Motion and Force - PHY111 - Chapter III - New Mansoura University - 18/10/2024 PDF

Motion and Force PHY111 Chapter III Hamdi M. Abdelhamid Physics Department Faculty of Science 18/10/2024 New Mansoura University Introduction Sir Isaac Newton ❑ 1642 – 1727 ❑ Formulated basic laws of...

Motion and Force PHY111 Chapter III Hamdi M. Abdelhamid Physics Department Faculty of Science 18/10/2024 New Mansoura University Introduction Sir Isaac Newton ❑ 1642 – 1727 ❑ Formulated basic laws of mechanics ❑ Discovered Law of Universal Gravitation ❑ Invented form of calculus ❑ Many observations dealing with light and optics 18/10/2024 22 Kinematics and Dynamics ❑ Kinematics: Describing object’s motion by answering: When? Where? How fast? How far? How long? without asking: Why is object moving in a certain way? 18/10/2024 33 Kinematics and Dynamics ❑ Dynamics: Describing object’s motion by answering: Why is the object moving in a certain way? What causes the object to change its velocity? ❑ Dynamics studies motion on a deeper level than kinematics: it studies the causes of changes in objects’ motion! 18/10/2024 44  Dynamics ❑ Describes the relationship between the motion of objects in our everyday world and the forces acting on them ❑ Language of Dynamics ◼ Force: The measure of interaction between two objects (pull or push). It is a vector quantity – it has a magnitude and direction ◼ Mass: The measure of how difficult it is to change object’s velocity (sluggishness or inertia of the object) 18/10/2024 55 Forces ❑ The measure of interaction between two objects (pull or push) ❑ Vector quantity: has magnitude and direction ❑ May be a contact force or a field force ◼ Contact forces result from physical contact between two objects ◼ Field forces act between disconnected objects ◼ Also called “action at a distance” 18/10/2024 66  Vector Nature of Force ❑ Vector force: has magnitude and direction ❑ Net Force: a resultant force acting on object      Fnet =  F = F1+ F2 + F3 +...... ❑ You must use the rules of vector addition to obtain the net force on an object 18/10/2024 77 Newton’s First Law ❑ An object at rest tends to stay at rest and an object in motion tends to stay in motion with the same speed and in the same direction unless acted upon by an unbalanced force ❑ An object at rest remains at rest as long as no net force acts on it ❑ An object moving with constant velocity continues to move with the same speed and in the same direction (the same velocity) as long as no net force acts on it - “Keep on doing what it is doing” 18/10/2024 88 Newton’s First Law ❑ An object at rest tends to stay at rest and an object in motion tends to stay in motion with the same speed and in the same direction unless acted upon by an unbalanced force ❑ When forces are balanced, the acceleration of the objection is zero ◼ Object at rest: v = 0 and a = 0 ◼ Object in motion: v = v0 and a = 0 ❑ The net force is defined as the vector sum of all the external forces exerted on the object. If the net force is zero, forces are balanced.      Fnet =  F = F1 + F2 + F3 +...... = 0 18/10/2024 99 Mass and Inertia ❑ Every object continues in its state of rest, or uniform motion in a straight line, unless it is compelled to change that state by unbalanced forces impressed upon it ❑ Inertia is a property of objects to resist changes in motion! ❑ Mass is a measure of the amount of inertia. ❑ Mass is a measure of the resistance of an object to changes in its velocity ❑ Mass is an inherent property of an object ❑ Scalar quantity and SI unit: kg 18/10/2024 10 10 Mass and Inertia ❑ Masses can be defined in terms of the accelerations produced by a given force acting on them m1 a2  m2 a1 ❑ The magnitude of the acceleration acting on an object is inversely proportional to its mass. 18/10/2024 11 11 Mass vs. Weight ❑ Mass and weight are two different quantities. ❑ Weight is equal to the