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Name: Form: Teacher: 1 TYPES OF FORCE Generally a force is a push, pull, twist or squeeze. Drag, Tension, Friction, Weight, Gravity, Thrust are applied to particular forces. For example, Thrust is used for the driving force provided by the engine of a car. However, despite all the names and situations where forces occur, outside the nucleus of atoms there are only 3 types of force! * Gravitational Force is a natural phenomenon by which all objects with mass attract each other, and is one of the fundamental forces of physics. In everyday life, gravitation is most commonly thought of as the agency that gives objects weight. It is responsible for keeping the Earth and the other planets in their orbits around the Sun; for keeping the Moon in its orbit around the Earth, for the formation of tides; for convection (by which hot fluids rise); for heating the interiors of forming stars and planets to very high temperatures; and for various other phenomena that we observe. Gravitation is also the reason for the very existence of the Earth, the Sun, and most macroscopic objects in the universe; without it, matter would not have joined into these large masses and life, as we know it, would not exist. Modern physics describes gravitation using the general theory of relativity, but the much simpler Newton's law of universal gravitation provides an excellent approximation in most cases. "Gravitation" is a general term describing the attractive influence that all objects with mass exert on each other. The gravitational force between an object and the Earth is weight. * Electrical Force. Matter is made of atoms. An atom is basically composed of three different components - electrons, protons, and neutrons. An electron can be removed easily from an atom. Normally, an atom is electrically neutral, which means that there are equal numbers of protons and electrons. Positive charge of protons is balanced by negative charge of electrons. It has no net electrical charge. When atoms gain or lose electrons, they are called "ions." 2 The electrical force is responsible for all interactions between objects. When atoms collide, they exert an electrical force on each other and as a result of electrical attraction between them they can chemically bond together. When the atoms in your foot meet those in the ground, when you walk for example, both sets of atoms exert an electrical force on each other – this force is sensed by your foot – “you have touched the ground”. All contact forces are electrical forces and all objects are affected by these contact forces. * Magnetic Force is the force between 2 magnetic objects. At an atomic level it is a force between moving charges. It is only important in examples using magnetised objects. 3 Newton's laws of motion Newton developed Galileo's ideas about motion and stated three laws (named after him). The laws are very nearly correct under all circumstances although more exact laws are required under certain extreme conditions (in particular, as the speed of an object approaches the velocity of light). Newton’s First Law: A body remains at rest or continues with uniform velocity (constant speed in a straight line) unless acted upon by a resultant force. The first law enables a force to be defined as that which changes the motion of a body (speeds the body up, slows the body down, causes the body to change direction). Another way of stating this is that a resultant force causes a body to accelerate. The first law also implies that matter has a reluctance to change its state of rest or motion. This property is called the inertia of the body. The mass of a body is a measure of its inertia. 4 Newton's second law Newton's second law indicates what happens when a resultant force is applied to a body. Experiments show that when the resultant force applied to a body of constant mass is changed then the acceleration produced is proportional to the resultant force. Experiments also show that when the resultant force is kept constant and the mass of the body changed then acceleration is inversely proportional to the mass. These experimental results are summarised by the equation which mathematically expresses the second law: aαF a α 1/m a α F/m or F α ma F = constant x ma Resultant force = Mass x Acceleration F = m x a Where F means resultant force Definition of the Newton A force of 1 Newton is the force which when applied to a body of mass 1 kg produces an acceleration of 1 ms-2. Combining the definition and the above equation, k = 1 Units In this equation the following units must be used: Mass must be expressed in kg, Acceleration will always be in ms-2 and Forces will always be in N. It is also important to realise that the net or resultant force is always in the same direction as the acceleration. For example for a car increasing in speed, the resultant force and acceleration are both forward. For a car decelerating right there is an acceleration left and a net force left. It is important to realise that in many questions, F = ma is used together with the equations of motion! 