4 s. a −9.8 We can also solve this problem treating the upward and downward motions separately. Consider first the upward motion. Here s=? u = 19.6 m s−1 v = 0 m s−1 a = −9.8 m s−2 t=? Using v = u + at we find that for the upward motion t = 2 s. The displacement may now be calculated from s = ut + 21 at2 giving s = 19.6 × 2 + 12 × (−9.8) × 22 = 19.6 m. For the downward motion, we have s = −19.6 m u = 0 m s−1 v=? a = −9.8 m s−2 t=? We can use s = ut + 21 at2 again with the above values to obtain −19.6 = 0 + 12 × (−9.8) × t2 , which gives t = 2 s. The total time taken is therefore 4 s. 2.4 Strategies for problem solving 1. Draw a diagram to represent the situation. A simple ‘block’ drawing using arrows to indicate the directions of the various velocities and or accelerations is often all that is needed to gain a good understanding of a problem. 2. Choose a set of coordinate axes and decide which directions are to be called positive and negative. It is often convenient to place the origin at the place where an object starts its motion. In problems involving one dimension, the positive axis is usually chosen to go from left to right. For motion in two directions (see Section 2.5.1) the vertical direction is usually chosen as the y direction with ‘up’ being positive. To avoid confusion, do not change your decision in the middle of a calculation. 3. Write down the available values for the kinematic variables (s, u, v, a and t). Be careful to assign the appropriate sign depending on the choice of coordinate axes made in 2. 4. At least three of the kinematic variables should have values. Be sure to read the question carefully. There may be implied data like ‘an object is accelerated from rest’, in which case we may write u = 0. 5. Often a problem is divided in parts. For instance, a car may accelerate for a period of time, travel at constant velocity for a distance and then slow down. In such a case, divide the problem into parts, bearing in mind that the initial values for each part are given by the final values of the previous part. 15 2.5 Motion in a plane 2.5.1 Projectile motion In this section we will consider the applications of the kinematic equations to the motion of projectiles. If a body is projected at an angle to the vertical it travels along a curved path. At first sight it may therefore appear that the equations describing uniformly accelerated motion in a single line may not be applicable to this type of motion. The equations for straight line motion may be applied to the motion of a projectile if the initial velocity is resolved into vertical and horizontal components. When the initial velocity is resolved into is vertical and horizontal components, we may treat these components independently. The vertical motion is treated as uniformly accel- erated motion in a straight line under gravity and the horizontal motion is treated as uniform motion (constant velocity) in which there is no acceleration. At any time t in the projectile’s motion, the actual velocity is then the vector sum of the separate horizontal and vertical velocities appropriate to the time t. u sin θ u h θ Q P u cos θ θ R Figure 7: Motion of a projectile. Figure 7 represents the motion of a projectile launched with an initial velocity u at an angle θ to the horizontal. We locate the origin of our coordinate system at the point P, with the positive x axis to the right and the positive y axis upwards. The vertical component of the velocity is uy = u sin θ, the acceleration in the y direction is ay = −g and the net displacement in the y direction is zero since we assume the projectile is launched over horizontal ground. We can determine the time of flight using Equation (15). Hence s = ut + 21 at2 which gives 0 = uy t − 21 gt2. Using uy = u sin θ and rearranging, we find 2u sin θ t=. (17) g 16  The total distance covered in the horizontal direction is known as the range of the projectile (R in Figure 7). There is no acceleration in the horizontal direction (ax = 0), hence the range is given by sx = ux t or   2u sin θ R = ux t = (u cos θ) ×. g Using the trigonometric relation sin 2θ = 2 sin θ cos θ, we obtain u2 sin 2θ R=. (18) g The maximum range occurs when sin 2θ = 1, which gives 2θ = 90° or θ = 45°. The height reached by the projectile may also be determined from Equation (15), using half the total time found in Equation (17). Thus h = sy = uy t + 21 ay t2    2 u sin θ g u sin θ = (u sin θ) × − × g 2 g which gives u2 sin2 θ h=. (19) 2g Example 13: Projectile motion An aeroplane flying 1000 m above level ground at120 m s−1 drops a relief package over a remote area. Ignoring air resistance, calculate (a) how long the package takes to hit the ground, (b) the horizontal and vertical components of the package’s velocity, and hence (c) the speed and the angle at which the package hits the ground. Solution: ux = 120 m s−1 1000 m R θ We take the origin of our coordinate system at the point where the package was released, the positive y axis points upwards and the positive x axis to the right. 