Chapter 4 Forces and Newton's Laws of Motion, Slides of Law

Download Chapter 4 Forces and Newton's Laws of Motion and more Slides Law in PDF only on Docsity! Chapter 4 Forces and Newton’s Laws of Motion continued 4.3 Newton’s Second Law of Motion Mathematically, the net force is written as where the Greek letter sigma denotes the vector sum of all forces acting on an object. ONE object! Net Force acting on ONE object 4.3 Newton’s Second Law of Motion ~ 5N = 1lb 4.3 Newton’s Second Law of Motion A free-body-diagram is a diagram that represents the object and the forces that act on it. 4.3 Newton’s Second Law of Motion The net force in this case is: 275 N + 395 N – 560 N = +110 N and is directed along the + x axis of the coordinate system. With the acceleration just calculated and starting at rest, how far has the car gone after 10s of pushing? Clicker Question 4.5 a) 0.059 m b) 10 m c) 0.59 m d) 3.0 m e) 0.3 m a = 0.059 m/s2 x = v0t + 1 2 at2 = 0+ 0.5(0.059 m/s2 )(10 s)2 = 3.0m 4.4 The Vector Nature of Newton’s Second Law The direction of force and acceleration vectors can be taken into account by using x and y components. is equivalent to & Net Force in x-direction = m times a in x-direction Net Force in y-direction = m times a in y-direction AND 4.4 The Vector Nature of Newton’s Second Law A) If mass of the object is known, and all forces acting on the object are known, then the acceleration vector can be calculated. How to use Newton’s 2nd law, B) If the acceleration vector and mass of an object are known, then the Net Force acting on the object can be calculated. It may surprise you! C) If the acceleration vector and mass of an object are known, but the calculated Net Force and the identified forces disagree, at least one additional force must act on the object. Find it! one object 4.4 The Vector Nature of Newton’s Second Law x = v0xt + 1 2 axt 2 = 48m y = v0 yt + 1 2 ayt 2 = 23m Acceleration after t = 0 Displacement at t = 65 s ax = Fx∑ m = +23 N 1300 kg = +0.018 m s2 ay = Fy∑ m = +14 N 1300 kg = +0.011 m s2 Predict the future: v0x = 0.15m/s v0 y = 0 Final displacement Starting point m = 1300 kg B) If the acceleration vector and mass of an object are known, then the Net Force acting on the object can be calculated. A paddle ball travelling horizontally bounces off a wall. The speed of the ball was 30 m/s before and after hitting the wall. If contact with the wall was for 0.02 s, what was the ball’s acceleration during the contact? v0 = +30m/s wall v = −30m/s If the paddle ball has a mass of 0.2 kg, what is the force that the wall applied to the ball? (1D vectors) F = ma = (0.2 kg)(–3000 m/s2 ) = −600 kg-m/s2 or − 600 N a = v − v0( ) t = −30− (+30)⎡⎣ ⎤⎦m/s .02 s = −3000 m/s2 Force on ball is to the LEFT Magnitude of 600N (~120 lbs) one object wall F = −600N C) If the acceleration vector and mass of an object are known, but the calculated Net Force and the identified forces disagree, at least one additional force must act on the object. Likely you will not know the origin of this force, but it must be there. A mass sliding on a table. A 2 kg mass slides on a table with an initial velocity of +1 m/s. It slows while sliding to +0.5 m/s, in 2 seconds. Table. v0 = +1 m/s 1) Calculate the acceleration vector 2) Use Newton’s 2nd law, to calculate the frictional force that must act on the mass. 4.5 Newton’s Third Law of Motion Newton’s Third Law of Motion Whenever one body exerts a force on a second body, the second body exerts an oppositely directed force of equal magnitude on the first body. There are 2 and ONLY 2 objects involved in applying Newton’s 3rd law. 4.5 Newton’s Third Law of Motion Suppose that the magnitude of the force is 36 N. If the mass of the spacecraft is 11,000 kg and the mass of the astronaut is 92 kg, what are the accelerations? 4.5 Newton’s Third Law of Motion On the spacecraft  F∑ =  P. ( one object) On the astronaut  F∑ = −  P. (another object) aS =  P mS = +36 N 11,000 kg = +0.0033m s2 Really tiny, and would not be noticed except over a very long time 4.5 Newton’s Third Law of Motion These two forces do not have a Net Force = 0! To use the Net force and Newton’s 2nd law, all the forces being summed must act on the same object. +P acts on the spacecraft –P acts on the astronaut Astronaut’s push Acting on spacecraft 4.5 Newton’s Third Law of Motion These two forces do not have a Net Force = 0! +P acts on the spacecraft –P acts on the astronaut Spacecraft’s push acting on the astronaut. To use the Net force and Newton’s 2nd law, all the forces being summed must act on the same object. 4.5 Newton’s Third Law of Motion Warning: Newton’s 3st law can appear to be violated if you can’t see the resulting movement (too small) of one of the two objects. Examples (clicker questions): Ball bouncing off a wall. Mass sliding on a table w/friction. Bat hitting a baseball Gun firing a bullet Clicker Question 4.7 a) F = −3000 N b) F = +3000N c) F = 0N d) F = 60N e) A ball cannot make a force. wall F = +3000 N on the wall. v0 F = −3000 N on the ball. ball point of contact with the wall. A ball heads horizontally toward a wall. While in contact with the wall, the wall applies a force, F = −3000 N on the ball, as shown. At the same time, the ball must apply what force on the wall? Clicker Question 4.7 These two forces do not result in a Net Force = 0. There is one force on the wall and one force on the ball. wall F = +3000 N on the wall. F = −3000 N on the ball. wall F = +3000 N on the wall. F = −3000 N on the ball. Simultaneously showing both forces that act the objects at the point of contact Clicker Question 4.8 While the mass is sliding, a friction force, F = −0.5N, acts on the mass. What friction force acts on the table? a) F = +0.5N b) F = −0.5N c) F = 0N d) F = 60N e) A mass cannot make a force. Table. v0 mon the mass F = −0.5N 4.6 Types of Forces: An Overview In nature there are two general types of forces, fundamental and nonfundamental. Fundamental Forces 1. Gravitational force 2. Strong Nuclear force 3. Electroweak force 4.6 Types of Forces: An Overview Examples of nonfundamental forces: friction tension in a rope normal or support forces