5 When Einstein came up with his special theory of relativity, which stated, amongst other things, that at speeds approaching light speed, objects increase in mass, it became clear that F = ma was invalid for high speeds. The difference between Newton’s theory and Einstein’s only amounts to a 1% discrepancy for an object travelling 1/10th speed of light. Nuclear physicists use Einstein’s theory, but the calculations used to launch NASA rockets were based on Newtons Laws. Task: Research in more detail, the limitations of F = ma. The use of theories, models and ideas to develop and modify scientific explanations can be discussed. Free body diagrams A free body diagram is one, which shows all the forces acting on a body in a particular situation. A free body diagram should show the forces acting on the body you are interested in and nothing else. The object can be drawn realistically or more commonly as a single point, with force arrows pointing out wherever possible. Normal reaction force Drag/Friction Forward push force Weight Normal Force: This is the reaction force at 90 degrees to a surface, pushing up on an object. 6 Draw free body diagrams for the following: (a) a person sitting on a stool (model the person as a particle). (b) a block of wood on an inclined plane which has a rough surface. (c) a cart being pulled along a level road by a tractor. (d) a tractor pulling a cart along a level road. In each case name the forces and state how they arise. Teaching Examples (a) A runner in a sprint race accelerates uniformly to reach 9 ms-1 in 3 s from the start of the race. If her mass is 50 kg, what force must she exert in order to do this? (b) An aeroplane lands with a velocity of 55 ms-1. Reverse thrust from the engines is used to slow it down to a velocity of 25 ms-1, in a time of 3 s. If the mass of the aeroplane is 3 x 104 Kg and the plane decelerates uniformly, what is the size of the reverse thrust supplied by the engines? 7 Year 12 Newton's Second Law Questions 1. A resultant force of 1.0 N acts on an initially stationary mass of 1.0 kg for a time of 1.0 s. How far will it move it? 2. Forces of 12 N and 16 N act on a body of mass 4.0 kg. If the forces are perpendicular to each other calculate the size of the acceleration of the body. 3. The speed of a locust at take off is 3.4 ms-1. If the body is accelerated through a distance of 40 mm and the mass of the locust is 3.0 g calculate the average force exerted by the ground on the hind legs of the locust. 4. A lift has a mass of 1.2 x 103kg. Draw a free-body diagram showing all the forces acting on the lift. Calculate the tension in the supporting cable when the lift is: (a) descending at uniform velocity (b) descending with downward acceleration of 2ms-2 (c) at rest (d) ascending with upward acceleration 2ms-2 (e) ascending with uniform speed. 5. A force of 8.0 N gives a mass m1 an acceleration of 12 ms-2, and a mass m2 an acceleration of 48 ms-2. Calculate the acceleration that the force would give the two masses when they are attached together. 8 6. A lift of mass 490kg is raised by means of a cable and carries a person of mass 84kg. (a) Draw the free body diagrams for (i) the lift. (ii) the person. Show on your diagrams the names and directions of the forces acting. (b) If the lift is moving upward with an acceleration of 0.20 ms-2, calculate the force on the person due to the floor of the lift. 7. A force of 30 N halves a body's velocity in 9.0m. If the mass of the body is 5.0Kg, calculate the original velocity and the time for which the force acts. 8. A spring balance carrying a mass of 20.0Kg in a lift registered 250 N. What was the acceleration of the lift? Draw free-body diagram from which you can calculate the balance reading for (a) free-fall, and (b) movement at constant velocity. 9. A force of 8.0 N gives a mass m1 an acceleration of 12 ms-2 and a mass m2, an acceleration of 48 m/s-2. Calculate the acceleration that this force would give the two masses when they are attached. 9 Forces questions page 31 – OCR Physics 10 Link Between Weight and Mass Suppose an object has a mass of m kg. If the object is released it falls to the ground with an acceleration due to the force of gravity of g ms-2 (acceleration of free fall). For this object: mass = m kg, acceleration = g ms-2 resultant force = WN (the weight of the object) Remember, weight is the force of gravity acting on an object! Applying resultant force = mass x acceleration W = m x g Therefore a body of mass m has a weight of mg The acceleration due to gravity g has a value of 9.81 ms-2 on Earth So, a mass of 1 kg has a weight 9.81N, a mass of 5kg has a weight of (5 x 9.81) N. To change Kg N x by 9.81 To change N Kg x by 9.81 For a mass of 1kg on the Earth its weight W will be given by W = mg W = 1 x 9.81 = 9.81N On the Moon g has a value of 1.6ms-2 A mass of 1kg has a force on it: W = 1 x 1.6 = 1.6N (1/6th of the weight on Earth). 