17  (a) The time the package takes to fall to the ground depends only on the vertical distance the package must fall. In the y direction, the initial velocity uy = 0 m s−1 , the displace- ment to the ground is sy = −1000 m and the acceleration is ay = −g = −9.8 m s−2. Putting u = 0 in Equation (15) gives s = 21 at2 , and hence r r 2s 2 × (−1000) t= = = 14.3 s. a −9.8 (b) There is no acceleration in the horizontal direction (since we are ignoring air resis- tance), hence the horizontal velocity is vx = 120 m s−1. The vertical velocity increases in the negative y direction. Putting uy = 0 in Equation (11) we have, vy = ay t = −gt = −9.8 × 14.3 = −140 m s−1. (c) The magnitude of the resultant velocity is obtained from Equation (3): q p v = vx2 + vy2 = 1202 + (−140)2 = 184 m s−1. The angle at which the package hits the ground is found using Equation (4):   −1 −140 θ = tan = −49°. 120 2.5.2 Uniform circular motion We now consider the problem of an object moving at constant speed v in a circle of radius r. Since the direction of the velocity is always changing, the object is, by definition, accelerating. It can be shown that the magnitude of this acceleration is v2 a= , (20) r and the direction is towards the centre of the circle. Since this acceleration is constant in magnitude but not in direction, we cannot use the constant-acceleration equations for circular motion. 3 Dynamics In Section 2 we studied kinematics. The motion of objects was described in terms of the observed quantities s, t, u, v, and a, but it was not considered what agent causes an object to move. The dynamics of an object is the study of the motion of an object under the action of forces. We are familiar with the notions of force and mass from everyday usage. A force might be described as a ‘push’ or a ‘pull’ and mass as a measure of the size of an object, or the quantity of matter. In the 17th century, Isaac Newton, building on the ideas of Galileo and others, developed laws and mathematical methods that enable us to define these concepts more rigorously, and treat the motions of objects under the action of forces in a mathematically consistent way. 18 3.1 Newton’s first law of motion Before Newton it was generally thought that a force was required to keep an object moving (see also Section 3.4.3). Newton instead postulated (on compelling experimental evidence) that an object would continue moving at constant speed in a straight line unless it was acted on by an unbalanced force. Newton’s first law of motion. A body will continue in a state of rest, or of constant speed along a straight line, unless compelled by an unbalanced force to change that state. Force is a vector quantity, and by the unbalanced or net force we mean the resultant force, or vector sum of all the forces acting on an object. Mathematically we write the net force X F= Fi , (21) P where is the mathematical symbol used to denote a sum of items indexed by the subscript i. Newton’s first law effectively sets the scene, because it defines the frame (or frames) of reference (i.e. coordinate system and clocks) with respect to which his remaining two laws of motion have meaning. The class of reference frames with respect to which Newton’s first law is valid are called inertial reference frames. When a net force acts on an object, the object’s velocity changes. The amount of change depends on the force as well as the mass of the object. If the same force acts on two objects of different mass, the more massive object will experience a smaller change in velocity. The tendency of an object with mass to resist a change in its state of motion is called the inertia of the object. The inertia of an object is measured by it’s mass. Inertia Inertia is that property of a body by virtue of which it tends to persist in a state of rest or uniform motion in a straight line. 3.2 Newton’s second law of motion Newton’s first law refers to a situation where there is no force. If a net force acts on an object, the object will change its state of motion according to Newton’s second law. Newton’s second law of motion If a net force acts on a body, the body will be accelerated; the magnitude of the acceleration is directly proportional to the magnitude of the net force and inversely proportional to the mass of the body, whilst the direction of the acceleration is in the direction of the net force. In mathematical terms Newton’s second law may be written as: F a∝ m or F ∝ ma, where a is the magnitude of the acceleration of the mass m produced by the net force F. In SI units, force is defined so as to make the constant of proportionality equal to 1. 19 Unit of force In the SI, the unit of force is the newton (N). The newton is defined as the net force which will give a mass of 1 kilogram an acceleration of 1 m s−2 in the direction of the force. Using the above definition, Newton’s second law takes the familiar form Fnet = ma (22) where F is the net force in newtons (N), m in kilograms and a in m s−2. 