PHY113 Chapter 3 Newton's Laws PDF

CHAPTER 3 Newton’s Laws of Motion: Dynamics Outcomes Force Newton’s First Law of Motion Mass Newton’s Second Law of Motion Newton’s Third Law of Motion Weight – the Force of Gravity; and the Normal Force Solving Problems with Newton’s Laws: Free-Body Diagrams Applications Involving Friction, Incline...

CHAPTER 3 Newton’s Laws of Motion: Dynamics Outcomes Force Newton’s First Law of Motion Mass Newton’s Second Law of Motion Newton’s Third Law of Motion Weight – the Force of Gravity; and the Normal Force Solving Problems with Newton’s Laws: Free-Body Diagrams Applications Involving Friction, Inclines Problem Solving – A General Approach 3.1 Force A force is a push or pull. – An object at rest needs a force to get it moving – A moving object needs a force to change its velocity. The magnitude of a force can be measured using a spring scale. 3.2 Newton’s First Law of Motion Often called the Law of Inertia Every object continues in its state of rest, or of uniform velocity in a straight line, as long as no net force acts on it. 3.2 Newton’s First Law of Motion Inertial reference frames: – An inertial reference frame is one in which Newton’s first law is valid. – This excludes rotating and accelerating frames. Example: – A school bus comes to a sudden stop, and all of the backpacks on the floor start to slide forward. Why? 3.3 Newton’s Second Law of Motion Mass is the measure of inertia of an object. In the SI system, mass is measured in kilograms. Mass is not weight: – Mass is a property of an object. – Weight is the force exerted on an object by gravity. If you go to the moon, whose gravitational acceleration is about 1/6 g, you will weigh much less. Your mass, however, will be the same. Mass—A Measure of Inertia CHECK YOUR NEIGHBOR If the mass of an object is halved, the weight of the object is A. B. C. D. halved. twice. depends on location. None of the above. Mass—A Measure of Inertia CHECK YOUR ANSWER If the mass of an object is halved, the weight of the object is A. B. C. D. halved. twice. depends on location. None of the above. 3.3 Newton’s Second Law of Motion Relationship between acceleration and force Consider an object of mass, m at rest – If you push the object with a force F the object will accelerate with an acceleration a – If you double the force to 2F the acceleration will also double to 2a Therefore:   aF Now half the mass to 1/2m and apply the same force F. The acceleration will double, therefore:  1 a m 3.3 Newton’s Second Law of Motion Newton’s second law states: – The acceleration of an object is directly proportional to the net force acting on it and is inversely proportional to its mass. The direction of the acceleration is in the direction of the net force acting on the object.   F a  NET m    FNET  ma Alternatively:   F  m a    Fx  ma x ;  Fy  ma y ;  Fz  ma z The unit of force in the SI system is the Newton (N). Newton’s Second Law of Motion CHECK YOUR NEIGHBOR Consider a cart pushed along a track with a certain force. If the force remains the same while the mass of the cart decreases to half, the acceleration of the cart A. B. C. D. remains relatively the same. halves. doubles. changes unpredictably. Newton’s Second Law of Motion CHECK YOUR ANSWER Consider a cart pushed along a track with a certain force. If the force remains the same while the mass of the cart decreases to half, the acceleration of the cart A. remains relatively the same. B. halves. C. doubles. D. changes unpredictably. Explanation: Acceleration = net force / mass Because, mass is in the denominator, acceleration increases as mass decreases. So, if mass is halved, acceleration doubles. Newton’s Second Law of Motion CHECK YOUR NEIGHBOR Push a cart along a track so twice as much net force acts on it. If the acceleration remains the same, what is a reasonable explanation? A. B. C. D. The mass of the cart doubled when the force doubled. The cart experiences a force that it didn’t before. The track is not level. Friction reversed direction. Newton’s Second Law of Motion CHECK YOUR ANSWER Push a cart along a track so twice as much net force acts on it. If the acceleration remains the same, what is a reasonable explanation? A. B. C. D. The mass of the cart doubled when the force doubled. The cart experiences a force that it didn’t before. The track is not level. Friction reversed direction. Explanation: Acceleration = net force / mass If force doubles, acceleration will also double, But it does not, so mass must also be doubling to cancel out effects of force doubling. 