11 Self Study Exercise: FRICTION Friction always occurs when there is relative motion e.g. an object falling through air, a fish swimming in the sea, an object being dragged along a surface. Prepare answers to the following questions: (a) What is friction? (b) How does it arise? (the theory) (c) How does it affect the motion of a body? (d) Give some examples of where friction is useful. (c) How may friction be overcome? (f) How may the friction force be measured? (g) What two types of friction are there? (h) What is meant by the coefficient of friction and how may it be measured? Friction in fluids: Viscosity Water resists the motion of boats, and air resists the motion of aircraft. This is fluid friction, which is called viscosity. Viscosity tends to prevent motion within the fluid – it is caused by internal friction between moving layers of fluid. The viscosity of fluids varies greatly. Water is much easier to pour than treacle. It is less viscous. Gases have a much lower viscosity than liquids. The viscosity of most fluids gets less as they get hotter. Warm treacle for example is quite “runny”. To demonstrate viscous force we can drop an object in air, water and in oil/washing up liquid. Viscosity changes with the shape of the object being dropped. The greater the Surface Area, the more the viscosity. Streamlining reduces viscosity. Air resistance is an example of viscosity. 12 Drag Air resistance or aerodynamic drag acts when a body moves through air or any fluid. It is caused by the fact that an object has to push air/fluid out of the way in order to move through it, and this requires a force. The force that is exerted by two surfaces rubbing together (friction) does not depend on the speed at which the two surfaces move over each other. However, the aerodynamic drag caused by an object moving through air does depend on speed - the faster the object moves, the greater the aerodynamic drag. In fact, the drag increases as the square of the velocity. D α v2 Drag depends on several things: *Velocity *Roughness of the surface passing through the fluid *Cross-sectional Area *Shape (streamlined or not) The physics of drag can be complicated. It is easy to understand that that a slow, thin, streamlined object will have less drag than a fast, fat, flat object. However, quantifying to what extent these issues affect drag, is difficult. A model of a plane undergoes tests in a wind tunnel to measure the planes drag coefficient Cd, in the following equation: F = ½ρCdAv2 F is the drag force, ρ is the density of air, A is the frontal area of plane, v is the velocity of the plane. We don’t need to remember this equation, only that Drag α v2 or D = K v2 Air resistance becomes more important the faster you want to go. Careful design can reduce aerodynamic drag, by producing shapes that can cut through the air/water and cause as little disturbance to it as possible. 13 Terminal Velocity Ignoring air resistance and resistive forces, if the force causing an object to accelerate increases, acceleration will increase. However, resultant force often decreases because resistive forces increase when the velocity increases, so acceleration will decrease. Because aerodynamic drag increases greatly as an object's velocity increases, objects tend to reach a terminal velocity as their velocity increases - whether they are a parachutist falling through air or a car travelling along a race track. Falling bodies Steady speed means forces are balanced - Viscosity up and Weight down for example, for a falling object. Steady speed of a falling object is called terminal velocity. In thick liquids Terminal Velocity reached very quickly but in less viscous liquids E.g. Air, this takes time to happen. As an object falls through a fluid, the force of gravity tries to increase its velocity. The viscous force or fluid friction, which increases as the object speeds up, acts so as to slow it down. At a certain velocity, which depends on both the viscosity of the fluid and the shape of the object, the two forces balance each other. The object now moves with a constant velocity, which is called the terminal velocity. (Remember, no resultant force means constant velocity). 14 Demonstration of viscous force Oil The tube contains engine oil. We will investigate the motion of a ball bearing falling through the oil. To do this, we will mark the position of the ball bearing every second after it is released at the surface of the oil. We will measure the distance fallen by the ball bearing after the times 1s. 2s. 3s. … As shown in the table. Time/seconds Distance/cm 0 1 2 3 4 5 Now plot a graph of distance (y-axis) against time (x-axis). What does the graph indicate? 