I.Newtons first law. II.Newtons second law. III.Particular forces: - Gravitational - Gravitational - Weight - Weight - Normal - Normal - Friction - Friction - Tension - Tension IV. Newtons third law. Chapter 5 Force and Motion I I. Newtons first law: If no net force acts on a body, then the bodys velocity cannot change; the body cannot accelerate v = constant in magnitude and direction. Principle of superposition: when two or more forces act on a body, the net force can be obtained by adding the individual forces vectorially. Newton mechanics laws cannot be applied when: 1) The speed of the interacting bodies are a fraction of the speed of light Einsteins special theory of relativity. 2) The interacting bodies are on the scale of the atomic structure Quantum mechanics Inertial reference frame: where Newtons laws hold. II. Newtons second law: The net force on a body is equal to the product of the bodys mass and its acceleration. The acceleration component along a given axis is caused only by the sum of the force components along the same axis, and not by force components along any other axis. System: collection of bodies. External force: any force on the bodies inside the system. III. Particular forces: Gravitational: pull directed towards a second body, normally the Earth Weight: magnitude of the upward force needed to balance the gravitational force on the body due to an astronomical body Normal force: perpendicular force on a body from a surface against which the body presses. Frictional force: force on a body when the body attempts to slide along a surface. It is parallel to the surface and opposite to the motion. Tension: pull on a body directed away from the body along a massless cord. Newtons third law: When two bodies interact, the forces on the bodies from each other are always equal in magnitude and opposite in direction. Newtons third law: When two bodies interact, the forces on the bodies from each other are always equal in magnitude and opposite in direction. QUESTIONS Q2. Two horizontal forces, pull a banana split across a frictionless counter. Without using a calculator, determine which of the vectors in the free body diagram below best represent: a) F 1, F 2 ; b) What is the net force component along (c) the x-axis, (d) the y-axis? Into which quadrant do (e) the net-force vector and (f) the splits acceleration vector point? QUESTIONS Q2. Two horizontal forces, pull a banana split across a frictionless counter. Without using a calculator, determine which of the vectors in the free body diagram below best represent: a) F 1, F 2 ; b) What is the net force component along (c) the x-axis, (d) the y-axis? Into which quadrant do (e) the net-force vector and (f) the splits acceleration vector point? Same quadrant, 4 Q1. The figure below shows overhead views of four situations in which forces act on a block that lies on a frictionless floor. If the force magnitudes are chosen properly, in which situation it is possible that the block is (a) stationary and (b) moving with constant velocity? a y 0 a=0 a y 0 a=0 Q5. In which situations does the object acceleration have (a) an x-component, (b) a y component? (c) give the direction of a. F net Q5. In which situations does the object acceleration have (a) an x-component, (b) a y component? (c) give the direction of a. Q. A body suspended by a rope has a weigh W of 75N. Is T equal to, greater than, or less than 75N when the body is moving downward at (a) increasing speed and (b) decreasing speed? FgFgFgFg FgFgFgFg (a) Increasing speed: v f >v 0 a>0 T < F g (b) Decreasing speed: v f < v 0 a F g Q8. The figure below shows a train of four blocks being pulled across a frictionless floor by force F. What total mass is accelerated to the right by (a) F, (b) cord 3 (c) cord 1? (d) Rank the blocks according to their accelerations, greatest first. (e) Rank the cords according to their tension, greatest first. T1T1T1T1 T2T2T2T2 T3T3T3T3 (a) F pulls m total = ( )kg = 20kg (b) Cord 3 T 3 m=(10+3+5)kg = 18kg (c) Cord 1 T 1 m= 10kg (d) F=ma All tie, same acceleration Q. A toy box is on top of a heavier dog house, which sits on a wood floor. These objects are represented by dots at the corresponding heights, and six vertical vectors (not to scale) are shown. Which of the vectors best represents (a) the gravitational force on the dog house, (b) on the toy box, (c) the force on the toy box from the dog house, (d) the force on the dog house from the toy box, (e) force on the dog house from the floor, (f) the