Newton s 1 st law: A body in uniform motion remains in uniform motion, and a body at rest remains at rest, unless acted on by a nonzero net force.

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1 PHYS1110H, 2011 Fall. Shijie Zhong 1. Newton s Laws Newton s 1 st law: A body in uniform motion remains in uniform motion, and a body at rest remains at rest, unless acted on by a nonzero net force. A body with zero net force is also called an isolated body. Therefore, the 1 st law can also be stated as An isolated body in uniform motion remains in uniform motion and at rest remains at rest. Motion is always measured with respect to a reference frame. A reference frame in which the Newton s 1 st law is valid is called as inertial reference frame. We may say that Newton s 1 st law defines inertial reference frame. Newton s 2nd law: Net force on a body is equal to the product of its mass multiplying its acceleration, or F = m a. (1.1) Remarks: 1) Newton s 2 nd law is valid only in an inertial reference frame (i.e., where the acceleration is measured). 2) Net force means that it could be a vector sum of multiple forces acting on the body. 3) Newton s 2 nd law is consistent with Newton s 1 st law. From the 2 nd law, for net force of zero (i.e., the isolated body), the acceleration is zero, or constant motion. However, Newton s 2 nd law cannot substitute the first law, because the 2 nd law is only valid for inertial systems which are defined by the 1 st law. 4) Another definition of Newton s 2 nd law is: The rate at which a body s momentum changes is equal to the net force acting on the body. d p dt = F, (1.2) where the moment of a body is defined as p = m v. (1.3) Note 2 1

2 More on the momentum is given in later sections. Easy to prove that (1.1) and (1.2) are the same, for mass m is constant. dp dt = d(m v ) dt Mass = m d v dt + v dm dt = m d v dt = m a = F. (1.4) What s mass? Newton s 2 nd law provides an operational definition of mass. Suppose that one applies the same amount of force to pull two bodies of different mass: m 1 and m 2, where m 1 could be one unit of mass. We can then measure accelerations of these two bodies (using ruler and watch) under the same force, a 1 and a 2. Then we can determine m 2, according to the Newton s 2 nd law, m 2 = a 1 a 2 m 1. (1.5) Larger mass has smaller acceleration, for a given force. Of course, in practice, we cannot use this operational definition to find mass for all the objects. For example, we cannot determine the mass for a mountain using this method. However, we can use other experiments and theories derived from the physics to determine the mass. Mass unit is kilogram (kg). Force In the last paragraph, we said that we applied the same amount of force to the two bodies of different mass. How do we know we apply the same Note 2 2

3 amount of force? We can use a spring to pull the body and use the stretch of the spring as a measure of force magnitude (i.e., the same amount of stretch means the same amount of force). Newton s 2 nd law can also be used to provide an operational definition of force, now that we define mass. One unit force (i.e., 1 Newton) is the force that causes one unit acceleration (i.e., 1 m/s 2 ) to a body of one unit mass (i.e., 1 kg). Gravitational force: for a body of mass m in a gravitational field (e.g., on the surface of Earth), W = mg, where g is gravitational acceleration (for Earth or Mars). The gravitational force is also called weight in physics, although in daily life, we often call mass m as weight. Other forces include electromagnetic forces, friction force, and so on. Newton s third law: Forces always appear as pairs. If body b exerts force F b a on body a, then there must be a force F a b on body b due to body a, such that F a b = F b a (i.e., these two forces are equal in magnitude but opposite in direction). There is no lone force! You stand on the surface of the Earth. Earth exerts a gravitational force on you (i.e., pull you down). The Earth s surface exerts a normal force to hold you up. Inertial systems. Example. Suppose that you, an astronaut driving a spaceship A, and your colleague driving another spaceship B, are chasing an enemy spaceship C. Suppose that the three ships move in the same direction. You want to figure out the force that the enemy ship C applies to itself. You and your colleague have devices on your ships to measure the distances of spaceship C to your ships at different times. So, you turn off engine and measure the distance between your ship and ship C at different time as X A (t). And you asked your colleague to do the same, and your colleague measured the distance between ships B and C as X B (t). Note 2 3

4 From the distance X A (t), you determine velocity V A (t)=dx A (t)/dt, and acceleration a A (t)=dv A (t)/dt for ship C. You found that V A (t) increases linearly with time and a A (t) is a constant at 1000 m/s 2. Therefore, you concluded that the force that ship C s engine exerts is F A =Ma A (t)=1000m Newton, where M is the mass of the ship C (you may not know, but it may not matter). Your colleague, after following the same procedures, found that a B (t) is a constant at 950 m/s 2. Therefore, your colleague concluded that the force for ship C is 950M Newton, different from yours. It turned out that your colleague did not completely shut down the engine and ship B actually accelerates with respect to you or an inertial system at an acceleration of 50 m/s 2. That is, ship B is not an inertial system, and Newton s second law does not really work for ship B (Your colleague can use a pencil to determine whether s/he is in an inertial system. How?). One may introduce a fictitious force to a non-inertial system. We may discuss this in the future. Limitations of Newton s laws. 1) Newtonian mechanics fails for a particle motion that is comparable to the speed of light and that is at atomic length scale. Under those conditions, relativity and quantum mechanics are needed. 2) Newton s laws in the forms given apply to particles and are inconvenient to describe a continuous system such as a fluid. However, the same principles can be extended to develop Newtonian mechanics for continuum. Note 2 4