15 Free fall and Terminal Velocity A man jumping out of a plane will accelerate towards the ground under the influence of the force of gravity from the Earth. The acceleration with which he falls is called the acceleration due to gravity or the acceleration of free fall. At the surface of the earth the approximate value of this acceleration is 10 ms-2. The acceleration is not uniform however… At the top of the jump, the man is instantaneously at rest so his air resistance is zero. The force acting on him is greatest at this his acceleration at this point has its maximum value. A little later, the man is moving more rapidly and his air resistance is now significant. The magnitude of his weight is still greater than his air resistance, so he is still accelerating downwards, but not as quickly as at first. Later still his velocity has reached a point where his air resistance is equal to his weight. Now the resultant force acting on him is zero - and he is no longer accelerating. The velocity at which this happens is called the terminal velocity. For a human being without a parachute, terminal velocity is about 56ms-1. On opening the parachute, the air resistance increases dramatically due to the parachute's large surface area. Now the air resistance is greater than the weight so the resultant force on the man is upwards. The man accelerates upwards and his velocity decreases. Eventually the man's velocity decreases to a new terminal velocity. This terminal velocity is much lower than the previous terminal velocity about 10ms-1. Hitting the ground at this speed still requires some care - it is like jumping off a wall 5 m high! 16 Air resistance of a falling body: Just out of 10 seconds Later on Parachute deployed Plane later AR AR AR mg mg mg mg Only force is Speed up = AR = mg. Bigger force up. Down. AR up Balanced force Speed drops–deceleration. Unbalanced = (AR opposite = steady speed TV reached again, at a Accelerate down. to motion). Called Terminal lower speed than before Bigger force Velocity. And man safely hits Still down. Ground. Still Bigger SA = Bigger AR. accelerates Down. AR still Increases Gravity is always constant, but Air resistance increases as speed increases until Terminal Velocity reached, where it stays constant. Taking down as positive, mg – AR = ma. As velocity increases, AR increases therefore, a decreases. When mg = AR, a = 0 as there is no resultant force. It can be noted here that, without the effect of air resistance, a falling body would continually accelerate and never reach a terminal velocity. 17 Graphs of Force, Acceleration and Velocity for a falling body (Take down as positive) Weight Force Acc’n Resultant Time Time A.R. Velocity Time 18 Falling Objects, Accelerating Objects 1. A free-fall sky-diver jumps from a high altitude balloon. The diagram shows the forces acting on her five seconds later. X Y a) Name the forces X and Y. b) What happens to the size of X as she continues to fall? c) Explain, in terms of X and Y, why she eventually reaches a steady terminal velocity. d) Explain how her motion changes when her parachute opens. e) As she nears the ground, she remembers the instruction to bend her knees on landing. Explain why this makes the landing “softer”. 19 2. A raindrop falling through air eventually travels at its terminal velocity. Use one of the statements below to complete the sentences that follow: Zero; downward and increasing; downward and decreasing; upward and constant; downward and constant; upward and decreasing. At its terminal velocity the (i). Velocity of the raindrop is _______________________________________ (ii). acceleration of the raindrop is ____________________________________ (iii). force of the air on the raindrop is __________________________________ (iv). weight of the raindrop is ________________________ 20 3. 4000 N Mass = 250 Kg 2500 N (weight) Use the information given in the diagram to calculate: a. the resultant upward force of the rocket b. the acceleration of the rocket Freefall and Drag questions page 33 – OCR Physics Project Work Write a short report on the following: How do fast moving jet aircraft use parachutes to decelerate? 21 Equilibrium Revision Adding and Subtracting Vectors Forces are vectors. Vectors can be represented using arrows. The length = the vector size. The sum can be found by joining the vector arrows together in any order tip to tail. The resultant is the single vector that could replace all the others and points in the direction in which they are going. This is a vector triangle: a b R R points start – finish The resultant (or sum) of two vectors is calculated using these vector triangles and Pythagoras theorem/trigonometry or scale drawing. If the vectors are at right angles use Pythagoras Theorem and Trigonometry to find the vector magnitude and vector angle. We can have more than two vectors being added together. We use the same method of scale drawing to find the resultant vector, and