force on the floor from the dog house? (g) Which of the forces are equal in magnitude? Which are (h) greatest and (i) least in magnitude? > Q. A toy box is on top of a heavier dog house, which sits on a wood floor. These objects are represented by dots at the corresponding heights, and six vertical vectors (not to scale) are shown. Which of the vectors best represents (a) the gravitational force on the dog house, (b) on the toy box, (c) the force on the toy box from the dog house, (d) the force on the dog house from the toy box, (e) force on the dog house from the floor, (f) the force on the floor from the dog house? (g) Which of the forces are equal in magnitude? Which are (h) greatest and (i) least in magnitude? (a) F g on dog house: 4 or 5 (c) F toy from dog house: 10 (b) F g on toy box: 2 (d) F dog-house from toy box: 9 (e) F dog-house from floor: 8 (f) F floor from dog house: 6 (g) Equal: 1=2, 1=5, 3=6 h) Greatest: 6,3 (h) Greatest: 6,3 (i) Smallest: 1,2,5 > 5. There are two forces on the 2 kg box in the overhead view of the figure below but only one is shown. The figure also shows the acceleration of the box. Find the second force (a) in unit- vector notation and as (b) magnitude and (c) direction. F2F2 F2F2 F2F2 F2F2 Rules to solve Dynamic problems Select a reference system. - Select a reference system. - Make a drawing of the particle system. - Isolate the particles within the system. - Draw the forces that act on each of the isolated bodies. - Find the components of the forces present. - Apply Newtons second law (F=ma) to each isolated particle. 23. An electron with a speed of 1.2x10 7 m/s moves horizontally into a region where a constant vertical force of 4.5x N acts on it. The mass of the electron is m=9.11x kg. Determine the vertical distance the electron is deflected during the time it has moved 30 mm horizontally. d x =0.03m dydydydy F v0v0 FgFg 23. An electron with a speed of 1.2x10 7 m/s moves horizontally into a region where a constant vertical force of 4.5x N acts on it. The mass of the electron is m=9.11x kg. Determine the vertical distance the electron is deflected during the time it has moved 30 mm horizontally. d x =0.03m dydydydy F v0v0 FgFg 13. In the figure below, m block =8.5kg and =30. Find (a) Tension in the cord. (b) Normal force acting on the block. (c) If the cord is cut, find the magnitude of the blocks acceleration. T N FgFg y x T N FgFg y x Friction Related to microscopic interactions of surfacesRelated to microscopic interactions of surfaces Friction is related to force holding surfaces togetherFriction is related to force holding surfaces together Frictional forces are different depending on whether surfaces are static or movingFrictional forces are different depending on whether surfaces are static or moving Kinetic Friction When a body slides over a surface there is a force exerted by the surface on the body in the opposite direction to the motion of the bodyWhen a body slides over a surface there is a force exerted by the surface on the body in the opposite direction to the motion of the body This force is called kinetic frictionThis force is called kinetic friction The magnitude of the force depends on the nature of the two touching surfacesThe magnitude of the force depends on the nature of the two touching surfaces For a given pair of surfaces, the magnitude of the kinetic frictional force is proportional to the normal force exerted by the surface on the body.For a given pair of surfaces, the magnitude of the kinetic frictional force is proportional to the normal force exerted by the surface on the body. Kinetic Friction The kinetic frictional force can be written as Where k is a constant called the coefficient of kinetic friction The value of k depends on the surfaces involved m = 20.0 kg F g = -mg F N = mg FpFpFpFp + + F fr = F N Static Friction A frictional force can arise even if the body remains stationary A block on the floor - no frictional force m F g = mg F N = -mg Static Friction If someone pushes the desk (but it does not move) then a static frictional force is exerted by the floor on the desk to balance the force of the person on the desk m F g = mg F N = -mg F fr = F N FpFp Static Friction If the person pushes with a greater force and still does not manage to move the block, the static frictional force still balances it m F g = mg F N = -mg F fr = F N FpFp Static Friction If the person pushes hard enough, the block will move. The maximum force of static friction has been exceededThe