magnitude of the gravitational force exerted on the object. ❑ Weight will vary with location. Example: wearth = 180 lb; wmoon ~ 30 lb mearth = 2 kg; mmoon = 2 kg 18/10/2024 12 12 Newton’s Second Law The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass   a= F = Fnet m m    Fnet =  F = ma 18/10/2024 13 13  Units of Force Newton’s second law:    Fnet =  F = ma SI unit of force is a Newton (N) kg m 1 N 1 2 s US Customary unit of force is a pound (lb) 1 N = 0.225 lb Weight, also measured in lbs. is a force (mass x acceleration). What is the acceleration in that case? 18/10/2024 14 14 More about Newton’s 2nd Law ❑ You must be certain about which body we are applying it to ❑ Fnet must be the vector sum of all the forces that act on that body ❑ Only forces that act on that body are to be included in the vector sum ❑ Net force component along an axis gives rise to the acceleration along that same axis Fnet , x = max Fnet , y = ma y 18/10/2024 15 15 Simple Problem One or two forces act on a puck that moves over frictionless ice along an x axis, in one-dimensional motion. The puck's mass is m = 0.20 kg. Forces F1 and F2 and are directed along the x axis and have magnitudes F1 = 4.0 N and F2 = 2.0 N. Force F3 is directed at angle  = 30° and has magnitude F3 = 1.0 N. In each situation, what is the acceleration of the puck? a) F1 = ma x F1 4.0 N ax = = = 20 m/s 2 m 0.2 kg b) F1 − F2 = ma x F1 − F2 4.0 N − 2.0 N ax = = = 10 m/s 2 m 0.2 kg c) F3, x − F2 = ma x F3, x = F3 cos  Fnet , x = max F3 cos  − F2 1.0 N cos 30 − 2.0 N ax = = = −5.7 m/s 2 m 0.2 kg 18/10/2024 16 16 Gravitational Force ❑ Gravitational force is a vector ❑ Expressed by Newton’s Law of Universal Gravitation: mM Fg = G 2 R ✓G – gravitational constant ✓M – mass of the Earth ✓m – mass of an object ✓R – radius of the Earth ❑ Direction: pointing downward 18/10/2024 17 17 Weight ❑ The magnitude of the gravitational force acting on an object of mass m near the Earth’s surface is called the weight w of the object: w = mg ❑ g can also be found from the Law of Universal Gravitation ❑ Weight has a unit of N mM Fg = G 2 w = Fg = mg R M g = G 2 = 9.8 m/s 2 R ❑ Weight depends upon location R = 6,400 km 18/10/2024 18 18 Normal Force ❑ Force from a solid surface which keeps object from falling through ❑ Direction: always perpendicular to the surface ❑ Magnitude: depends on situation N − Fg = ma y N − mg = ma y N = mg 18/10/2024 19 19 Tension Force: T ❑ A taut rope exerts forces on whatever holds its ends ❑ Direction: always along the cord (rope, cable, string ……) and away from the object T1 ❑ Magnitude: depend on situation T1 = T = T2 T2 18/10/2024 20 20 Gravitational Mass vs. Inertial Mass ❑ In Newton’s Laws, the mass is the inertial mass and measures the resistance to a change in the object’s motion. ❑ In the gravitational force, the mass is determining the gravitational attraction between the object and the Earth. ❑ Experiments show that gravitational mass and inertial mass have the same value. Inertial mass is measured by measuring an object's resistance to changes in velocity; while gravitational mass describes the force on an object in a gravitational field. 18/10/2024 21 21 Newton’s Third Law ❑ If object 1 and object 2 interact, the force exerted by object 1 on object 2 is equal in magnitude but opposite in direction to the force exerted by object 2 on object 1   Fon A = − Fon B ❑ Equivalent to saying a single isolated force cannot exist 18/10/2024 22 22 Newton’s Third Law cont. ❑ F12 may be called the action force and F21 the reaction force ❑ Actually, either force can be the action or the reaction force ❑ The action and reaction forces act on different objects 18/10/2024 23 23 Action and Reaction Force ❑ If a bird collides with the windshield of a fast- moving plane, which experiences an impact force with a larger magnitude? A. The bird. B. The plane. C. The same force is experienced by both. D. Not enough information is given 18/10/2024 24 24 Action and Reaction Force ❑ Which experiences greater acceleration? A. The bird. B. The plane. C. The same acceleration is experienced by both the bird and plane. D. Not enough information is given 18/10/2024 25 25 Some Action-Reaction Pairs ❑ An apple rests on a table. Identify the forces that act on it and the action-reaction pairs. 