advertisement

Lecture Outline Chapter 5 Physics, 4th Edition James S. Walker Copyright © 2010 Pearson Education, Inc. Chapter 5 Newton’s Laws of Motion Copyright © 2010 Pearson Education, Inc. Units of Chapter 5 • Dynamics • Force and Mass • Newton’s 1st, 2nd and 3rd Laws of Motion • The Vector Nature of Forces: Forces in Two Dimensions • Weight • Normal Forces Copyright © 2010 Pearson Education, Inc. Newton’s Concepts 1) m (mass): How much matter something contains 2) v (velocity): A body’s speed and direction 3) a (acceleration): The change in a body’s velocity per unit time 4) F (force): any external agent that causes a change in the motion of a free body Copyright © 2010 Pearson Education, Inc. 5-1 Force and Mass Force: push or pull Force is a vector – it has magnitude and direction Copyright © 2010 Pearson Education, Inc. 5-1 Force and Mass Mass is the measure of how hard it is to change an object’s velocity. Mass can also be thought of as a measure of the quantity of matter in an object. Copyright © 2010 Pearson Education, Inc. 5-2 Newton’s First Law of Motion If you stop pushing an object, does it stop moving? Only if there is friction! In the absence of any net external force, an object will keep moving at a constant speed in a straight line, or remain at rest. This is also known as the law of inertia. Copyright © 2010 Pearson Education, Inc. 5-2 Newton’s First Law of Motion In order to change the velocity of an object – magnitude or direction – a net force is required. An inertial reference frame is one in which the first law is true. Accelerating reference frames are not inertial. Copyright © 2010 Pearson Education, Inc. 5-3 Newton’s Second Law of Motion Observations show that the acceleration of an object is proportional to the net force exerted on it and inversely proportional to its mass: Or, more familiarly, Copyright © 2010 Pearson Education, Inc. 5-3 Newton’s Second Law of Motion An object may have several forces acting on it; the acceleration is due to the net force: (5-1) Copyright © 2010 Pearson Education, Inc. 5-3 Newton’s Second Law of Motion An object may have several forces acting on it; the acceleration is due to the net force: (5-1) In terms of vector components : ∑F x = ma x Copyright © 2010 Pearson Education, Inc. ∑F y = ma y ∑F z = ma z 5-3 Newton’s Second Law of Motion Free-body diagrams: A free-body diagram shows every force acting on an object. • Sketch the forces • Isolate the object of interest • Choose a convenient coordinate system • Resolve the forces into components • Apply Newton’s second law to each coordinate direction Copyright © 2010 Pearson Education, Inc. 5-3 Newton’s Second Law of Motion Example of a free-body diagram: Copyright © 2010 Pearson Education, Inc. 5-3 Newton’s Second Law of Motion Example 5-1: mboat = 753 kg, F = 80.5N. What are magnitudes and directions of accelerations in (a) and (b) Copyright © 2010 Pearson Education, Inc. 5-3 Newton’s Second Law of Motion Example 5-1: mboat = 753 kg, F = 80.5N. What are magnitudes and directions of accelerations in (a) and (b) (a) Create free - body diagram, isolate source of interest and sketch the external forces on this body and their components. 3(80.5N) ∑ Fx = ma x ⇒ FL + FM + FC = ma x ⇒ a x = 753kg = 0.321m/s2 (b) ∑ Fx = ma x ⇒ −FL + FM − FC = ma x ⇒ a x = Copyright © 2010 Pearson Education, Inc. € (−80.5N) = −0.107m/s2 753kg 5-3 Newton’s Second Law of Motion Example 5-2: mball = 0.15 kg, v = 90 mi/h and Δx = 2m. Estimate the force on the ball. Copyright © 2010 Pearson Education, Inc. 5-3 Newton’s Second Law of Motion Example 5-2: mball = 0.15 kg, v = 90 mi/h and Δx = 2m. Estimate the force on the ball. Equation of motion along x - direction : ⎛ mi ⎞ 2 ⎛ 1609.3m ⎞ 2 ⎟ ⎜ 90 ⎟ ⎜90 2 v ⎠ ⎠ ⎝ ⎝ h 3600s = = = 400m/s2 v 2 = v 20 + 2a x Δx = 2aΔx ⇒ a = 2Δx 2(2m) 4m Newton's Second Law : ∑F x € = ma x ⇒ Fx = (0.15kg)(400m/s2 ) = 60N Copyright © 2010 Pearson Education, Inc. 5-4 Newton’s Third Law of Motion Forces always come in pairs, acting on different objects: If object 1 exerts a force on object 2, then object 2 exerts a force – on object 1. These forces are called action-reaction pairs. NOTE: The action and reaction forces would appear in the free-body diagrams of