3.2.1 Free-body diagrams A free-body diagram is a diagram that represents an object and the forces acting on the object. The forces are represented by arrows (since force is a vector quantity) with length proportional to the magnitude of the force and the direction of the arrow indicating the direction of the force. A free-body diagram should always be drawn when a problem involves Newton’s second law. Example 14: Free-body diagram Suppose two people push a car along a horizontal road. One person applies a force of 300 N and the other a force of 200 N. The car has a mass of 1200 kg and the total force due to resistance is 400 N and acts in the opposite direction to the forces exerted by the people pushing. Draw a free-body diagram that shows the horizontal forces on the car and find the acceleration of the car. Solution: We need to use Newton’s second law. However we first need to find the resultant force. The free-body diagram below shows the car as a square and the arrows representing the forces. resistance humans 400 N 300 N 200 N Since the forces all act along one direction (the x direction say), the resultant force will also be in this direction, as will the acceleration. We choose the positive direction to the right, thus the net force is X F = Fi = +300 N + 200 N − 400 N = +100 N. The positive sign indicates that the force, and hence the acceleration, is in the direction we chose to be to the right. The acceleration may now be determined from Equation (22): F 100 N a= = = 0.083 m s−2. m 1200 kg 20 3.3 Newton’s third law Newton’s third law If body A exerts a force on body B, body B exerts a force equal in magnitude and opposite in direction on body A. Mathematically: FAB = −FBA (23) If two people pull on spring scales hooked together, no matter how hard each person tries to pull, the readings on the two scales will be the same. Likewise, if a ball is hit with a bat, there is not only a force exerted by the bat on the ball, but also a force, which is the same in magnitude but opposite in direction, exerted by the ball on the bat. To walk, a person exerts a force on the earth. Consistent with Newton’s third law, the earth exerts an oppositely directed force of equal magnitude on the person’s foot and causes the person to move forward. Note that in Newton’s third law and in all the above examples: the two forces act on different bodies. 3.4 Types of forces There are four known types of forces in Nature. 1. Gravitational force Gravitation is the weakest of the known forces. It is also the only force that is purely attractive. Weight is a gravitational force (see Section 3.4.1). 2. Electromagnetic force Most forces we encounter in daily life are electromagnetic forces, they arise from the interaction of the electrically charged particles that make up atoms and molecules. Im- portant examples are: Normal contact forces like collisions and throwing of objects. Tension forces such as surface tension and the tension in stretched strings. Compressive forces such as those in springs or in a hydraulic press. Frictional forces such as air drag on a skydiver and the grip on shoes (see Sec- tion 3.4.3). Many, if not most forces we encounter are combinations of forces. 3. Weak nuclear force A manifestation of the electromagnetic force that plays a role in the radioactive decay of atoms. 4. Strong nuclear force Plays a role in the interactions in nuclei of atoms. 21 3.4.1 The gravitational force and weight Every particle in the universe attracts every other particle. The magnitude of the force acting on each of two particles of mass m1 and m2 , separated by a distance r, is given by Gm1 m2 F = , (24) r2 where G = 6.67 × 10−11 N m2 kg−2 is the universal gravitational constant. By a particle we understand something so small that it may be regarded as a mathematical point. Although Equation (24) is for ‘point’ particles, it can be used with good accuracy when the masses are small compared to the distance separating them. For objects that are not particles, r in Equation (24) is the distance between the centres of the objects. The weight of an object is the gravitational force the earth exerts on it. The weight always acts downward, towards the centre of the earth. An object will not neces- sarily weigh the same on another planet. If the mass of an object is m and the acceleration due to gravity is g, then its weight on earth is given by GME m W = mg = , RE2 where ME and RE are the mass and radius of the earth respectively. Here we used Newton’s second law (Equation (22)) with the acceleration a = g. Since the mass m of the object appears on both sides of the equation, the acceleration due to gravity on a planet of mass M and radius R is given by GM g=. (25) R2 Example 15: The mass of the earth Calculate the mass of the earth given that the radius of the earth is RE = 6.38 × 106 m, g = 9.8 m s−2 and G = 6.67 × 10−11 N m2 kg−2. Solution: We rearrange Equation (25) with M the subject. Then