3.3 Newton’s Second Law of Motion Examples: – What average net force is required to bring a 1500 kg car to rest from a speed of 28 m/s within a distance of 55 m? FNET = ma; a= -282/110 = - 7.1 m/s2 FNET = 1500 x -7.1 = - 10690 N – A net force of 256 N accelerates a bike and rider at 2.30 m/s2. What is the mass of the bike and rider together? mass = 111`kg – A 0.140 kg baseball traveling 35.0 m/s strikes the catcher’s mitt, which, in bringing the ball to rest, recoils backward 11.0 cm. What was the average force applied by the ball on the glove? a =- 5568m/s F = - 779.55N 3.3 Newton’s Second Law of Motion Examples: 1) A horizontal cable pulls a 200-kg cart along a horizontal frictionless track. The tension in the cable is 500 N. Starting from rest, (a) how long will it take the cart to reach a speed of 8 m/s? (b) How far will it have gone? 2) A force acts on a 2-kg mass and gives it an acceleration of 3 m/s2. What acceleration is produced by the same force when acting on a mass of (a) 1 kg? (b) 4 kg? (c) How large is the force? 3) What average force is needed to accelerate a 7 gram pellet from rest to 125 m/s over a distance of 0.8m along the barrel of a rifle? 3.4 Newton’s Third Law of Motion What causes an object to accelerate? Forces exerted on the object by other objects Newton’s third law: – Whenever one object exerts a force on a second object, the second object exerts an equal force in the opposite direction on the first object. 3.4 Newton’s Third Law of Motion A key to the correct application of the third law is that the forces are exerted on different objects. Make sure you don’t use them as if they were acting on the same object. Rocket propulsion can also be explained using Newton’s third law: hot gases from combustion spew out of the tail of the rocket at high speeds. The reaction force is what propels the rocket. 3.4 Newton’s Third Law of Motion   FPG  FGP FPG  FGP Helpful notation: the first subscript is the object that the force is being exerted on; the second is the source. 3.5 Weight and Normal Force Weight is the force exerted on an object by gravity. Close to the surface of the Earth, where the gravitational force is nearly constant, the weight is:   FG  mg An object at rest must have no net force on it. If it is sitting on a table, the force of gravity is still there; what other force is there? The force exerted perpendicular to a surface is called the normal force. It is exactly as large as needed to balance the force from the object (if the required force gets too big, something breaks!) 3.6 Problem Solving – Free Body Diagrams Draw a sketch. For one object, draw a free-body diagram, showing all the forces acting on the object. Make the magnitudes and directions as accurate as you can. Label each force. If there are multiple objects, draw a separate diagram for each one. Resolve vectors into components. Apply Newton’s second law to each component. Solve. Free Body Diagrams Free Body Diagrams Example The beast pulls on Tarzan with a force of 180 N and Tarzana pulls with a force of 150 N. What is Tarzan’s acceleration? Given Assume the standard reference system. FT = 150 N; FB = - 180 N; mT = 90 kg Construct a free-body diagram (consider the horizontal direction only). NII: F =ma; FT – FB = ma = 150-180 = -30 a = -30/90 = - 0.33 m/s2 Exercise problem 10 How much tension must a rope withstand if it is use to accelerate 1200 kg car vertically upwards at 0.8 m/s2? Construct a free-body diagram (consider the vertical direction only). NII: F =ma; FT – Fg = ma FT = Fg + ma = m (g+a) = 1200 ( 9.8 +0.8) = 1200 (10.6) = 12720 N 3.6 Problem Solving – Free Body Diagrams A friend has given you a special gift, a box of mass 10.0 kg with a mystery surprise inside. The box is resting on the smooth (frictionless) horizontal surface of a table. – Determine the weight of the box and the normal force exerted on it by the table. – Now your friend pushes down on the