18/10/2024 26 26  Applying Newton’s Third Law II ❑ A person pulls on a block across the floor. Identify the action-reaction pairs. 18/10/2024 27 27 A paradox? ❑ If an object pulls back on you just as hard as you pull on it, how can it ever accelerate? 18/10/2024 28 28 Free Body Diagram ❑ The most important step in solving problems involving F hand on book Newton’s Laws is to draw the free body diagram ❑ Be sure to include only the forces acting on the object of interest F Earth on book ❑ Include any field forces acting on the object ❑ Do not assume the normal force equals the weight 18/10/2024 29 29 Objects in Equilibrium ❑ Objects that are either at rest or moving with constant velocity are said to be in equilibrium ❑ Acceleration of an object can be modeled as zero:  a=0 ❑ Mathematically, the net force acting on the object is zero  F =0 ❑ Equivalent to the set of component equations given by F x =0 F y =0 18/10/2024 30 30 Equilibrium, Example 1 ❑ A lamp is suspended from a chain of negligible mass ❑ The forces acting on the lamp are ▪ the downward force of gravity ▪ the upward tension in the chain ❑ Applying equilibrium gives F y = 0 → T − Fg = 0 → T = Fg 18/10/2024 31 31 Equilibrium, Example 2 ❑ A traffic light weighing 100 N hangs from a vertical cable tied to two other cables that are fastened to a support. The upper cables make angles of 37° and 53° with the horizontal. Find the tension in each of the three cables. ❑ Conceptualize the traffic light ◼ Assume cables don’t break ◼ Nothing is moving ❑ Categorize as an equilibrium problem ◼ No movement, so acceleration is zero ◼ Model as an object in equilibrium 18/10/2024 Fx =0 F y =0 32 32 Equilibrium, Example 2 ❑ Need 2 free-body diagrams Apply equilibrium equation to light  Fy = 0 → T3 − Fg = 0 T3 = Fg = 100 N Apply equilibrium equations to knot  x 1x 2 x 1 F = T + T = −T cos 37  + T2 cos 53 =0 F y = T1 y + T2 y + T3 y = T1 sin 37  + T2 sin 53 − 100 N = 0  cos 37   T2 = T1     = 1.33T1  cos 53  T1 = 60 N T2 = 1.33T1 = 80 N 18/10/2024 33 33 Accelerating Objects ❑ If an object that can be modeled as a particle experiences an acceleration, there must be a nonzero net force acting on it ❑ Draw a free-body diagram ❑ Apply Newton’s SecondLaw in component form   F = ma F x = ma x F y = ma y 18/10/2024 34 34  Accelerating Objects, Example 1 ❑ A man weighs himself with a scale in an elevator. While the elevator is at rest, he measures a weight of 800 N. What weight does the scale read if the elevator accelerates upward at 2.0 m/s2? a = 2.0 m/s2 What weight does the scale read if the elevator accelerates downward at 2.0 m/s2? a = - 2.0 m/s2 ❑ Upward: F y = N − mg = ma N N = mg + ma = m( g + a ) N = 80(2.0 + 9.8) = 944 N w 800 N N m= = = 80 N g 9.8 m/s 2 N  mg ❑ Downward:N = 80(−2.0 + 9.8) = 624 N N  mg mg mg 18/10/2024 35 35 Forces of Friction ❑ When an object is in motion on a surface or through a viscous medium, there will be a resistance to the motion. This resistance is called the force of friction ❑ This is due to the interactions between the object and its environment ❑ We will be concerned with two types of frictional force ▪ Force of static friction: fs ▪ Force of kinetic friction: fk ❑ Direction: opposite the direction of the intended motion ▪ If moving: in direction opposite the velocity ▪ If stationary, in direction of the vector sum of other forces 18/10/2024 36 36 Forces of Friction: Magnitude ❑ Magnitude: Friction is proportional to the normal force Static friction: Ff = F  μsN Kinetic friction: Ff = μkN μ is the coefficient of friction ❑ The coefficients of friction are nearly independent of the area of contact (why?) 18/10/2024 37 37 Some Coefficients of Friction 18/10/2024 38 38 Static Friction ❑ Static friction acts to keep the object from moving ❑ If F increases, so does ƒ s ❑ If F decreases, so does ƒ s ❑ ƒs  µs N ❑ Remember, the equality holds when the surfaces are on the verge of slipping 18/10/2024 39 39 Kinetic Friction ❑ The force of kinetic friction acts when the object is in motion ❑ Although µk can vary with speed, we shall neglect any such variations ❑ ƒk = µk N 18/10/2024 40 40  Explore Forces of Friction ❑ Vary the applied force ❑ Note the value of the frictional force ◼ Compare the values ❑ Note what happens when the can starts to move 18/10/2024 41 41 Hints for Problem-Solving ❑ Read the problem carefully at least once ❑ Draw a picture of the system, identify the object of primary interest, and indicate forces with arrows ❑ Label each force in the picture in a way that will bring to mind what physical quantity the label stands for (e.g., T for tension) ❑ Draw a free-body diagram of the object of interest, based on the labeled picture. If additional objects are involved, draw separate free- body diagram for them ❑ Choose a convenient coordinate system for each object ❑ Apply Newton’s second law. The x- and y-components of Newton second law should be taken from the vector equation and written individually. This often results in two equations and two unknowns ❑ Solve for the desired unknown quantity, and substitute the numbers Fnet , x = max Fnet , y = ma y 18/10/2024 42 42 Objects in Equilibrium ❑ Objects that are either at rest or moving with constant velocity are said to be in equilibrium ❑ Acceleration of an object can be modeled as zero: ❑ Mathematically, the net force acting on the object is zero ❑ Equivalent to the set of component equations given by F x =0 F y =0 18/10/2024 43 43 Equilibrium, Example 1 ❑ What is the smallest value of the force F such that the 2.0-kg block will not slide down the wall? The coefficient of static friction between the block and the wall is 0.2. ? f N F F mg 18/10/2024 44 44 Accelerating Objects ❑ If an object that can be modeled as a particle experiences an acceleration, there must be a nonzero net force acting on it ❑ Draw a free-body diagram ❑ Apply Newton’s Second Law in component form    F = ma F x = ma x F y = ma y 18/10/2024 45 45 Inclined Plane ❑ Suppose a block with a mass of 2.50 kg is resting on a ramp. If the coefficient of static friction between the block and ramp is 0.350, what maximum angle can the ramp make with the horizontal before the block starts to slip down? 18/10/2024 46 46 Inclined Plane ❑ Newton 2nd law: F x = mg sin  −  s N = 0 F y = N − mg cos  = 0 ❑ Then N = mg cos  F y = mg sin  −  s mg cos  = 0 ❑ So tan  = s = 0.350  = tan −1 (0.350) = 19.3 18/10/2024 47 47 Multiple Objects ❑ A block of mass m1 on a rough, horizontal surface is connected to a ball of mass m2 by a lightweight cord over a lightweight, frictionless pulley as shown in figure. A force of magnitude F at an angle θ with the horizontal is applied to the block as shown and the block slides to the right. The coefficient of kinetic friction between the block and surface is μk. Find the magnitude of acceleration of the two objects. 