2 gRE2 9.8 m s−2 × 6.38 × 106 m ME = = G 6.67 × 10−11 N m2 kg−2 = 5.98 × 1024 kg. For distances greater than the earth’s radius (as in the case of a satellite), we must add the height above the earth’s surface to the radius. 22 Example 16: Acceleration due to gravity for a satellite in orbit Determine the acceleration due to gravity for a satellite in orbit 200 km above the surface of the earth. Use the same data given in Example 15. Solution: We again use Equation (25) with the following data: M = ME = 5.98 × 1024 kg R = RE + h = (6.38 × 106 + 200 × 103 ) m and G = 6.67 × 10−11 N m2 kg−2. Hence GM 6.67 × 10−11 × 5.98 × 1024 g= = = 9.2 m s−2. R2 (6.38 × 106 + 200 × 103 )2 3.4.2 The Normal force Normal in a mathematical sense means perpendicular. An object resting on a table for instance exerts a force equal to it’s weight on the surface of the table. By Newton’s third law, the table exerts an equal and opposite force on the object. The normal force FN is the force, or component of a force, that a surface exerts on an object in contact with it. The normal force N is often equal to the weight W of a body — but it is not necessarily so. Consider the following cases: CASE III CASE I CASE II N N b Wk N F θ W⊥ b b W θ W W W⊥ = W cos θ Wk = W sin θ N =W N = W − F cos θ N = W cos θ Example 17: Weight and the normal force in an elevator Find the apparent weight of a person whose mass is 60 kg in an elevator, when the elevator (a) is stationary, (b) accelerating upward at 2 m s−2 , (c) accelerating downward at 2 m s−2 , and (d) in free-fall. 23  N N N N =0 a = 0 m s−2 a = 2 m s−2 a = −2 m s−2 a = −9.8 m s−2 W W W W (a) (b) (c) (d) Solution: Imagine a person standing on a scale in the elevator. The apparent weight of the person is the normal force exerted by the scale on the man (the reading on the scale). Hence we must find the net force in each case and use Newton’s second law with the acceleration a given by the acceleration of the lift. In the diagrams above, W is the weight of the person (unchanged in each case) and N the normal force. The net force on the person is the vector sum of the normal force and the weight. If we regard the upward direction as positive, then F = −W + N, where W = mg = 60 × 9.8 = 588 N. Using Newton’s second law (F = ma), the normal force is N = ma + W. (a) The acceleration is zero, hence N = ma + W = 0 + 588 = 588 N. (b) Here a = 2 m s−2 , therefore N = ma + W = 60 × 2 + 588 = 120 + 588 = 708 N. (c) The acceleration is now a = −2 m s−2 , hence N = ma + W = 60 × (−2) + 588 = −120 + 588 = 468 N. (d) In free-fall, the lift (and person inside) is accelerating downwards at 9.8 m s−2. Thus a = −9.8 m s−2 and N = ma + W = 60 × (−9.8) + 588 = −588 + 588 = 0 N. 24 3.4.3 Friction The force that opposes the motion of one surface moving over another, with which it is in contact, is called the force of friction. Its magnitude depends on the materials of which the two surfaces are made, as well as on the force pressing them together. Some of the energy put into machines is transformed into heat energy because of the friction between moving parts. The heat may cause serious damage in addition to being wasteful. We cannot get rid of friction entirely but we can reduce it considerably by suitable choice of surfaces and by using lubrication. Consider a force F applied to a block B on a horizontal N surface S. If F is slowly increased from zero, the body remains at rest until F reaches a certain value, after which B F B accelerates in the direction of F (to the right in the b f diagram alongside). As the force F increases, the force of static friction fs increases and ‘adjusts itself’ to always exactly cancel W F. The maximum value of the force of static friction is fs (max). Experimentally we find that fs (max) = µs N , (26) where N is the normal force between the two surfaces in contact and µs , the coefficient of static friction, is a physical constant for the pair of surfaces in contact. When the applied force F is larger than µs N , there is a resultant force in the direction of F and the body will start to slip and accelerate in the direction of F. Once the block starts to slide, the frictional force drops below µs. We now talk about the force of kinetic (or sliding) friction fk. Experimentally it is found that fk = µk N , (27) where µk , the coefficient of kinetic friction, is usually slightly less than µs. It is an experimental fact that µs depends only on the nature of the sliding surfaces; it is independent of the area of contact or the relative speed of the surfaces. Example 18: Block on a horizontal surface with friction A block which has a mass of 100 kg rests on a rough horizontal floor. The coefficient of sliding friction between the block and the floor is 0.25. Calculate the horizontal force FH which would be required