5 Day 5: Newton s Laws and Kinematics in 1D

Transcription

1 5 Day 5: Newton s Laws and Kinematics in 1D date Friday June 28, 2013 Readings Knight Ch 2.4-7, Ch 4.6, 4.8 Notes on Newton s Laws For next time: Knight lecture demo car on a track, free-fall in vacuum two force plates class activities Knight 4.6, 4.8 Knight Learning Objectives: In covering the material for this day s lecture, students will: 1. Begin the process of understanding the connection between force and motion. 2. Understand the relationship between forces between two di erent objects. 3. Begin learning how to explain an observation on the basis of physical principles. 4. Develop their understanding of the kinematic variables position, velocity, and acceleration. 5. Be able to solve basic problems of motion in one dimension, including objects falling under the influence of gravity. 5.2 Common Misconceptions The prevailing student belief is that motion requires a force. This belief is based on much common-sense evidence, and it is a belief that is highly resistant to change. More specifically, the student version of the laws of motion is: If there s no force on an object, the object is at rest or will immediately come to rest. The converse is not true. An object at rest does not automatically imply no net force. Motion requires a force or, alternatively, force causes motion. In general, force is proportional to velocity. 28

2 Halloun and Hestenes (1985) have characterized student beliefs about interactions in terms of a dominance principle: The larger (or faster or more active) object exerts a larger force than the smaller (or slower or less active) object. Students tend to view an interaction as a conflict in which the stronger wins. It s not hard to understand how this common-sense view comes about. After all, the e ect of the collision on the compact car is much larger than its e ect on the truck. Di erent e ects would seem to require di erent causes, hence di erent amounts of force. The di erence in the masses does not appear to students as a significant factor in drawing conclusions about forces. This basic misconception about interaction forces is likely the most persistent and hard to change of all the student misconceptions in mechanics. Some of the more specific di interacting systems are: culties students have with Newton s third law and with Students don t believe Newton s third law. It s too contrary to common sense. Students have di culty identifying action/reaction force pairs: They match two forces on the same object. They place forces on the wrong objects. They don t believe that long-range forces (e.g., gravity) have reaction forces. Students confuse equal force with equal acceleration. 5.3 Mini-lecture 9: Newton s Laws We saw before than a net external force causes a change in momentum with time. We can write this statement (Newton s second law) as: ~F net = ~p t We can substitute the definition of momentum, ~p = m~v into the equation above. If the mass of the system does not change, any and all changes in momentum will be due to changes in velocity: ~F net = ~p t = (m~v) t = m ~v t 29

3 We define the change in velocity with respect to time as acceleration. Substituting the symbol for acceleration, ~a into the equation above, we obtain the most common expression for Newton s second law: ~F net = m ~v t = m~a It is important to keep in mind the assumption we made in deriving this equation: we assumed that the total mass of the system did not change. If we ever encounter a situation in which the mass of the system changes, then ~ F net = m~a will not be valid. The original mathematical statement of Newton s second law, that a net external force causes a change in momentum over time, is still true. Note that both force and acceleration are vectors. Since mass is not a vector, then the direction of the net force on a system is the same as the direction of the acceleration. It is important to repeat this note from the textbook: When several forces act on an object, be careful not to think that the strongest force overcomes the others to determine the motion on its own. It is ~ F net, the sum of all the forces, that determines the acceleration ~a. Motion in the real world often involves two or more objects interacting with each other. An interaction is the mutual influence of two objects on each other. Two objects interact by exerting an action/reaction pair of forces on each other. For example, if we consider a hammer hitting a nail, the hammer exerts a force on the nail, but the nail also exerts a force on the hammer. 30

4 Newton s third law is usually stated as: for every action there is an equal and opposite reaction. How can two things be simultaneously equal and opposite? A more accurate way of stating Newton s third law is: Every force occurs as one member of an action/reaction pair. The two members of this pair act on two di erent objects, point in opposite directions, and are of equal magnitude. Demo: pushing the TA with force plates In this demonstration, Beatriz and a TA each had a force plate which is like a bathroom scale and they pushed each other using the force plates. The values of force measured by the force plates were displayed in real time on the projector. No matter how hard they pushed, or in which direction they moved, the force on the force plates was almost exactly the same (give or take a few Newtons, due to this not being a well-controlled experiment). Demo: tug-of-war In this demonstration, two teams participated in a tug of war. One team won. Who pulled harder? The pull force of Team A was the same as the pull force of Team B. Then why did they all move? The answer to that question depends on how the team members interact with the floor. When the previously winning team was asked to put plastic bags on their feet, and the tug-of-war rematch happened, they lost! So what is going on? Lets start by drawing the situation. Since the motion is in the x-direction only, we will only include the horizontal forces, though the team s weight and the normal force from the ground are also there. If we define the system or interest as one of the teams, we see that there are two horizontal forces on the system: the tension on the rope, and the static friction force of their feet on the ground. We will talk later in the course exactly how static friction works. For now we can say that, if the system is not moving, then the tension and the static friction 31

