Chapter 4 Forces in Action 4.1 Force, Mass and Weight 4.2 Centre of Mass 4.3 Free-Body Diagrams 4.4 Drag and Terminal Velocity 4.5 Moments and Equilibrium 4.6 Couples and Torques 4.7 Triangle of Forces 4.8 Density and Pressure 4.9 p = hïēg and Archimedes’ Principle 1 4.1 FORCE, MASS & WEIGHT (specification reference 3.2.1) Mass is one of the fundamental properties of nature, along with charge and spin (which is used in quantum Physics). As of 2018 the unit of mass – the kilogram, is no-longer defined in relation to an ‘artefact’, the IPK, held in a vault near Paris (as is stated in the textbook published in 2015) but is defined by a constant of nature. The Kilogram like all other S.I. base units is now defined in terms of a universal physical constant. The kilogram is defined by Planck’s constant as measured by a Kibble Balance You do not need to know about these but can read about it (and other interesting experiments) in the articles linked below: http://www.nature.com/news/frontier-experiments-tough-science-1.9723 https://www.youtube.com/watch?v=Oo0jm1PPRuo Force, mass and acceleration At IGCSE, you have learned Newton’s second law can be expressed as the equation: ð = ðð At A level, we will redefine Newton’s second law in terms of rate of change of momentum. However, in the case where mass is constant (most of the time), the above equation still holds true. Rearranging the above equation as ð = ð/ð implies that acceleration is directly proportional to force and inversely proportional to mass. This can be represented graphically as follows: 2 You may recall this from an experiment we did at iGCSE with the air track How can we change the axes on the acceleration vs mass graph in order to create a straight line going through the origin? Remember a force of 1 Newton will give a mass of 1 Kg an acceleration of 1 ms-2. Coincidentally, an apple has an approximate weight force of 1 Newton. A typical problem in this area is to calculate an acceleration using SUVAT and then use F=ma to calculate a force: Worked example A car of mass 1,500Kg decelerates from 50kmh-1 to rest in 5 seconds, calculate the deceleration and thus the force of the brakes on the car. ð = (ðĢ − ðĒ)/ðĄ V = (50 x 1000) / (60 x 60) = 14ms-1, u=0, t=5 Therefore a = (14 - 0) / 5 = 2.8 ms-2 As F = ma, F = 2.8 x 1500 = 4200 Newtons Example to try A force of 100N brings a bullet of mass 10g to rest in 0.02 seconds. Calculate the initial velocity of the bullet. 3 Mass and weight Gravity is more properly considered as a field. Fields are areas where forces act on one of the fundamental properties of nature. Gravitational fields cause forces to act on masses (electrical fields act on charge). All objects with a mass have a gravitational field around them, the greater the mass the greater the gravitational field around it. Near celestial bodies, we term the force due to the gravitational field weight. The relationship between mass and weight is given by the following equation: ðū = ðð Weight is a force and thus a vector with the direction always pointing to the centre of mass between the two objects. If one body is far more massive than the other (e.g the earth and a person) the centre can be assumed to be the centre of the massive object. This definition of weight gives ‘g’ the units N/Kg and these are the units of gravity considered as a field. As the masses in F=ma and W=mg are the same, we can show that gravity causes the same acceleration to all masses: F=ma=mg. Therefore a=g This is the acceleration of free fall that you studied earlier and means that we can also give ‘g’ as an acceleration with the units ms-2. Now try summary questions 1-6 page 48. Mass and Weight – A-level Physics online: https://www.youtube.com/watch?v=r3lvzIKKaJg The difference between mass and weight – Veritasium: https://www.youtube.com/watch?v=_Z0X0yE8Ioc Only go here if you really want to blow your mind! What the heck is mass? – Science Asylum: https://www.youtube.com/watch?v=XkPudRiWspc 4 4.2 CENTRE OF MASS (specification reference 3.2.3) Weight acts on all masses. So far, we have dealt with point masses. On an object composed of atoms, clearly weight acts on every single atom. Where weight acts If we apply a force to an object at any point where the line of action of the force does not go through the centre of mass, it will both move in a straight line and rotate. If we hang an object then it has to hang with the centre of its mass directly below the point of rotation (pivot). The object behaves as if all its mass is acting at a single point: the centre of mass or gravity. The centre of mass can be defined as that point through which the entire weight of an object appears to act. This makes the mathematics much easier and is a weighted average of the location of all the masses. For instance, in the above diagram, the centre of mass (x) is closer to B as B has the greater mass. Centre of mass of regular objects This can be found at the intersection of lines of symmetry of the object. 