different objects so they do not cancel out. Copyright © 2010 Pearson Education, Inc. 5-4 Newton’s Third Law of Motion Some action-reaction pairs: Copyright © 2010 Pearson Education, Inc. 5-4 Newton’s Third Law of Motion Example 5-3: m1 = 150 kg, m2 = 250 kg. The force of person in canoe 1 on canoe 2 is F = 46N. (a) acceleration of each canoe? (b) separation of canoes after t = 1.2s? Copyright © 2010 Pearson Education, Inc. 5-4 Newton’s Third Law of Motion Example 5-3: m1 = 150 kg, m2 = 250 kg. The force of person in canoe 1 on canoe 2 is F = 46N. (a) acceleration of each canoe? (b) separation of canoes after t = 1.2s? Apply Newton's second law on each canoe : canoe 1: canoe 2 : ∑F ∑F 1x = m1a1 2x = m 2a 2 Apply Newton's third Law : F1 = (- 46 N)xˆ F2 = (46 N)xˆ F1 −46 N F 46N = = −0.31m/s2 a 2 = 2 = = 0.18m/s2 m1 150 kg m2 250 kg Canoe 1 with the smallest mass accelerates more. a1 = 1 2 1 a1t = (−0.31m/s2 )(1.2s) 2 = −0.22m 2 2 Δx = x 2 - x1 = 0.35m x1 = Copyright © 2010 Pearson Education, Inc. € x2 = 1 2 1 a 2t = (0.18m/s2 )(1.2s) 2 = 0.13m 2 2 5-4 Newton’s Third Law of Motion Contact forces: The force exerted by one box on the other is different depending on which one you push. Copyright © 2010 Pearson Education, Inc. 5-4 Newton’s Third Law of Motion Apply Newton's 2nd Law on both boxes : F F = (m1 + m2 )a ⇒ a = (m1 + m2 ) case1: Force on small box 2. Box 2 exerts a force F1 on box 1 and box 1 exerts an equal and opposite force F2 on box 2. Contact force = F1 = F2 Apply Newton's 2nd law on box 1 (only one external force in x direction on box 1) : m1F Contact Force : F1 = m1a = (m1 + m2 ) Copyright © 2010 Pearson Education, Inc. € 5-4 Newton’s Third Law of Motion Apply Newton's 2nd Law on both boxes : F = (m1 + m2 )a ⇒ a = F (m1 + m2 ) case2 : Force on large box 1. Box 2 exerts a force F1 on box 1 and box 1 exerts an equal and opposite force F2 on box 2. Contact force = F1 = F2 Apply Newton's 2nd law on box 2 (only one external force in x direction on box 2) : m2F Contact Force : F2 = m2a = (m1 + m2 ) The contact force is larger when you push on the smaller box! Copyright © 2010 Pearson Education, Inc. € Newton’s Laws of Motion 1) An object remains at rest, or moves in a straight line at a constant speed, unless acted upon by a net outside force. 2) A body’s acceleration is proportional to the net outside force acting on the object, and will be in the direction of that force. Its resistance to acceleration depends on its mass. In equation form, this is F=ma m = mass of object (kg) a = acceleration of object (m/s2) F = net external force (kg m/s2 = Newton) 3) Whenever one object exerts a force on a second object, the second object exerts an equal and opposite force on the first object. Copyright © 2010 Pearson Education, Inc. 5-5 The Vector Nature of Forces: Forces in Two Dimensions The easiest way to handle forces in two dimensions is to treat each dimension separately, as we did for kinematics. Copyright © 2010 Pearson Education, Inc. 5-5 Forces in Two Dimensions Example 5-5: Jack and Jill. mp = 1.30 kg F1 = 7.0 N, F2 = 11 N, W = 12.8 N and φ=28°. (a) θ = ? For ax = 0 (b) ay =? Copyright © 2010 Pearson Education, Inc. 5-5 Forces in Two Dimensions Example 5-5: Jack and Jill. mp = 1.30 kg F1 = 7.0 N, F2 = 11 N, W = 12.8 N and φ=28°. (a) θ = ? For ax = 0 (b) ay =? Use component form of Newton's seconds law : ∑F x = ma x ∑F y = ma y a x = 0 ⇒ ∑ Fx = 0 ⇒ −F1sinθ + F2sin28 = 0 ⇒ sinθ = F2 sin28 F1 θ = 48 F1cosθ + F2cos28 - W ∑ Fy = F1cosθ + F2cos28 - W ⇒ a y = mp (7.0 N)cos48 + (11.0 N)cos28 -12.8 N ay = = 1.2m/s2 1.30 kg Copyright © 2010 Pearson Education, Inc. 5-6 Weight The weight of an object on the Earth’s surface is the gravitational force exerted on it by the Earth. Copyright © 2010 Pearson Education, Inc. 5-6 Weight Example 5-6 : Where’s the fire? mf = 97 kg, h = 3.0m, t = 1.2s. What is the upward force from the pole on the fireman? Copyright © 2010 Pearson Education, Inc. € 5-6 Weight Example 5-6 : Where’s the fire? mf = 97 kg, h = 3.0m, t = 1.2s. What is the upward force from the pole on the fireman? Apply Newtons' Second Law : ∑F y = mf a y ⇒ F − W = mf a y ⇒ F - mf g = mf a y 1 2h 2(3.0m) 2 y = h + a yt 2 ⇒ a y = − 2 = − 2 = −4.2m/s 2 t (1.2s) Note