maximum force of static friction has been exceeded m F g = mg F N = -mg F fr = F N FpFpFpFp Static Friction The maximum force of static friction is given byThe maximum force of static friction is given by Static friction can take any value from zero to s F NStatic friction can take any value from zero to s F N In other words Forces on an incline Often when solving problems involving Newtons laws we will need to deal with resolving acceleration due to gravity on an inclined surfaceOften when solving problems involving Newtons laws we will need to deal with resolving acceleration due to gravity on an inclined surface Forces on an incline x y W = mg mgsin mgcos What normal force does the surface exert? Forces on an incline Equilibrium If the car is just stationary on the incline what is the (max) coefficient of static friction?If the car is just stationary on the incline what is the (max) coefficient of static friction? Equilibrium problems ( F = 0) Equilibrium problems ( F = 0) Connected Object problems One problem often posed is how to work out acceleration of a system of masses connected via strings and/or pulleysOne problem often posed is how to work out acceleration of a system of masses connected via strings and/or pulleys for example - Two blocks are fastened to the ceiling of an elevator. Connected masses Two 10 kg blocks are strung from an elevator roof, which is accelerating up at 2 m/s 2.Two 10 kg blocks are strung from an elevator roof, which is accelerating up at 2 m/s 2. m 1 =10kg m 2 =10kg a 2 m/s 2 T1T1T1T1 T2T2T2T2 Connected masses Two 10 kg blocks are strung from an elevator roof, which is accelerating up at 2 m/s 2.Two 10 kg blocks are strung from an elevator roof, which is accelerating up at 2 m/s 2. m 1 =10kg m 2 =10kg a 2 m/s 2 T1T1T1T1 T2T2T2T2 Connected masses Two 10 kg blocks are strung from an elevator roof, which is accelerating up at 2 m/s 2.Two 10 kg blocks are strung from an elevator roof, which is accelerating up at 2 m/s 2. m 1 =10kg m 2 =10kg a 2 m/s 2 T1T1T1T1 T2T2T2T2 m 1 a + m 2 a = T 1 T 2 m 1 g + T 2 m 2 g T 1 = (m 1 +m 2 ) (a+g)=(10+10)(2+9.8)=236N T 2 = m 2 (a+g)=10(2+9.8)=118N Connected masses What is the acceleration of the system below, if T is 1000 N?What is the acceleration of the system below, if T is 1000 N? What is T*?What is T*? m 2 =10kg m 1 =1000kg T T* a Connected masses What is the acceleration of the system below, if T is 1000 N?What is the acceleration of the system below, if T is 1000 N? What is T*?What is T*? m 2 =10kg m 1 =1000kg T T* m 1 a= T -m 2 a a = T / (m 1 +m 2 )=.99m/s 2 T* = m 2 a=9.9N a Connected masses What are T* and a now?What are T* and a now? m 2 =10kg m 1 =1000kg T T* a Connected masses What are T* and a now?What are T* and a now? m 2 =10kg m 1 =1000kg T T* a Connected masses Add in friction: = 0.4 Add in friction: = 0.4 What are T* and a now? What are T* and a now? m 2 =10kg m 1 =1000kg T T* a Connected masses Add in friction: = 0.4 Add in friction: = 0.4 What are T* and a now? What are T* and a now? m 2 =10kg m 1 =1000kg T T* a Motion over pulleys We know that the magnitude and sense of the acceleration should be the same for the whole systemWe know that the magnitude and sense of the acceleration should be the same for the whole system 1000kg 10kg a a T = 30 0 = 0.4 Motion over pulleys We know that the magnitude and sense of the acceleration should be the same for the whole systemWe know that the magnitude and sense of the acceleration should be the same for the whole system 1000kg 10kg a a T = 30 0 = 0.4 . The figure below gives as a function of time t, the force component F x that acts on a 3kg ice block, which can move only along the x axis. At t=0, the block is moving in the positive direction of the axis, with a speed of 3m/s. What are (a) its speed and (b) direction of travel at t=11s? 55. The figure below gives as a function of time t, the force component F x that acts on a 3kg ice block, which can move only along the x axis. At t=0, the block is moving in the positive direction of the axis, with a speed of 3m/s. What are (a) its speed and (b) direction of travel at t=11s? Two bodies, m 1 = 1kg and m 2 =2kg are connected over a massless pulley. The coefficient of kinetic friction between m 2 and the incline is 0.1. The angle of the incline is 20. Calculate: a) Acceleration of the blocks. (b) Tension of the cord. T N m1m1 m2m2 20 m1gm1g T m2gm2g f Two bodies, m 1 = 1kg and m 2 =2kg are connected over a massless pulley. The coefficient of kinetic friction between m 2 