5 Finding the centre of mass of an irregular object If we suspend an object freely, it will hang with its centre of mass directly below the pivot. If we repeat this process more than once the intersection of these lines will give the exact point of the centre of mass. Now carry out the following experiment to determine the centre of mass of an irregular shape: ï· ï· ï· ï· Cut an irregular shape out of card. Make a small hole in the card. Suspend it freely by the hole from a suitable pivot. Use the plumb lines (masses attached to thread) provided to draw a vertical line beneath the pivot. The centre of mass/gravity must lie on this line. ï· Repeat this with two other points in the card. The three lines should all intersect at the centre of mass. Balance the shape on your finger (or a pin) to confirm that you have correctly found the centre of mass. Try and do the above accurately and minimising experimental error. Write down sources of experimental error and how you have tried to minimise them. Keep the shapes and stick them in your notes (do 2 per pair). Now try summary questions 1-5 page 50 Centre of mass and gravity – A-level Physics online: https://www.youtube.com/watch?v=7fCb3dPRAnQ&list=PLNHM5YgkZWDFrsTdbG1YU6xXzb_hIy 7Nd&index=3&app=desktop Measuring C.O.M. in practise – VTPhysics: https://www.youtube.com/watch?v=vDqcEIdhJ_8 6 4.3 FREE-BODY DIAGRAMS (specification reference 3.2.1) Free body diagrams are representations of all the forces acting on an object. Some important forces Weight This is the force due to a gravitational field (=mg). It is always attractive and acts towards the centre of mass of the objects. Friction This force opposes motion and is due to two surfaces rubbing against one another. Drag This is a form of friction when a solid body is travelling through a fluid. It always acts in the opposite direction to the velocity. This is the force within a stretched cable or rope. Tension 7 Upthrust An upward buoyancy force acting on an object in a fluid (liquid or gas). Normal contact force A force that acts at 90 degrees to a surface on an object resting upon it. Representing forces We have already done this a lot in the vectors topic. ï· ï· Each force is represented with a line with an arrow on it. Each line is drawn to scale. Solving force problems ï· ï· ï· Draw a free body diagram Choose two convenient directions at 90 degrees to resolve the forces into. Generally, this will be ‘up/down’ and ‘left/right’. However, for ramp problems, it will generally be parallel and perpendicular to the slope (as the normal reaction force is perpendicular to the slope and the motion/friction is parallel to the slope). If the system is in equilibrium (not moving or moving at constant speed), the forces have to be balanced in both directions that you have chosen. If it is only accelerating in one direction, the forces have to be in equilibrium in the other direction. Now try summary questions 1-4 on page 53. Free Body Diagrams and Objects on an Inclined Plane - A Level Physics online: https://www.youtube.com/watch?v=IBDk8Opqn8E 8 4.4 DRAG AND TERMINAL VELOCITY (specification reference 3.2.2) Falling without drag As you already know, a falling object will be acted upon by a gravitational force (weight) given by the product of its mass and the gravitational field (W=mg). The free force diagram will merely be one force pulling towards the centre of the earth (on planet earth). The acceleration does not depend on the mass. The velocity/time graph will be a straight line going through the origin and the distance/time graph will be a curve with increasing gradient (a parabola). 9 Adding in drag Drag is a form of friction which acts in the opposite direction to the velocity of the object. It depends on two factors: ï· ï· Velocity Surface area It is important to remember that it does not depend on mass so, when we use F=ma, mass does not cancel out and heavier objects fall faster than lighter objects. In general, the drag is proportional to the velocity squared. Fαv2 F=kv2 (f=ma=mg-kv2 ï a=g-kv2/m. As you can see, as m increases, the second term in this expression decreases and the acceleration increases). Worked example: If the drag on a falling person is 50N when he is falling at 10ms -1, what is it at 50ms-1? 