add them tip to tail as before: B A C Resultant 22 Sometimes the resultant is almost zero because the forces can cancel out: A Resultant B Equilibrium One condition for equilibrium is as follows: An object is in equilibrium if the forces acting are balanced (zero resultant force. Static solid objects are always in equilibrium. If you add a number of vectors tail to nose, the resultant vector needed to keep the body in equilibrium can be found by joining the vectors start to the finish. The resultant vector closes the loop and points in the opposite direction from the other vectors (it balances the overall vector that replaces the vectors!). R a b R a c b c If the vectors are already balanced you end up with a closed polygon. The arrows will lead back to the start because the resultant is zero. Look at the diagram below. Forces A, B and C are acting at a point on an object. They add together to give a zero resultant force. This triangle is called the triangle of forces A B A B C C 23 Force C cancels out the resultant of A and B. Zero resultant means zero acceleration. An object that has zero resultant force is said to be in equilibrium. An object can be in equilibrium if it is at rest or moving with constant velocity. Examples 1. A climber on a rock face is in equilibrium under the action of 3 forces – his weight, the force the rock face exerts on the climber (50 degrees west of the vertical) and the force the rope exerts on the climber (40 degrees east of the vertical). The diagram is as follows: Force of rope Force of wall on on climber Y climber X Weight of climber Calculate force X and Y, given that the mass of the climber is 80 kg. a work out the weight of the climber: W = m x g = 80 x 9.81 = 785 N b next draw a sketch of the forces, with angles and then draw a triangle of forces: Y X Y 785 N 785 N X c we have a right angled triangle which makes calculations easier: X = 785 cos 50 = 505 N Y = 785 cos 40 = 601 N 24 Questions 1. A 20 N weight is hanging by a string from the roof. There is tension T in the string. A horizontal force H is applied to hold the string at an angle of 30 degrees to the vertical. By drawing a diagram, obtain the value of the force H and Tension T. 2. Three strings are attached to a small metal ring. Two of the strings make an angle of 70 degrees to each other and each is pulled with a force of 7 N. What force must be applied to the third string to keep the ring stationary? 3. A body is in equilibrium under the action of three forces. One force is 6.0 N acting due East and one is 3.0 N, in a direction 60 degrees, North of East. What is the magnitude and direction of the third force? 4. An acrobat stands on a tightrope. The rope makes an angle of 7degrees with the horizontal on each side. If his weight is 800 N what is the tension in each half of the rope? 5. A bob of mass 0.2Kg is pulled aside by a horizontal force so that the string makes an angle of 25 degrees with the vertical. What is the new tension in the string? 6. A sphere of mass 150 g is suspended by a thread which will break under a tension of 2 N. If the sphere is slowly pulled aside by a force which is always acting horizontally, at what angle to the vertical will the string break? What is the value of the horizontal force when this happens? 7. A mass of 20 kg is pulled up a smooth plane, which makes an angle of 30 degrees with the horizontal, by a force F acting on it. F is of value 200 N and always acts horizontally. Calculate the reaction of the plane on the mass and the acceleration of the body. 8. A mass of weight 5 N is attached to a point B of a string ABC. AB is at 30 degrees to the vertical, BC is at 45 degrees to the vertical and AC is horizontal. Determine the pull in each string. 9. A boy and girl, raising a bucket of water, pull at an angle to each other. Each exerts a force of 150N and their effective pull is 260N. What is the angle between their arms? (Scale drawing) 10. A 100N lamp is suspended over a road by two straight wires, each making an angle of 20 degrees with the horizontal and attached to posts on opposite sides of the road. Find the tension in each wire. 11. A kite weighing 50N flies at the end of a string, which makes an angle of 30 degrees with the horizontal. The force of the wind on the kite makes an angle of 45 degrees with the horizontal. Determine the pull on the string and also the force of the wind. 25 Equilibrium questions page 35 – OCR Physics 26 Centre of Gravity When we considered the climber in the last section of work, climbing up a rock face, we took him to be a Point Mass. This makes life a lot easier! In problems involving solid objects, whenever we use the idea of a point mass we often draw the weight of an object as acting through a single point. This point is called the centre of gravity. It is the point at which the whole weight of the body may be considered to act. For a regular object the centre of gravity