that the total acceleration of the fireman is negative because it is pointed in the - y direction. F = mf (g + a y ) = 97 kg(9.81m/s2 − 4.2m/s2 ) = 540N Copyright © 2010 Pearson Education, Inc. 5-6 Weight Apparent weight: Your perception of your weight is based on the contact forces between your body and your surroundings. If your surroundings are accelerating, your apparent weight may be more or less than your actual weight. Copyright © 2010 Pearson Education, Inc. 5-6 Weight Apparent weight : The force from the floor on person. Newton's Second Law on person : Wa - W = ma y ⇒ Wa = m(g + a y ) Case 1: At rest a y = 0 and the apparent weight is equal to our weight Case 2 : For elevator accelerating upward ay > 0 and Wa > W Case 3 : For elevator accelerating downward ay < 0 and Wa < W € Copyright © 2010 Pearson Education, Inc. 5-6 Weight Example 5-7 : Don’t buy Salmon in an upward accelerating elevator. What is the scale weight (a) at rest (b) upward acceleration of 2.5m/s2 (c) downward acceleration of 3.2m/s2. Copyright © 2010 Pearson Education, Inc. 5-6 Weight Example 5-7 : Don’t buy Salmon in an upward accelerating elevator ∑F y = ma y Fy = Wa − W = Wa − mg Wa = m(g + a y )yˆ (a) At rest a y = 0 and Wa = mgyˆ = 5.0kg(9.81m/s2 )yˆ = 49Nyˆ (b) Wa = 5.0kg(9.81m/s2 + 2.5m/s2 )yˆ = 62Nyˆ (your're paying too much) (c) Wa = 5.0kg(9.81m/s2 - 3.2m/s2 )yˆ = 33Nyˆ € Copyright © 2010 Pearson Education, Inc. Wa = m(g + a) For a = - g one has zero apparent weight Copyright © 2010 Pearson Education, Inc. 5-7 Normal Forces The normal force is the force exerted by a surface on an object. Copyright © 2010 Pearson Education, Inc. 5-7 Normal Forces The normal force may be equal to, greater than, or less than the weight. Copyright © 2010 Pearson Education, Inc. 5-7 Normal Forces Example 5-8: Ice Block F1 = 13 N, F2 = 11 N (a) What is the acceleration of the ice? (b) What is the normal force on the ice? Copyright © 2010 Pearson Education, Inc. 5-7 Normal Forces Example 5-8: Ice Block F1 = 13 N, F2 = 11 N, m = 6.0 kg (a) What is the acceleration of the ice? (b) What is the normal force on the ice? ∑F = ma x ∑F = F1x + F2x = F1 cos60 − F2 cos 30 = (13N)0.5 - (11N)0.87 = 6.5N - 9.5N = -3N x x ∑F y = ma y (a) ma x = - 3 N ⇒ a x = ∑F y -3 N = −0.5m/s2 a = - 0.5 m/s2 xˆ 6.0kg = 0 ⇒ FN − W − F1y − F2y = 0 ⇒ FN − mg − F1 sin60 − F2 sin 30 = 0 FN = mg + F1 sin60 + F2 sin 30 = (6.0kg)(9.81m/s2 ) + (13 N)(0.87) + (11 N)(0.5) FN = 75N FN = 75N yˆ € Copyright © 2010 Pearson Education, Inc. 5-7 Normal Forces The normal force is always perpendicular to the surface. Copyright © 2010 Pearson Education, Inc. 5-7 Normal Forces Example 5-9: Toboggan, m and θ known (a) What is the acceleration of the child? (b) What is the normal force on the child? Copyright © 2010 Pearson Education, Inc. 5-7 Normal Forces Example 5-9: Toboggan, m and θ known (a) What is the acceleration of the child? (b) What is the normal force on the child? ∑F x = ma x ∑F y = ma y Wx mgsin θ ∑ Fx = Wx = ma x ⇒ a x = m = m = gsinθ ∑ Fy = FN − Wcosθ = ma y = 0 ⇒ FN = mgcosθ FN = mgcosθyˆ € Copyright © 2010 Pearson Education, Inc. Summary of Chapter 5 • Force: a push or pull • Mass: measures the difficulty in accelerating an object • Newton’s first law: if the net force on an object is zero, its velocity is constant • Inertial frame of reference: one in which the first law holds • Newton’s second law: • Free-body diagram: a sketch showing all the forces on an object Copyright © 2010 Pearson Education, Inc. Summary of Chapter 5 • Newton’s third law: If object 1 exerts a force on object 2, then object 2 exerts a force – on object 1. • Contact forces: an action-reaction pair of forces produced by two objects in physical contact • Forces are vectors • Newton’s laws can be applied to each component of the forces independently • Weight: gravitational force exerted by the Earth on an object Copyright © 2010 Pearson Education, Inc. Summary of Chapter 5 • On the surface of the Earth, W = mg • Apparent weight: force felt from contact with a floor or scale • Normal force: force exerted perpendicular to a surface by that surface • Normal force may be equal to, lesser than, or greater than the object’s weight Copyright © 2010 Pearson Education, Inc.