1) Calculate K K = F/v2 K = 50/102 = 50/100 = 0.5 2) Substitute the new velocity in the equation that you have derived. F = 0.5v2 F = 0.5 x 502 = 0.5 x 2500 = 1250 N Example for you to do: If the drag on a parachutist is 200 N when he is falling at 5 ms-1, what is it at 10 ms-1? Draw below the free body diagram of a body falling under gravity with air resistance (drag)? 10 Velocity of person falling under gravity with drag Initially, when the body has just been dropped (zero velocity), its acceleration will be equal to the acceleration of free fall (9.81 ms-2). As it falls, the drag force will increase proportionally with the square of the velocity. The resultant force will thus decrease and, as F=ma (constant mass), its acceleration will also decrease. Remember, the body is still increasing in velocity but at a slower rate. Ultimately, the drag force will become equal to the weight of the body and there will be no resultant force. At this time, the body will fall at a constant velocity, termed the terminal velocity. The terminal velocity will depend on two factors: ï· ï· Surface area Mass Graphs of a body falling with drag 11 The displacement time graph is less intuitive than the velocity time graph. It is important to remember that the gradient of this graph represents the velocity. This continues to increase (steeper line) until the body has reached terminal velocity. It then becomes a straight line with the gradient representing the terminal velocity. Example: Dr Price is dropped out of an aircraft. If the drag when she is travelling at 10 ms-1 is 20 N, calculate her terminal velocity (Dr Price has a mass of 65 Kg – optimistically speaking!). Hints: Firstly, calculate k (the drag constant) as in the examples above. Use this k in the formula and calculate the velocity at which the drag is equal to her weight. When drag is an advantage and when it is a disadvantage Add force arrows to the adjacent picture. Explain below why cars have a top speed. Why is the above car the shape that it is? What is the name for this? ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ 12 Add force arrows to the adjacent picture. Explain below why people use parachutes. Draw the velocity time graph of the parachutist from the time he jumps from the plane (freefalling first) to just after he releases his parachute. ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ Now try summary questions 1-5 on page 56. Drag, Air Resistance and Water Resistance - A Level Physics online: https://www.youtube.com/watch?v=x-ly8BXpPxE Terminal Velocity isn't just for falling things - A Level Physics online: https://www.youtube.com/watch?v=MNfANj2vkUU 13 4.5 MOMENTS AND EQUILIBRIUM (specification reference 3.2.3) Forces can cause objects to accelerate and also to rotate. The ability of a force to cause an object to rotate is called the moment of the force. The moment of a force is defined as the force multiplied by the perpendicular distance of the line of action of the force from the axis or point of rotation. It is important, when you take moments, to consider (and clearly state) the axis or point about which you are taking moments. The SI unit for the moment of a force is the Newton Metre (Nm). Moments cause objects to rotate in either a clockwise or anticlockwise direction. They thus have directions and are vectors. Meaning of perpendicular distance If you look at the above two examples, you can see that the perpendicular distance is that distance found by extending the force arrow until you can draw a line at right angles to it which joins up to the pivot. In the hinge example, the perpendicular distance d, will be equal to the length of the door multiplied by sin(30). Hopefully you can see that these are equal to the perpendicular component of the force multiplied by the actual distance from the pivot. Both these methods are equivalent, but I find the second one far more intuitive and comes naturally from what you have learned in the vectors topic. 