will be at the geometrical centre of the body. Finding the C og G for a cylinder, sphere and uniform beam can be done easily This will not be so if for an irregular object. Indeed, centre of gravity can be outside the body (boomerang)! If a body is supported on a pivot directly below the centre of gravity the body will balance. It will be in equilibrium. The weight of the body is taken to act at the centre of gravity and is therefore marked using an arrow starting at the c of g. C of G Weight Stability is important in everyday objects – otherwise, things topple over. Toppling occurs when the centre of gravity of an object falls outside its base. A low C of G and a wide base helps make an object more stable, by making as large as possible the angle through which the object must tilt in order to cause the centre of gravity to fall outside the base. 27 Experiment to find the C of G of an irregular object 1. Cut the cardboard into a strange shape. 2. Punch a hole near the edge of the cardboard and hang it suspended from a nail. 3. Make a plumb line by tying a weight to a thread and attaching the thread to the nail. 4. Mark a line on the cardboard where the thread touches it. 5. Repeat steps 2, 3, and 4, in two more spots on the side of the cardboard. 6. Remove the thread from the cardboard. 7. Balance the cardboard on the end of a pencil. How do you know where to put the pencil? 8. Describe the centre of gravity for the piece of cardboard. 9. Explain how this principle might help a tightrope walker in a circus. Step 3: Suspend the shape from another point Step 2: Suspend the not too close to the first. shape from a point near Drop a plumb line again Step 1: Cut any 2D shape. an edge. Drop a plumb line and mark. The and mark on the object. intersection of the two lines is the centre of gravity. 28 Moments The turning effect of a force is called its moment. The moment of a force about a pivot is the turning effect or rotation effect the force produces about that pivot (fulcrum). Moment of a force (Nm) = Force (N) x perpendicular distance (m) from the line of action of the force to the pivot point For a body in equilibrium, sum of clockwise and anticlockwise moments must be equal about each and every point. This is called the PRINCIPLE OF MOMENTS Weight of arm The forearm is an example of a pivot system. The Elbow is the pivot point The weight of forearm x distance from elbow plus the carried weight x distance from elbow supplies the clockwise moment The muscle x distance from the elbow the anticlockwise moment 29 Moment Experiment (Ruler and weights) Distance 2 Ruler Distance 1 Weight 2 Weight 1 Clamp and Stand Get a metre ruler with a hole drilled in the centre and pivot it on a clamp and stand. Hang the correct weight on the left, in the correct position, and balance with the named weight on the right. Note down the position on the right for balancing to occur. Calculate force x distance on both sides (Remember, 100g = 0.981N). Left Hand Side Right Hand Side W1 (N) D1 (m) W1 x D1 W2 (N) D2 (m) W2 x D2 1wt 0.4 4wt 3wt 0.4 4wt 6wt 0.2 3wt 2wt 0.45 3wt 2wt 0.3 3wt When the lever balances, what is the connection between W1 x D1 and W2 x D2? 30 Couples A single force can cause a body to turn but it will also cause linear motion. A couple is a pair of equal and opposite forces along different lines of action (not in the same straight line). Couples cause rotation but no linear motion/acceleration because the upward force balances the downward force. The resultant is zero, but they do tend to cause rotation. Two equal, but oppositely directed forces acting simultaneously on opposite sides of an axis of rotation. Since the translatory forces (forces that produce linear motion) cancel out each other, a force couple produces torque (rotatory forces) only. The magnitude of the force couple is the sum of the products of each force and its moment arm. Force couple The turning effect of a couple, called its Torque is calculated by multiplying one of the forces by the perpendicular distance between the forces. Torque is measured in Newton metres. Electric Torque Wrenches can be used to give the correct torque or turning effect on a nut. 31 Stability, Moments and Centre of Gravity Pivot point If the line of action of the centre of gravity is to the left of the pivot point, there will be an anticlockwise moment, which will return the bus to its upright position. If the line of action of the centre of gravity is to the right of the pivot point, there will be a clockwise moment, which will cause the bus to topple over. A low centre of gravity and a wide object will be more stable because the angle through which you need to tilt before the line of action of the C of G falls to the right of the pivot point will be greater! 