to move the block along the floor with constant velocity. Solution: N When the block moves with constant velocity a = 0. Therefore, since F = ma, and a = 0, the total un- FH balanced force F must be zero. Taking the positive Ff direction to the right, FH − Ff = 0. W Also, Ff = µN = mgµ, therefore FH = mgµ = 100 × 9.8 × 0.25 = 245 N. 25 Example 19: Block on an inclined plane with friction A body of mass 20 kg rests on a plane surface AB inclined at 10° to the horizontal, B being lower than A. The mass is connected by a light string which passes over a pulley at B, to another mass of 20 kg that hangs freely below B. If the coefficient of sliding friction between the body and the surface of the plane is 0.40, calculate the acceleration with which these bodies would move, and the tension in the string connecting them. Solution: N T A Ff Wk b T b 10° W⊥ W W B A B Consider the acceleration of both masses and choose the direction of motion as positive. The net force in the direction of motion is then W − T + T + Wk − F f , where the frictional force Ff = µN = µmg cos 10°. The total mass m = 2m = 40 kg. Newton’s second law then gives mg + mg sin 10° − 0.4mg cos 10° = 2ma which gives a = 3.82 m s−2. (Note that because the masses are equal, we do not need to know the mass for this calculation.) The tension T can be obtained by considering the motion of the mass B. Newton’s second law gives W − T = ma, hence T = W − ma = 20 × 9.8 − 20 × 3.82 = 119.6 N. A considerably smaller force called rolling friction is sufficient to keep one body moving against another if there are hard rollers or balls between two surfaces. It is important that the metal surfaces of roller or ball bearings that come into contact should be really hard. If one of the surfaces is not hard then the rolling friction might well be more than the sliding friction; it is for this reason that aircraft landing on soft snow fit skis in place of wheels. An important example of Newton’s first law is the case of an object falling through a medium. When it first starts to fall, it speeds up because its weight is bigger than the upthrust on it. But the dragging force on it increases as its velocity increases, and a stage can be reached when the upthrust plus dragging force (upwards) is as large as the weight (downwards). There is then no unbalanced force and the object continues to fall with the velocity it had reached — a constant velocity known as the terminal velocity. Example 20: Velocity time graph for a skydiver A skydiver jumps from an aeroplane. Sketch a velocity–time graph for the vertical motion of the skydiver indicating where the skydiver reaches terminal velocity, opens her parachute and lands. 26  velocity A B C D E time Solution: The origin represents the moment the skydiver jumps from the aeroplane. Her initial ve- locity in the vertical direction is zero at this point. Her acceleration on the other hand is equal to the acceleration due to gravity. Since we take the upward direction as positive, the acceleration is negative. (The slope of the graph gives the acceleration. If we draw a tangent to the curve at the origin, the slope of this line is negative). As the downward velocity increases, the magnitude of the acceleration decreases as the frictional drag due to the air increases. The force due to friction is in the opposite direction to the velocity. At A the skydiver reaches terminal velocity — here the force due to the air resistance is equal and opposite to the force due to gravity. The slope of the graph between A and B is zero and the velocity remains the same. At B she opens her parachute. Her acceleration here is positive and is a maximum as she opens her parachute, decreasing to zero as she again reaches terminal velocity at C. The segment between D and E is where she reaches the ground. 3.4.4 Tension If two people pull on either end of a rope there will be a certain tension in the rope. The force experienced by each person will be the same and will equal the tension in the rope. Figure 8 depicts the free-body diagram for this scenario. Each end of the rope provides the reaction −T T −T T b b Figure 8: Tension in a rope force on the person pulling at that end, as required by Newton’s third law. The force of the people pulling on the rope in effect gets transmitted through the rope. 3.5 Application of Newton’s laws In this section we discuss applications of Newton’s laws to various systems. In Section 3.5.2 we consider systems that are in equilibrium and in Section 3.5.3 some examples of non-equilibrium situations are discussed. 27 3.5.1 Guidelines for solving problems involving Newton’s laws The following points must be remembered when the relationship F = ma is used: 1. The mass in Newton’s second law (F = ma) represents the total mass accelerated by the net force. 