18/10/2024 48 48 Multiple Objects ❑ m1: F x = F cos  − f k − T = m1a x = m1a F y = N + F sin  − m1 g = 0 ❑ m2: T = m2 (a + g ) N = m1 g − F sin  f k = k N = k ( m1 g − F sin  ) F cos − k ( m1 g − F sin  ) − m2 (a + g ) = m1a F (cos  +  k sin  ) − (m2 +  k m1 ) g a= m1 + m2 18/10/2024 49 49  Uniform Circular Motion: Definition Uniform Circular Motion: Definition Uniform circular motion Constant speed, or, Motion along a circle: constant magnitude of velocity Changing direction of velocity 18/10/2024 50 50  Uniform Circular Motion: Observations ❑ Object moving along a curved path with constant speed ◼ Magnitude of velocity: same ◼ Direction of velocity: changing ◼ Velocity : changing ◼ Acceleration is NOT zero! ◼ Net force acting on an object is NOT zero   ◼ “Centripetal force” Fnet = ma 18/10/2024 51 51 Uniform Circular Motion ❑ Magnitude: vi        v v f − vi Δv = vf - vi r = rf − ri a= = t t vf v r vr vi y B = so, v = A v r r vf v r v v 2 Δr R = = t t r r ri rf v v 2 ar = = O t r x ❑ Direction: Centripetal 18/10/2024 52 52 Uniform Circular Motion ❑ Velocity:   ac ⊥ v ◼ Magnitude: constant v ◼ The direction of the velocity is tangent to the circle ❑ Acceleration: v2 ac = ◼ Magnitude: r ◼ directed toward the center of the circle of motion ❑ Period: ◼ time interval required for one complete revolution of the particle 2r T= v 18/10/2024 53 53 Centripetal Force ❑ Acceleration: v2     ac = ac ⊥ v Fnet ⊥ v ◼ Magnitude: r ◼ Direction: toward the center of the v2 circle of motion ac = r ❑ Force: Fnet Fnet ◼ Start from Newton’s 2nd Law   Fnet = ma Fnet ◼ Magnitude: mv 2 Fnet = mac = r   ◼ Direction: toward the center of the ac || Fnet circle of motion 18/10/2024 54 54  What provides Centripetal Force ? ❑ Centripetal force is not a new kind of force ❑ Centripetal force refers to any force that keeps an object following a circular path mv 2 Fc = mac = r ❑ Centripetal force is a combination of ◼ Gravitational force mg: downward to the ground ◼ Normal force N: perpendicular to the surface ◼ Tension force T: along the cord and away from object ◼ Static friction force: fsmax = µsN 18/10/2024 55 55 What provides Centripetal Force ? Fnet = N − mg = ma v2 N = mg + m r N Fnet = T = ma a v 2 mv T= r mg 18/10/2024 56 56  Problem Solving Strategy ❑ Draw a free body diagram, showing and labeling all the forces acting on the object(s) ❑ Choose a coordinate system that has one axis perpendicular to the circular path and the other axis tangent to the circular path ❑ Find the net force toward the center of the circular path (this is the force that causes the centripetal acceleration, FC) ❑ Use Newton’s second law ◼ The directions will be radial, normal, and tangential ◼ The acceleration in the radial direction will be the centripetal acceleration ❑ Solve for the unknown(s) 18/10/2024 57 57 The Conical Pendulum ❑ A small ball of mass m = 5 kg is suspended from a string of length L = 5 m. The ball revolves with constant speed v in a horizontal circle of radius r = 2 m. Find an expression for v and a. T θ mg 18/10/2024 58 58 The Conical Pendulum ❑ Find v and a m = 5 kg L =5m r =2m mv 2 T sin  = F y = T cos  − mg = 0 r T cos  = mg T cos  = mg v2 tan  = 2 mv  Fx = T sin  = r gr r v = rg tan  sin  = = 0.4 L v = Lg sin  tan  = 2.9 m/s r tan  = = 0.44 v2 L2 − r 2 a= = g tan  = 4.3 m/s 2 r 18/10/2024 59 59 Level Curves ❑ A 1500 kg car moving on a flat, horizontal road negotiates a curve as shown. If the radius of the curve is 35.0 m and the coefficient of static friction between the tires and dry pavement is 0.523, find the maximum speed the car can have and still make the turn successfully. 18/10/2024 60 60 Level Curves ❑ The force of static friction directed toward the center of the curve keeps the car moving in a circular path. 2 vmax f s ,max =  s N = m r  Fy = N − mg = 0 N = mg  s Nr  s mgr vmax = = =  s gr m m = (0.523)(9.8m / s 2 )(35.0m) = 13.4m / s 18/10/2024 61 61 Banked Curves ❑ A car moving at the designated speed can negotiate the curve. Such a ramp is usually banked, which means that the roadway is tilted toward the inside of the curve. Suppose the designated speed for the ramp is to be 13.4 m/s and the radius of the curve is 35.0 m. At what angle should the curve be banked? 18/10/2024 62 62 Banked Curves v = 13.4 m/s r = 35.0 m mv 2  Fr = n sin  = mac = r  Fy = n cos  − mg = 0 n cos  = mg v2 tan  = rg 13.4 m/s  = tan −1 ( ) = 27. 6  (35.0 m)(9.8 m/s 2 ) 18/10/2024 63 63