32 YEAR 12 MOMENTS 1. The bar in the diagram is in equilibrium. X 1m 2m 3m O 6N F What is the moment of the 6N force about O? What is the moment of the force F about 0? Calculate F and the reaction force X. Now calculate the moments of F, X and the 6N force about P. What is the total clockwise moment about P? What is the total anticlockwise moment about P? 2. In the diagram a light plank on two trestles is supporting a man and a block of concrete. X Y 1m 1m 1m A B 400N 700N Write down the moment of each force about A. Use the principle of moments to calculate the reaction force Y. What is the total upward force X + Y? What is the size of the reaction force X? In the diagram the man walks past A towards the end of the plank. What is the reaction force at Y at the instant the plank starts to tip? How far is the man from A as the plank starts to tip? 3. The diagram shows a wheel of mass 15 kg and radius 1 m being pulled by a horizontal force F against a step 0.4 m high. F 1m 0.4m 33 What initial force is just sufficient to turn the wheel so that it will rise over the step? What happens to the size of this horizontal force as the wheel rises? 4. A car bonnet weighs 200 N and is 1.8 m long. The bonnet's weight acts at its midpoint. If the bonnet is freely hinged at one end and is supported by a vertical force at the other end, what size must this force be in order to support the bonnet at 30 degrees to the horizontal? 5. A car maintenance manual specifies that the cylinder head bolts on an engine should be tightened to a torque of 69 Nm. What force must be applied to the end of the spanner 30 cm long in order to achieve this? What might be the result of using a higher torque than this to tighten the nut? 6, A railway engine weighing 7.5 x 105 N is 40 m across the centre span of a bridge consisting of an iron girder 120 m long, weighing 1.0 x 107N. Calculate the load on each support at each end of the bridge. 80m 40m 34 Moments questions etc. pages 40 and 41 – OCR Physics 35 YEAR 12 EQUILIBRIUM 1. An object is held in the position shown by two strings. If the force of friction at the pulley is negligible find the values of the forces T1 and T2. T1 T2 20 50 250 N 2. A beam of length l and weight w is supported horizontally from two spring balances A and B attached to its ends as shown. The reading of A is twice that of B. A B (i) Find the position of the centre of gravity of the beam. (ii) Explain whether you consider the beam to be uniform. 3. The diagram shows a body of weight 120 N supported by two wires from fixed points A and B. The angles made by the wires with the horizontal are shown in the diagram. Calculate the tensions in the wires. A B 60 30 36 5. An aerial attached to the top of a radio mast 20 m high exerts a horizontal force on it of 3 X 102 N. A stay-wire from the mid-point of the mast to the ground is inclined at 60 degrees to the horizontal. Assuming the action of the ground on the mast can be regarded as a single force, find: (a) the force exerted on the mast by the stay-wire, (b) the magnitude and direction of the action of the ground. 6. A uniform ladder 5.0 m long and having a mass 40 kg rests with its upper end against a smooth vertical wall and with its lower end 3.0 m from the wall on rough ground. Find the magnitude and direction of the force exerted at the bottom of the ladder. 7. A uniform heavy rod OA of mass 12 kg is pivoted freely at O and held in the position shown by a force F acting at right angles to the rod and applied at A. Calculate the magnitude of F. A O 60 8. Draw a diagram of a cube of side 1m and clearly indicate the position of its centre of gravity. The cube is then placed on a rough surface, which is gradually tilted. To what angle can the surface be tilted before the cube topples? Assume that the surface is rough so that no slipping can occur. 9. A rectangular box 2 m tall and with base 1m x 1m is filled with matter of uniform density. It stands on the back of a flat bed truck. What is the maximum acceleration that the truck can have if the box is not to topple over (Assume that the friction between box and truck is such that the box does not slide)? Start by drawing a free body diagram for the box showing the situation when it begins to topple and so only makes contact with the truck along its rear edge. 10.a. Figure i shows a uniform cylinder whose parallel end faces are not perpendicular to its axis. Indicate clearly on the diagram the precise position of its centre of gravity, G. b. A student places a 2p coin on a table as shown in Fig ii. A second 2p coin is places on top of the first one, but slightly displaced to one side. Further 2p coins are added, each one being given the same displacement relative to the previous coin. 37 The line through the centres of the coin makes an angle of 10 degrees with the vertical. If each coin has a diameter of 25 mm and a thickness of 2.0 mm, show that the student is able to add no more than 70 coins before the pile topples. Explain your reasoning. 10 38 39

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