2. F represents the net force in the direction of motion. If the accelerated mass is acted upon by a number of forces, the total net component of the forces in the direction of the motion must be calculated. The following systematic approach to problems in which the relationship F = ma has to be used may be useful: 1. Draw a diagram representing the general situation. 2. Select one object from the situation whose motion is to be analysed and draw a free- body diagram for this object. For this, the object is removed from its environment, together with all the forces exerted on it by bodies with which it interacts. 3. Select a convenient origin and orientation of the coordinate axes. 4. Write an expression for the net force. 5. Apply Newton’s second law. 3.5.2 Equilibrium applications Equilibrium An object is in equilibrium when it has zero acceleration. When the acceleration of an object is zero, the net force on the object is zero by Newton’s first law. Thus when an object is in equilibrium in two dimensions, we must have X Fx = 0 (28a) X Fy = 0. (28b) Examples of systems in equilibrium include a book lying on a table, a lamp hanging from a cord or a vehicle moving at constant velocity. Example 21: Tension in a cord, one dimension (equilibrium case) A lamp is suspended from the ceiling by a cord. If the lamp has a mass of 5 kg, determine the tension in the cord. Solution: T Since the system is in equilibrium, we use Equation (28b) to find the net force. T − W = 0 gives T = W = mg = 5 kg × 9.8 m s−2 = 49 N. W = mg 28 Example 22: Tension in a cord, two dimensions (equilibrium case) A lamp is suspended by three cords as depicted in the diagram below. The cord attached to the ceiling makes an angle of 60° with the ceiling and the cord attached to the wall is stretched horizontally. If the lamp has a mass of 5 kg, determine the tensions in the cords. 60° Solution: Since the forces (tensions in the cords) do not act in the A same direction, we will need to resolve the components of the forces in the x and y directions. First we construct a free-body diagram representing the forces acting at the intersection of the cords (the magnitudes of the forces are labelled A, B and C). C 60° The tension B is simply equal to the weight of the lamp. Hence B = W = mg = 5 × 9.8 = 49 N. Equating the x and y components of A, B and C according to Equations (21), we have B Ax + Bx + Cx = 0 and Ay + By + Cy = 0, where Ax = A cos 60° and Ay = A sin 60°, Bx = 0 and By = −B, Cx = −C and Cy = 0. Thus, equating first the y components: A sin 60° − 49 + 0 = 0, which gives A = 56.6 N. Equating the x components: A cos 60° + 0 − C = 0, which gives C = 28.3 N. 29 3.5.3 Non-equilibrium applications When an object is not in equilibrium, there are unbalanced forces acting on the object and hence the net force is non-zero. The approach to solving non-equilibrium problems is almost identical to the approach used to treat equilibrium problems. Instead of equating the net force to zero as in Equations (28), we must use Newton’s second law. Thus for an accelerating object in two dimensions X Fx = max (29a) X Fy = may. (29b) Example 23: The tension in a rope (non-equilibrium) Suppose the magnitude of the net force accelerating a car and trailer is F = 3000 N. The mass of the car is 1000 kg and the mass of the trailer is 500 kg. Determine the acceleration of the car and trailer, and the tension in the rope. Assume the mass of the rope is negligible. T −T 1000 kg 500 kg F Solution: The acceleration may be determined by applying Newton’s second law to the whole system (the car and trailer). We are given the net force and we know the total mass of the system. Thus F 3000 a= = = 2 m s−2. m 1000 + 500 Newton’s second law may also be applied to the trailer by itself. Here the net force in the horizontal direction is the tension T and the mass of the system is the mass of the trailer. The acceleration was found above. Hence F = T = ma = 500 × 2 = 1000 N. T Example 24: Objects connected by a rope 10 kg A block of mass 10 kg on a table is attached to a block of mass 30 kg by a rope passing over a pulley as shown T in the diagram alongside. Ignoring all frictional effects and assuming the pulley to be massless, find (a) the 30 kg acceleration of the two blocks and (b) the tension in the cord. (Take g = 10 m s−2.) 30 Solution: (a) The net force available to accelerate the system is due to the 30 kg mass (the forces on the 10 kg block in the y direction are equal and opposite). Hence F = W = −mg = −30 kg × 10 m s−2 = −300 N. The total mass of the system is the mass of the two blocks, mTOT = 10 kg + 30 kg = 40 kg. We can now use Newton’s second law to find the acceleration. F −300 N a= = = −7.5 m s−2. m 40 kg (b) the only unbalanced force on the 10 kg mass is due to the tension T in the rope. Using Newton’s second law with a = 7.5 m s−2 , since the 10 kg block is accelerated in the positive x direction, T = ma = 10 kg × 7.5 m s−2 = 75 N. 3.5.4 Motion on a smooth inclined plane When a block of mass m is placed on a smooth frictionless inclined plane, as shown