box with a force of 40.0 N. Again determine the normal force exerted on the box by the table. – If your friend pulls upward on the box with a force of 40.0 N, what now is the normal force exerted on the box by the table? APPROACH Construct a free-body diagram (consider the vertical direction only). No Forces in the horizontal direction a) NII: F =ma;  Fy = FN – F g = 0 FN = Fg = 10 (9.8) = 98 N b) NII: F =ma;  Fy = FN – Fg - 40= 0 FN = Fg + 40 = 98+40 = 138 N c) NII: F =ma;  Fy = FN – Fg +40= 0 FN = Fg - 40 = 58 N 3.6 Problem Solving – Free Body Diagrams What happens when a person pulls upward on the box with a force greater than the box's weight, say Fp = 100.0 N?  Fy = FN – Fg + Fp = 0 FN = Fg – Fp = - 2N; a = F/m = 0.2 m/s/s A 65 kg woman descends in an elevator that briefly accelerates at 0.20g downwards when leaving a floor. She stands on a scale that reads in kg. (a) During this acceleration, what is her weight and what does the scale read? (b) What does the scale read when the elevator descends at a constant speed of 2.0 m/s? 3.6 Problem Solving – Free Body Diagrams A boy pulls a 10 kg box by the attached cord along a smooth surface of the table. The magnitude of the force exerted by the boy is Fb = 40.0 N and it exerted at 30.00 angle as shown. Calculate (a) the acceleration of the box, and (b) the magnitude of the upward force FN exerted by the table on the box. Assume friction can be neglected. 3.6 Problem Solving – Free Body Diagrams When a cord or rope pulls on an object, it is said to be under tension, and the force it exerts is called a tension force. 3.6 Problem Solving – Free Body Diagrams Elevator and counterweight (Atwood’s machine): – A system of two objects suspended over a pulley by a flexible cable, as shown in the figure, is sometime referred to as Atwood’s machine. In real-life: application of an elevator. The cable’s mass, mass of the pulley and any friction is negligible. Note that FT is the same both sides of the pulley. Find the acceleration and FT. 3.7 Applications Involving Friction and Inclines On a microscopic scale, most surfaces are rough. The exact details are not yet known, but the force can be modeled in a simple way. Static friction is the frictional force between two surfaces that are not moving along each other. Static friction keeps objects on inclines from sliding, and keeps objects from moving when a force is first applied. Ffr  s FN Where s is the coefficient of static friction 3.7 Applications Involving Friction and Inclines The static frictional force increases as the applied force increases, until it reaches its maximum. Then the object starts to move, and the kinetic frictional force takes over. Ffr   k FN Where k is the coefficient of kinetic friction Note: The direction of the frictional force is always opposite to the direction of motion 3.7 Applications Involving Friction and Inclines 3.7 Applications Involving Friction and Inclines (A) A 10 kg box rests on a horizontal floor. The coefficient of static friction s = 0.40 and the coefficient if kinetic friction k = 0.30. Determine the force of friction Ffr acting on the box if the applied force F exerted on it has a magnitude: (a) 0 N, (b) 10 N, (c) 20 N, (d) 38 N and (e) 40 N. (B) A 10.0-kg box is pulled along a horizontal surface by a force FP = 40.0 N applied at an angle of 30. The coefficient of kinetic friction is equal to 0.30. Calculate the acceleration of the box. A. Construct a free-body diagram a) NII: F =ma;  Fx = Fp – FFr = ma = 0, Static Friction FFr = sFN = 0.4 (98) = 39.2 N  Fy = FN – Fg = 0, No motion vertically. FN = 98 N b) NII: F =ma;  Fx = Fp – FFr = 0, Static Friction FFr = 10 N c) NII: F =ma;  Fx = Fp – FFr = ma = 0, Static Friction FFr = 20 N d) NII: F =ma;  Fx = Fp – FFr = ma = 0, Static Friction FFr = 38 N e) NII: F =ma;  Fx = Fp – FFr = ma = 0, Static Friction FFr = 40 N box will start moving Since it exceeds static Friction. Therefore; e)  Fx = Fp – FFr = ma , Kinetic Friction FApplied 40 - kFN = 40 - 0.3(98) = 11 N = 10 a a = 11/10 = 1.1 m/s2 FFr B) A 10.0-kg box is pulled along a horizontal surface by a force FP = 40.0 N applied at an angle of 30. The coefficient of kinetic friction is equal to 0.30. Calculate the acceleration of the box. Construct a free-body diagram NII: F =ma;  Fy = FN + Fpsin30 – Fg = 0, FN = 98 – 20 =78 N  Fx = Fpcos30 – FFr = ma 40cos3 – 0.3 (78) = 10a 34.64 – 23.4 = 10a a = 1.124 m/s2 No motion vertically. 