14 Examples of moments Calculate the following moments: Distance =1.5m Angle =30 degrees Force=10N The principle of moments For a body in equilibrium (not accelerating or turning), there is no net moment about any point. This means that the sum of the clockwise moments must be equal to the sum of the anticlockwise moments. You already know that, for a body in equilibrium, there can be no resultant force on the object. The above two principles allow you to solve many problems. If a body is in equilibrium, the moments about any point will sum to zero. However, to solve problems easily, we take advantage of the fact that the moment of a force at zero distance from this point is obviously zero. We thus tend to choose the point where the unknown force acts, which allows us to solve a simple equation in one variable. Worked example 1 If the above system is in equilibrium, clockwise moments have to be equal to anticlockwise moments. 15 Clockwise moment=force x distance = 8F Anticlockwise moment = force x distance =12 x 20 = 240 Nm 8F = 240 F = 30 N Worked example 2 A wooden pole is held in an upright position by two wires in tension. These forces are shown in the diagram below 600N 2m T 35o 3.5m Calculate the tension T in the wire. Taking moments about the bottom of the pole: Clockwise moment = 600 x 5.5 = 3300 Nm Anticlockwise moment = force x perpendicular distance = T x 3.5sin35 = 2T If the system is in equilibrium 2T = 3300 T = 1150N. What other forces need to act on this pole if it is in equilibrium? Study the examples in the textbook and attempt summary questions 1-4 on page 59 The Principle of Moments - A Level Physics online: https://www.youtube.com/watch?v=7x6SpYdkjC4 16 4.6 COUPLES AND TORQUES (specification reference 3.2.3) As you know by now, a force can produce a combination of translational and rotational movement. In order to produce purely rotational movement, the forces themselves must be balanced, but applied so that the line of action of the force does not go through the centre of mass. One way of doing this is two equal and opposite forces on different lines. This is called a couple. Calculate the moment of the upward force about the axis of rotation (goes through centre of circle): Calculate the moment of the downward force about the axis of rotation: Thus, calculate the total moment of the two forces: 17 The moment of a couple is known as a torque. A torque produces a rotational acceleration. The torque of a couple is defined as the product of one of the forces and the perpendicular separation between the forces. Moment of a couple = torque = Fd Example F=50N d=20cm Two parallel 50N forces act at an angle to the horizontal of 60 degrees, across a circle of diameter 20cm. Calculate the torque of the couple. Study the examples in the textbook and attempt summary questions 1-4 on page 61 Torque of a Couple - A Level Physics online: https://www.youtube.com/watch?v=_LbeGj68atU&feature=emb_logo 18 4.7 TRIANGLE OF FORCES (specification reference 3.2.3) Conditions for equilibrium An object is in equilibrium when the resultant force on it is zero. The pendulum bob in the figure will not stay in its current position if you remove your finger, because the tension, T, and weight, W, have a resultant force, R, that will return the string to a vertical position. When the force from the finger, F, is equal and opposite to the resultant, R, then the bob is in equilibrium and the three forces, T, F, and W, form a triangle of forces. If three forces are acting on an object and it is in equilibrium, then the forces will always form a closed triangle of forces. By drawing the triangle of forces, you can find the magnitude and direction of one of the forces if you know the other two. In solving problems with triangles, remember the angles in a triangle add up to 180°. In a rightangled triangle this means the other two angles add up to 90°. 19 Worked example Question A box weighing 100 N is on a slope with angle 30° to the horizontal. Friction prevents the box sliding down the slope. a Sketch a triangle of forces for the forces acting on the box. Answer a Step 1 First sketch the situation: the box on the slope of 30° to the horizontal, the weight of the box, the normal contact force perpendicular to the slope, and the friction parallel to and up the slope. Decide how to draw the triangle of forces. The three forces must be in the directions shown in the diagram below and you must put them together in an order so that they all follow round, in a clockwise, or anti-clockwise, direction. You cannot have one going in the opposite direction. Sketch the triangle and mark the angle or angles you know. Now try summary questions 1-4 on page 64 Resolving Forces – Physics online: https://www.youtube.com/watch?v=2kHCvtTjOJs Triangle of Forces – Maths videos Australia: https://www.youtube.com/watch?v=5WE36XM5knc 20 4.8 DENSITY AND PRESSURE (specification reference 3.2.4) The density of a substance is defined as its mass