5 on our system add up to zero. As the other team pulls harder on the rope, the tension increases. If our system is still not moving, this means that friction increased as well. Friction can only increase up to a certain point; if the other team pulls hard enough to overcome the maximum friction on the system, they win! We can rig the tug-of-war game by decreasing the friction between one of the teams and the floor. In class, we accomplished this by making one team wear plastic bags on their feet. 5.4 Mini-lecture 10: Kinematics in 1 dimension So far we have only considered uniform motion motion with a constant velocity. In this section, we will consider motion in which the velocity is changing at a constant rate, that is, motion with constant acceleration. We will then consider the special case of free-fall, which is vertical motion of objects that experience only the force of gravity. Acceleration is defined as the rate of change of velocity with respect to time. We can write this definition mathematically, and re-arrange the terms to find the velocity equation of an object with constant acceleration: a x = v x t a x = (v x) f (v x ) i t (v x ) f =(v x ) i + a x t Here we expressed the result for the x-axis, but the result is valid for any axis. We can use the fact that the velocity versus time graph for an object with constant acceleration is linear together with the relationship between the position versus time graph and the velocity versus time graph (The displacement is the area under the curve of the velocity versus time graph.) to obtain the position equation for an object with constant acceleration. x f = x i +(v x ) i t a x( t) 2 Again, we have expressed the result for the motion in x-direction, but the result is valid for any axis. Example 2.12 Finding the braking distance A car is traveling at a speed of 30 m/s, a typical highway speed, on wet pavement. The driver sees an obstacle ahead and decides to stop. From this instant, it takes him 0.75 s to begin applying the brakes. Once the brakes are applied, the car experiences an acceleration of 6.0 m/s 2. How far does the car travel from the instant the driver notices 32

6 the obstacle until stopping? Prepare Lets begin by drawing a picture of the car, and a motion diagram of the situation. You might also want to draw a velocity versus time graph representing the motion. Before making a list of known quantities, we notice that this problem has two parts: the first part is from the moment the driver notices the obstacle to when he applies the brakes and the second part is from the moment he applies the breaks to the moment he comes to a full stop. The first motion is a constant velocity motion, the second motion is a constant acceleration motion. There are three important moments in the motion as a whole: 1. The moment the driver notices the obstacle. 2. The moment the driver applies the brakes. 3. The moment the car stops. We will use subscripts 1, 2, and 3 to label the motion variables at these three important moments. We can now make a list of the known quantities, with the appropriate labels. We are asked to find the total distance traveled. We will need to make two calculations, one for the constant velocity motion and one for the constant acceleration motion. 33

7 Solve From t 1 to t 2 the velocity stays constant at 30 m/s. This is uniform motion, so the position at time t 2 can be calculated as follows: x 2 = x 1 +(v x ) 1 (t 2 t 1 ) = 0 m + (30 m/s)(0.75 s) = 22.5 m At t 2, the velocity begins to decrease at a steady 6.0 m/s 2 until the care comes to rest at t 3. This time interval can be computed using the following formula, and solving for t. (Pro tip: mistakes are easier to find if you do algebra first, and plug in numbers last) (v x ) 3 =(v x ) 2 + a x t t = t 3 t 2 = (vx) 3 (v x) 2 0m/s 30 m/s a x = =5.0 s 6.0 m/s 2 We can now compute the position at time t 3 by taking point 2 as the initial point and point 3 as the final point. Remember that t in this case is the time di erence between the final moment (t 3 ) and the initial moment (t 2 ) of this portion of the motion. x 3 = x 2 +(v x ) 2 t a x( t) 2 = 22.5 m + (30 m/s)(5.0 s) ( 6.0 m/s2 )(5.0 s) 2 = 98 m x 3 is the position of the car at the end of the problem and so the car travels 98 m before coming to rest. 34

8 Analyze The numbers for the reaction time and the acceleration on wet pavement are reasonable ones for an alert driver in a car with good tires. The final distance is quite large it is more than the length of a football field. Free fall Free fall is a special case for motion in one dimension at a constant acceleration. Here is a list of the main points of to keep in mind when thinking about free fall: Any two objects in free fall, regardless of their mass, have the same acceleration. We saw this in the recommended video from Veritasium, the video of astronaut David Scott who dropped a hammer and a feather on the moon, and the demo during class (dropping a coin and a feather inside a tube with no air). The free fall acceleration always points down. The standard value for the acceleration of an object in free fall close to the Earth s surface is 9.80 m/s 2. This value is always positive. 35