3.7 Applications Involving Friction and Inclines Two boxes are connected by a cord running over a pulley. The coefficient of kinetic friction between box A and the table is 0.20. We ignore the mass of the cord and the pulley and any friction in the pulley. Find the acceleration of the system and the tension in the cord as box B moves down, box A moves to the right.. 3.7 Applications Involving Friction and Inclines An object sliding down an incline has three forces acting on it: the normal force, gravity, and the frictional force. – The normal force is always perpendicular to the surface. – The friction force is parallel to it. – The gravitational force points down. If the object is at rest, the forces are the same except that we use the static frictional force, and the sum of the forces is zero. 3.7 Applications Involving Friction and Inclines The skier in the figure has just begun descending the 30° slope. Assuming the coefficient of kinetic friction is 0.10, calculate (a) her acceleration and (b) the speed she will reach after 4.0 seconds. 3.7 Applications Involving Friction and Inclines General problem solving approach – – – – – – – – Read the problem carefully; then read it again. Draw a sketch, and then a free-body diagram. Choose a convenient coordinate system. List the known and unknown quantities; find relationships between the knowns and the unknowns. Estimate the answer. Solve the problem without putting in any numbers (algebraically); once you are satisfied, put the numbers in. Keep track of dimensions. Make sure your answer is reasonable. 3.7 Applications Involving Friction and Inclines A force F of magnitude 10 N is applied to a block of mass 2 kg sliding down an inclined plane as shown in the diagram. Calculate – The acceleration and the normal reaction on the block due to the plane assuming the surface to be frictionless. – The acceleration of the block if the coefficient of friction between block an plane is 0.2. Three blocks on a horizontal, frictionless table are connected as shown in the diagram and pulled to the right with a force T3 = 60 N. If m1 = 10 kg, m2 = 20 kg and m3 = 30 kg, calculate the tension T1 and T2. 3.7 Applications Involving Friction and Inclines A person pushes a 14.0-kg lawn mower at constant speed with a force of 88 N directed along the handle, which is at an angle of 45.0º to the horizontal. (a) Draw the free-body diagram showing all forces acting on the mower. Calculate (b) the horizontal friction force on the mower, then (c) the normal force exerted vertically upward on the mower by the ground. (d) What force must the person exert on the lawn mower to accelerate it from rest to 1.5 m/s in 2.5 seconds, assuming the same friction force? 3.7 Approach (a) Assume that the mower is being pushed to the right. (b) NII: For Horizontal Direction. The forces must sum to 0 since the mower is not accelerating NII: F = ma = 0 , no accelerating Fx = Fpcos45 – Ffr = 0 Ffr = Fpcos45 = 62.2 N (c) NII: For Vertical Direction. The forces must sum to 0 since the mower is not accelerating NII: F = ma = 0 , no accelerating Fy = FN – mg - Fpsin45 = 0 FN FN = 199 N Fp (d) Eq. of motion to find acceleration V = Vi + at; a = (1.5 - 0)/2.5 = 0.6 m/s2 FFr NII: For the x direction to find pushing force. NII: F = ma Fx = Fpcos45 – Ffr = ma Fg Fp = 99.9 N 3.7 Applications Involving Friction and Inclines A window washer pulls herself upward using the bucket–pulley apparatus shown in Figure. – How hard must she pull downward to raise herself slowly at constant speed? – If she increases this force by 15%, what will her acceleration be? The mass of the person plus the bucket is 65 kg. FBD FT FT a) NII: Vertical Direction, with up as positive NII: F =ma = 0 , constant speed F = FT+FT-mg =0, 2FT = mg FT = 320 N b) Net force increased by 15%, so Fg FT = 320N(1.15) = 368 NII: F = FT+FT-mg =ma a = (2FT - mg)/m = 1.5m/s2 3.8 Work, Energy and Power The work done by a constant force is defined as the distance moved multiplied by the component of the force in the direction of displacement: W  Fd cos  In the SI system, the units of work are joules (J), where 1 J = 1 N.m 3.8 Work, Energy and Power A person pulls a 50-kg crate 40 m along a horizontal floor by a constant force Fp = 100 N which acts at an angle of 37. The floor is rough and exert a friction force Fr = 50 N. Determine the work done by each force on the crate and the net work done on the crate. 