in Figure 9, the block will be accelerated down the plane. The force in the direction of motion which gives rise to the acceleration of the mass is the component of the weight of the body down the plane. The weight W of the body (which is a force acting vertically downwards) can be resolved into components acting along the plane and perpendicular to the plane (see also Section 1.3.5). m W sin θ W cos θ θ W Figure 9: An object on a smooth inclined plane. If the plane is inclined at an angle θ to the horizontal, then the component of the weight parallel to the plane is Wk = W sin θ, and the component perpendicular to the plane is W⊥ = W cos θ. Example 25: Motion on an inclined plane Show that the acceleration of a sliding body down a frictionless plane is independent of the mass of the body. 31 Solution: Consider a body with mass m which is placed on a frictionless plate inclined at an angle θ to the horizontal as in Figure 9. The net force in the direction of the motion is Fk = W sin θ. The total mass accelerated is m. Using Newton’s second law, we have F = ma = Fk = W sin θ = mg sin θ, and hence a = g sin θ, which is independent of the mass m. 3.6 The centripetal force For an object moving at constant speed v in a circle of radius r the centripetal acceleration is given by v2 a= r (see Section 2.5.2). By Newton’s first law, if an object is accelerating, there must be a force acting on it. For circular motion this force is known as the centripetal force. The centripetal force may take different forms. For instance, for a car travelling in a circular path on a horizontal surface the centripetal force is the frictional force between the tyres and the surface; for a satellite in orbit around a planet the centripetal force is the gravitational force; for an object at the end of a string spun in a circular path, the tension in the string provides the centripetal force. Newton’s second law takes the form X v2 Fnet =m , (30) r P where the resultant force causing the circular motion Fnet is called the centripetal force. The centripetal force is directed towards the centre of rotation (in the same direction as the acceleration). 3.7 Satellites in circular orbits Consider an object of mass m (e.g. a satellite) moving in a circular orbit (radius r) at a constant speed v0 around the earth. Newton’s law of gravitation, the second law F = ma with a = v02 /r (see Equations (24), (22) and (20)), then gives  2 GmM v0 = m , r2 r or GM v02 =. (31) r 32 The period T of this object is the time to complete one full orbit. Time = circumference/speed, so 2πr T =. (32) v0 Squaring (32) and substituting from (31) gives 4π 2 r2 4π 2 r2 T2 = = v02 GM/r or 4π 2 r3 T2 =. GM This important result, which shows that the square of the satellite’s orbit is proportional to the cube of its orbital radius, is known as Kepler’s third law. You should remember that T 2 ∝ r3. (33) 3.7.1 Geostationary orbits As long ago as 1945 the scientist and science fic- tion writer Arthur C. Clarke had suggested that, if a satellite were placed above the equator at a height 3.6 × 104 km such that its orbital period was equal to the rota- tional period of the Earth, it would appear stationary from a point on the Earth’s surface. This character- istic would enable the satellite to provide permanent coverage of a given area. A geostationary orbit By substituting the value of T as 8.64 × 104 s (i.e. 42 hours) the value of R is found to be 4.23 × 107 m. Taking the Earth’s radius as 6.37 × 106 m, the height of the orbit is 3.59 × 107 m or 3.6 × 104 km. This is called a geostationary orbit. Clarke had further suggested that if three such satel- lites were equally spaced in geostationary positions above the equator then communication coverage of most of the world would be possible except for the World-wide coverage with polar regions. geostationary satellites 4 Hydrostatics The science of hydrostatics is the study of fluids at rest. For our purposes we take a fluid to be a liquid or a gas. 4.1 Density The density ρ of a substance is its mass per unit volume. 33  m ρ=. (34) V The density of a substance changes with both temperature and pressure. Therefore when the density of a substance is given, the temperature should also be given. The density of pure water at 4 ◦C is 1000 kg per cubic metre, i.e. 1000 kg m−3 (or 1 g cm−3 ). 4.2 Relative density The relative density (RD) of a substance is defined as the ratio of the density of the substance and the density of water. Thus density of substance RD =. (35) density of water (at same temp.) Relative density is a ratio and therefore has no units. Note that since density = mass/volume, then if we consider equal volumes of the substance and water, the expression for RD becomes mass of a given substance RD =. mass of an equal volume of water (at same temp.) Relative density is sometimes known as the specific gravity. It may be determined for both solids and liquids — see Practical Manual. Example 26: Relative density of aluminium. The density of aluminium is 2700 kg m−3. Find the relative density of aluminium given that the density of water is 1000 kg m−3. Solution: We can apply Equation (35) directly. Hence 2700 kg m−3 RD = = 2.7. 