and the incline is 0.1. The angle of the incline is 20. Calculate: a) Acceleration of the blocks. (b) Tension of the cord. T N m1m1 m2m2 20 m1gm1g T m2gm2g f The three blocks in the figure below are connected by massless cords and pulleys. Data: m 1 =5kg, m 2 =3kg, m 3 =2kg. Assume that the incline plane is frictionless. (i) Show all the forces that act on each block. (i) Show all the forces that act on each block. (ii) Calculate the acceleration of m 1, m 2, m 3. (ii) Calculate the acceleration of m 1, m 2, m 3. m1m1 m2m2 m3m3 30 m3gm3g T2T2 T1T1 m2gm2g T2T2 T1T1 N F g2x F g2y (iii) Calculate the tensions on the cords. (iv) Calculate the normal force acting on m 2 m1gm1g The three blocks in the figure below are connected by massless cords and pulleys. Data: m 1 =5kg, m 2 =3kg, m 3 =2kg. Assume that the incline plane is frictionless. (i) Show all the forces that act on each block. (i) Show all the forces that act on each block. (ii) Calculate the acceleration of m 1, m 2, m 3. (ii) Calculate the acceleration of m 1, m 2, m 3. m1m1 m2m2 m3m3 30 m3gm3g T2T2 T1T1 m2gm2g T2T2 T1T1 N F g2x F g2y (iii) Calculate the tensions on the cords. (iv) Calculate the normal force acting on m 2 F g2y =m 2 gcos F g2x =m 2 gsin30 m1gm1g The three blocks in the figure below are connected by massless cords and pulleys. Data: m1=5kg, m2=3kg, m3=2kg. Assume that the incline plane is frictionless. (i) Show all the forces that act on each block. (i) Show all the forces that act on each block. (ii) Calculate the acceleration of m1, m2, m3. (ii) Calculate the acceleration of m1, m2, m3. m1m1 m2m2 m3m3 30 m3gm3g T2T2 T1T1 m2gm2g T2T2 T1T1 N F g2x F g2y (iii) Calculate the tensions on the cords. (iv) Calculate the normal force acting on m 2 m1gm1g Block 1: m 1 g-T 1 =m 1 a Block 2: m 2 g(sin30) +T 1 -T 2 =m 2 a Block 3: T 2 -m 3 g = m 3 a Adding (1)+(2)+(3): g(m m 2 -m 3 )=a(m 1 +m 2 +m 3 ) a=4.41m/s2 (ii) T 1 =m 1 (g-a)= 5kg(9.8 m/s m/s 2 ) = 26.95N (iii) T 2 =m 3 (g+a)= 2kg(9.8 m/s m/s 2 )= 28.42N (iv) N 2 = F g2y = m 2 gcos30 = 25.46N 1B. (a) What should be the magnitude of F in the figure below if the body of mass m=10kg is to slide up along a frictionless incline plane with constant acceleration a=1.98 m/s 2 ? (b) What is the magnitude of the Normal force? y x 30 F N FgFg 20 1B. (a) What should be the magnitude of F in the figure below if the body of mass m=10kg is to slide up along a frictionless incline plane with constant acceleration a=1.98 m/s 2 ? (b) What is the magnitude of the Normal force? y x 30 F N FgFg 20 Three forces, given by,, and, act on an object to give it an acceleration of magnitude 3.75 m/s 2. (a) What is the direction of the acceleration? (b) What is the mass of the object? (c) If the object is initially at rest, what is its speed after 10.0 s? (d) What are the velocity components of the object after 10.0 s? where represents the direction of a at Three forces, given by,, and, act on an object to give it an acceleration of magnitude 3.75 m/s 2. (a) What is the direction of the acceleration? (b) What is the mass of the object? (c) If the object is initially at rest, what is its speed after 10.0 s? (d) What are the velocity components of the object after 10.0 s? (a) counterclockwise from the x-axis (b) (d) (c) A bag of cement of weight 325 N hangs from three wires as suggested in Figure. Two of the wires make angles 1 = 60.0 and 2 = 25.0 with the horizontal. If the system is in equilibrium, find the tensions T 1, T 2, and T 3 in the wires. A block is given an initial velocity of 5.00 m/s up a frictionless 20.0 incline. How far up the incline does the block slide before coming to rest? After it leaves your hand, the blocks speed changes only because of one component of its weight: A woman at an airport is towing her 20.0-kg suitcase at constant speed by pulling on a strap at an angle above the horizontal. She pulls on the strap with a 35.0-N force, and the friction force on the suitcase is 20.0 N. Draw a free-body diagram of the suitcase. (a) What angle does the strap make with the horizontal? (b) What normal force does the ground exert on the suitcase? (a) (b) Three objects are connected on the table as shown in Figure. The table is rough and has a coefficient of kinetic friction of The objects have masses 4.00 kg, 1.00 kg and 2.00 kg, as shown, and the pulleys are frictionless. Draw free-body diagrams of each of the objects. (a) Determine the acceleration of each object and their directions. (b) Determine the tensions in the two cords. For m 1,