5 2. Applications of Newton s Laws. There are some simple rules that may help us use Newton s laws to solve problems. 1) Divide the system into smaller systems, each of which can be treated as a point mass. 2) Draw a force or free-body diagram for each mass. 3) Introduce a coordinate system (i.e., positive x and y axis). 4) Apply Newton s third law. 5) Apply constraints to different bodies. 6) Set up Newton s 2 nd law for each mass, and solve the equations. We will use some example problems to demonstrate these procedures. Example 1: The astronauts tug-of-war. Two astronauts, A and B with mass Ma and Mb, initially at rest in free space, pull on either end of a rope that has mass Mp (but negligibly small). Astronaut A is stronger than Astronaut B. Find their motion, if each pulls the rope as hard as they can. The diagram shows that the system is divided into three smaller systems, with three point masses: Ma, Mb, and Mp, and also how the forces acting on each point mass. For the coordinate system, 1-D is sufficient, and only x axis is introduced in the diagram. In this coordinate system, some forces are positive (along x axis), but others are negative (opposite to the x axis). According to the Newton s third law, we may state that F a = F a ', and F b = F b '. Note 2 5

6 For the rope, assuming that its acceleration is a p in positive x direction, the Newton s 2 nd law: F b + F a = M p a p, (2.1) in vector form. Considering the directions of these vectors, the scalar form: F b F a = M p a p. (2.2) Since the mass of the rope, Mp, is negligibly small, this leads to F b =F a, and F a '= F b '. We do not need to consider the rope, and we can reduce the number of point masses from 3 to 2. (Note that in many problems, the mass of rope (or pulley) can be ignored, and we do not need to consider it in analyses. However, if the mass of the rope or pulley is given and you are not told that they can be ignored, then you will need to include them in the analysis). Now consider Newton s 2 nd law for masses Ma and Mb, in vector form: F a '= M a a a, F b '= M b a b. Or in scalar form: F a '= M a a a, F a '= M b a b, where we consider the directions of accelerations and F a '= F b ' from the Newton s third law. a a = F a '/ M a, and a b = F a '/ M b. The negative sign for a b means that the acceleration of mass B has an opposite direction from mass A. Note that although astronaut A is stronger than astronaut B, we still have F a '= F b ' from the Newton s third law, i.e., the force exerting on B by A is the same as the force on A by B. Example 2: Freight train. Three freight cars, each with mass M, are pulled by a force F. Ignore the friction. Determine the force on each car and their acceleration. Note 2 6

7 The system is divided into 3 smaller systems with three point masses, and force diagram is given in the figure for each point mass. Also, positive x and y axes are defined as in the figure. Since the cars remain on the ground all the time, there cannot be any motion in y direction at any time, and consequently a y =0. Therefore, normal force on a car cancels/balances the weight of the car. There is no need to consider the vertical direction. According to the Newton s third law, we have F 1 = F 1 ', and F 2 = F 2 '. The cars move together, therefore a 1 = a 2 = a 3 = a. We can now consider Newton s second law for each car. F 1 = Ma, (2.3) F 2 + F 1 = Ma, (2.4) F + F 2 = Ma. (2.5) In their scalar form: F 1 = Ma, (2.6) F 2 F 1 = Ma, (2.7) F F 2 = Ma. (2.8) We can solve these three equations (2.6), (2.7), and (2.8), for three unknowns F 1, F 2 and a. a=f/(3m), F 1 =F/3, and F 2 =2F/3. Constraints: For the last problem, the relation between point mass accelerations is straightforward they are the same because they are constrained to move together. However, this is not always true. Consider the following problem. Example 3. Two masses are connected by a string that passes over a pulley. The two masses are different and are therefore not balanced. The pulley itself is pulled upward with some non-zero acceleration. How are the accelerations of masses 1 and 2, and pulley related to each other? Set up a coordinate system with y axis pointing upward. The length of the string is fixed at L. Based on the diagram, we have: Note 2 7

8 L = πr + (y p y 1 ) + (y p y 2 ), (2.9) where R is the radius of the pulley. Differentiate equation (2.9) with respect to time t twice (i.e., second order derivative), and lead to 0 = 2 d 2 y p dt 2 d 2 y 1 dt 2 d 2 y 2 dt 2, (2.10) 2a p = a 1 + a 2. (2.11) When the pulley is fixed or moved at a constant velocity (i.e., a p =0), the accelerations of the two masses are equal in magnitude and opposite in directions. Example 4. Two pulleys are connected by a string as in the diagram. The lower pulley with a hanging block 1 can move vertically, while the other pulley is fixed to the ceiling and is connected to another block 2. How are the accelerations for the two blocks related? Use the coordinate system. Block 1 has the same y acceleration as the lower pulley. Similar to the last problem, the length of the string is constant. L = πr + (y 2 h) + (y 1 h) + πr + y 1, or L = 2πR + y 2 + 2y 1 2h, (2.12) where R is the radius of the pulleys and h is the distance of pulley 2 to the ceiling, both constant. Differentiate equation (2.12) with respect to time t twice (i.e., second order derivative), and lead to a 2 = 2a 1. (2.13) The acceleration of the second block is twice of that of the first block, but they have opposite directions. Note 2 8

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