1000 kg m−3 Example 27: Relative density of a mixture. 10 cm3 of a liquid A whose relative density is 0.8 is mixed with 15 cm3 of a liquid B whose RD is 1.2. If there is no contraction on mixing, find the relative density of the mixture. Solution: To calculate the RD of the mixture, we need to find the density of the mixture. We are given the volume of each liquid, thus we first find the mass of each liquid. Suppose the density of water is ρ, and let the density and mass of liquid A and B be ρA , mA and ρB , mB respectively. For 10 cm3 of A ρA mA /10 RD = = = 0.8, ρ ρ which gives mA = 8ρ. 34  Similarly for liquid B ρB mB /15 RD = = = 1.2, ρ ρ which gives mB = 18ρ. The density of the liquid is therefore mA + mB 26ρ ρAB = =. 10 + 15 25 Finally, we can determine the density of the mixture: ρAB 26ρ RD = = = 1.04. ρ 25ρ 4.3 Pressure Pressure If a force F acts over an area A perpendicular to the force, then the pressure P is the force per unit area. F P =. (36) A The SI unit of pressure is the N m−2 or the pascal (1 Pa = 1 N m−2 ). 4.3.1 Summary of some laws of pressure in fluids at rest A fluid is a liquid or a gas. 1. The pressure at a depth h in a fluid at rest, due to the fluid itself, is hρg pascals where ρ is the density of the fluid. Let O be a point in the surface of a fluid and let P be a point in the fluid a distance h vertically below O. b O Imagine a cylinder of cross-sectional area A having OP as its axis, as in the diagram. The whole weight of the cylinder of fluid acts on the base around P. h Weight of cylinder = mass × g = volume × density × g = (A × h) × ρ × g Substituting the above expression for the weight in Equa- area b P tion (36), we find A (A × h) × ρ × g P = , A 35  which gives P = hρg. (37) The pressure hρg is sometimes called the ‘hydrostatic pressure’. So the absolute pres- sure at P equals (the pressure at the surface O) + (the hydrostatic pressure). If the pressure at O is the atmospheric pressure P0 , then P = P0 + hρg. (38) In other words, in a fluid at rest, the pressure increases linearly with distance below the surface, assuming that the density of the liquid remains constant throughout. 2. At any point in a liquid which is at rest the total pressure is the pressure on the surface of the liquid plus the pressure due to the liquid itself 3. At any two points in the same horizontal plane in any one liquid which is at rest, the pressures are the same. (Otherwise the liquid would flow.) 4. Pressure applied to the surface of a liquid is transmitted equally throughout the liquid in every direction. (This is called Pascal’s Law.) This principle is used in the Hy- draulic Press (or the Bramah Press – after Joseph Bramah (1748–1814), locksmith and inventor). 4.3.2 Gauge pressure As the name implies, this is the pressure recorded by a pressure gauge, and is frequently the difference between absolute pressure and atmospheric pressure. The equation ∆P = hρg indicates why it is convenient to refer to pressures by heads of liquid. A unit commonly used for gas pressures is the atmosphere (atm), which is defined to be 101 325 Pa. It is essentially equivalent to that exerted by 760 mm of mercury (mmHg) of specified density under standard gravity. Note that 1 mmHg exerts a pressure of 133 Pa. Pressure can be measured by various means. Two ways are described below. The simple barometer Atmospheric pressure P0 ‘balances’ the pressure due to the mercury column of height h. The pressure at A or B therefore equals the pressure at X. Hence P0 = hρg, h where ρ = 13 600 kg m−3 is the density of mercury. P0 P0 Standard atmospheric pressure corresponds to a mercury height of 0.76 m, i.e. X A B b b b P0 = 0.76 × 13 600 × 9.8 = 1.01 × 105 N m−2. This pressure is often stated simply as ‘76 centimetres of mer- cury’. 36 Another unit of pressure often used is the bar, it is related to pascals in the following way: 1 bar ≡ 105 N m−2 = 105 Pa = 100 kPa −3 2 −2 1 mbar = 10 bar ≡ 10 N m = 100 Pa. Garage pressure gauges read tyre pressure in bars in excess of atmospheric pressure. Standard atmospheric pressure is approximately 1013 millibars (mbar). Near the ground, atmospheric pressure decreases by about 1 cm of mercury per 120 m above sea level. Thus, in Pietermaritzburg (about 600 m above sea level), atmospheric pressures are usually about 71 cmHg, or 950 mbar. Example 28: A water barometer. Calculate the height of a water barometer corresponding to standard atmospheric pressure. Solution: We need to calculate the height of a column of water that will give a pressure of 1.01 × 105 N m−2. Using Equation (37) with ρ = 1000 kg m−3 : P0 1.01 × 105 N m−2 h= = = 10.3 m. ρg 1000 kg m−3 × 9.8 m s−2 Example 29: Pressure due to a column of