per unit volume. ð = ð/ð Where m is mass and V is volume. The SI unit of density is Kgm-3. Volumes of regular solids Worked example: A 60 cm3 volume of steel has a mass of 0.48 kg. Calculate the density of steel. Firstly, we need to convert cm3 into m3. We do this by dividing by (100)3 = 1x106 Thus: Volume = 60/106 = 6x10-5 m3 We then use our formula ð = ð/ð Density = (4.8 x 10-1) / (6 x 10-5) = 8 x 103 Kgm-3 21 Example: A lady’s necklace has a hanging gold ball, which is a sphere of radius 2 cm. If the density of gold is 19300 kgm-3 and the cost of gold is £25000 per kilogram, then what is the cost of the gold ball? Experimental determination of density To determine the density of a substance, one needs to measure both a mass and a volume. The mass can be determined directly using a mass balance. For regular solids, the volume can be calculated by using a ruler or digital callipers, or a micro-meter. The volume of irregular solids can be determined by displacement. If the water level increases by 30 ml then the solid’s volume must be 30 ml 22 Pressure in solids Pressure is the normal force exerted per unit cross sectional area. ð· = ð/ðĻ Force is in Newtons and area in m2, so the units of pressure are Nm-2 or Pascals. Example: Estimate the pressure exerted by a woman in high heels and an elephant? Estimates are rough calculations, not guesses, so the numbers used should be realistic. Now try summary questions 1-6, page 66 Density – Physics online: https://www.youtube.com/watch?v=vodt8kgshbU Pressure – Physics online: https://www.youtube.com/watch?v=4Vn07qFWcO0 23 4.9 p = hïēg AND ARCHIMEDES’ PRINCIPLE (specification reference 3.2.4) Figure 1 Figure 2 As fluids (liquids and gases) have a weight, they exert a force over an area and thus exert a pressure. As they can flow, however, the pressure is exerted equally in all directions. How pressure varies with depth Consider the diagram above: P = F/A The force exerted by the column of fluid is its weight. W = mg But mass is equal to volume multiplied by density: W = ρVg The volume of the column is the cross-sectional area multiplied by the height of the column W = ρghA Pressure = Force/Area = ρghA/A = ρgh. 24 So, the pressure exerted by a fluid at a depth h only depends upon its density and the height of the column. If you look at figure 1, this can easily be shown to be true by puncturing a canister at different depths and seeing how far the water travels. Example: Calculate the pressure exerted on a swimmer 5m underwater. The density of water is approximately 1000 kgm-3. What is this pressure in additional to? Atmospheric pressure As can be seen in figure 2, our atmosphere is a fluid and thus exerts a pressure upon us. At sea level, this is approximately 100 KPa. Small variations in this pressure are the cause of our weather. Our atmosphere has an average height of 12 Km. Calculate the average density of air in the atmosphere: Calculate how far one would need to swim underwater to be subject to a total of twice atmospheric pressure: Upthrust and Archimedes principle Upthrust is the buoyancy force experienced by a body within a fluid. If we consider the above block in a fluid its upper surface experiences a downward force of F1=PA where the pressure is equal to ρg(h2-h). So, the force is ρgA(h2-h). On the other-hand the downward surface experiences (by the same method) an upward force of F2 = ρgh2A. 25 The block thus experiences a net upward force of F2-F1 = ρgAh2-ρgA(h2-h) = ρgAh But as Ah is the volume of the block, this is equivalent to ρgV or the weight of the fluid that it has displaced! This is Archimedes principle. Archimedes principle states that the upthrust exerted on a body immersed in a fluid, whether fully or partially submerged, is equal to the weight of the fluid that the body displaces. An object will sink if the upthrust is less than the weight. For a body that floats, the upthrust (and thus the weight) must be equal to the weight of fluid displaced: Example: An iceberg has a density of 900 kgm-3 whereas water has a density of 1000 kgm-3. How much of an iceberg of volume 40000 m3 is above the water. (i) Calculate the weight of the Iceberg. (ii) Calculate the volume of water it will need to displace to be equal to its weight. (iii) Subtract this from the initial volume to see how much is above the water. Now try questions 1-5 page 69 Pressure in Liquids and Fluids – Physics online: https://www.youtube.com/watch?v=Eo0sGKYDz-o Upthrust and Archimedes' Principle – Physics online: https://www.youtube.com/watch?v=o7EbeVyqdko 26