3.8 Work, Energy and Power Energy was traditionally defined as the ability to do work. We now know that not all forces are able to do work In this course – only deal with mechanical energy If we write the acceleration in terms of the velocity and the distance, we find that the net work done: Wnet 1 1 2  m2  m12 2 2 3.8 Work, Energy and Power We define the kinetic energy: KE  1 m2 2 Work-energy principle: The net work done on an object is equal to the change in the object’s kinetic energy Wnet  KE  KE2  KE1 If the net work is positive, the kinetic energy increases. If the net work is negative, the kinetic energy decreases. Si unit for (kinetic) energy is also the Joule (J) How much work is required to accelerate a 1000 kg car from 20 m/s to 30 m/s? 3.8 Work, Energy and Power An object can have potential energy by virtue of its surroundings. Familiar examples of potential energy: A wound-up spring, a stretched elastic band and an object at some height above the ground In raising a mass m to a height h, the work done by the external force is: Wext  Fextd cos 0o  mgh  Wext  mg ( y 2  y1 ) The work done by gravity is: Wext  FG d cos 180o  mgh  Wext  mg ( y 2  y1 ) 3.8 Work, Energy and Power We define the gravitational potential energy of an object (where y is taken as some distance above some reference) as: PEG  mgy Wext  mg ( y 2  y1 )  PE 2  PE1  Wext  PE WG  mg ( y 2  y1 )  PE1  PE 2  WG  PE 3.8 Work, Energy and Power A 1000 kg rolar-coaster car moved from point 1 to Point 2 and then to point 3. (a) What is the gravitational potential energy at 2 and 3 relative to point 1, that is take y = 0 at point 1. (b) What is the change in potential energy when the car goes from point 2 to pint 3. (c) Repeat parts (a) and (b), but take the reference point (y = 0) to be at point 3. 3.8 Work, Energy and Power The work done against gravity in moving an object from one point to another does NOT depend on the path taken Therefore, gravity is a conservative force If friction is present, the work done depends not only on the starting and ending points, but also on the path taken. Therefore, friction is a nonconservative force. Potential energy can only be defined for conservative forces. 3.8 Work, Energy and Power Therefore, we distinguish between the work done by conservative forces and the work done by non-conservative forces. We find that the work done by non-conservative forces on an object is equal to the total change in kinetic and potential energies: Wnet  WC  WNC  Wnet  KE and WC  PE  WNC  KE  PE 3.8 Work, Energy and Power When no non-conservative forces are acting on an object, then: KE  PE  0  KE2  KE1  PE 2  PE1  0  KE1  PE1  KE2  PE 2 We now define the total mechanical energy (E) as: E  KE  PE  E1  E 2  const. Principal of conservation for conservative forces: – If only conservative forces are acting; the total mechanical energy of a system neither increases nor decreases in any process. It remains constant – it is conserved! 3.8 Work, Energy and Power Power is the rate at which work is done: Work done Energy transformed P  time time The SI unit for power is the Watt (W), where 1 W = 1 J/s The difference between walking and running up these stairs is power – the change in gravitational potential energy is the same. Power is also needed for acceleration and for moving against the force of gravity. The average power can be written in terms of the force and the average velocity: P W Fd   F t t Tut Problem A 330-kg piano slides 3.6 m down a 28º incline and is kept from accelerating by a man who is pushing back on it parallel to the incline (Fig. 1-5-1). The effective coefficient of kinetic friction is 0.40. Calculate: (a) the force exerted by the man, (b) the work done by the man on the piano, (c) the work done by the friction force, (d) the work done by the force of gravity, and (e) the net work done on the piano. Fig.1-5-1 End