1. Objectives

2. Objective The purpose of this tutorial is to teach the viewer how to solve for the acceleration of each mass in a standard two- body pulley system (Atwood machine). Such problems are typical in introductory level physics courses and are used as an example of the application of Newton’s Laws. This tutorial assumes basic knowledge of kinematics and dynamics as would be typical in the first few months of a high school physics course. Familiarity with solving algebraic systems of equations is also assumed. Click on the next slide button to see instructions and an animated example of such a pulley system. Next

3. Quick Review: Newton’s Laws (1) A body at rest tends to remain at rest; a body in motion tends to remain in motion. (2) ΣF= m a (3) For every action force, there is an equal and opposite reaction force. m 1m 2 Next

4. Example of a Two-Body Pulley Problem In a standard two-body pulley problem, the student is given the mass of each of the two masses that hang from a pulley and asked to calculate the acceleration. Usually, the mass of the pulley, itself, is neglected and the system is assumed to be frictionless. Clicking on the link below will take you to an applet on the web which simulates such a two-body pulley system. You may vary the mass of each of the hanging masses. But you should choose zero for the mass of the pulley (mass #3) and also make the system frictionless. Try some different values for the different masses to get a feel for how the system behaves. Applet Next

5. Structure and Navigation of this Tutorial In order to learn how to calculate the acceleration of the masses in the pulley system, you will follow a series of steps (six steps total) that will be taught to you on a series of slides. These slides ask you to proceed in a specified order through the steps. At any time you may go back to the previous slide or jump back to the Six Steps Summary Slide After viewing the last of the six step slides, you will have a chance to try an example calculation for yourself to see if you can apply the six steps. 6-Step Summary

6. Six Steps Summary Slide Step 1: Draw a Picture of the SituationDraw a Picture of the Situation Step 2: Draw and Label all Force Vectors According to Standard ConventionDraw and Label all Force Vectors According to Standard Convention Step 3: Decide on a Direction for the Acceleration of Each MassDecide on a Direction for the Acceleration of Each Mass Step 4: Write Newton’s Law for Each MassWrite Newton’s Law for Each Mass Step 5: Solve the System of Simultaneous Equations that Results from Step 4Solve the System of Simultaneous Equations that Results from Step 4 Step 6: Examine the Answer to See that it Makes Sense PhysicallyExamine the Answer to See that it Makes Sense Physically Step 1

7. Step 1: Drawing a Picture Let’s assume that you have been given the following problem to solve: A 2kg mass and a 4kg mass hang on opposite sides of a mass- less, frictionless pulley. Determine the acceleration of the masses. The first thing you should do is to draw a picture of the situation. Like this... Notice that we have drawn the less massive block a little smaller than the more massive one. This is not essential but it is good practice. Anything you can do to represent things realistically will help you understand what is happening in the long run. 4kg2kg 6-Step Summary Step 2

8. Step 2: Draw and Label Force Vectors According to Standard Convention We have drawn the two vectors for the force of gravity on each mass (their “weights”). Since we already know that the force of gravity is the mass times the gravitational field strength at the Earth’s surface (9.8m/s 2 which we will round to 10m/s 2 here for simplicity) we write those values right on the diagram. W = mg and we get the two values 40 Newtons and 20 Newtons. Note that we have tried to draw the 20 N force vector about half as long as the 40 N one. The two upward force vectors represent the force of string tension on each mass. The force is equal on each mass – that’s from Newton’s third law for equal and opposite forces. Since we don’t know the value for this tension force we just label it T. All of these force vectors have been drawn to standard convention. Click on this button to see an example and explanation of an incorrect drawing. 4kg2kg T T 40 N 20 N 6-Step Summary Step 3Step 1 Example

9. Incorrect Drawing of Force Vectors Here we have just been sloppy with our force vectors. 1. We have not drawn the forces of gravity (the weights) as originating from the objects’ centers of mass 2. We have drawn the weight of the 2 kg mass as long as (if not longer than) the weight of the 4 kg mass. 3. The tension force vectors are off center and not originating from the point of contact. Such a poor drawing may not cause you to obtain an incorrect answer for this problem. But it will make it harder for someone else looking at your work to understand your thinking. And perhaps, worse, it allows bad habits to develop which will ultimately make things harder for you in the long run. 4kg2kg Back

10. Step 3: Decide on the Direction of Each Mass’ Acceleration This is an important step. By examining the situation you can tell that the 4 kg mass will accelerate down and the 2 kg mass will accelerate up. We put arrows alongside the masses to indicate this. The direction of acceleration is taken to be the positive direction for forces acting on that mass when we write Newton’s Law in the next step. It is very important to observe this convention or your answer will be incorrect. 4kg2kg a a 6-Step Summary Step 4Step 2

11. Step 4: Write Newton’s Law for Each Mass For the mass on the left Since the acceleration is downward (see last slide) the downward force is positive and the upward tension force is negative. Filling in Newton’s Law we get: For the mass on the right Since the acceleration is upward (see last slide) the upward tension force is positive and the downward force of gravity is negative. Filling in Newton’s Law we get: 2kg 4kg T 40 N T 20 N 6-Step Summary Step 5Step 3

12. Step 5: Solve Equations for Two Variables (a and T) Equation 1: Equation 2: Solution: Here we are most concerned with the value of the acceleration and not the value for the tension force since it was not asked for in the original problem statement. On the next slide we’ll take a moment to consider whether or not these answers make any physical sense to help us determine whether or not we have made any mistakes. The solution on the left is simply presented to you assuming that you have some familiarity with solving a system of two equations with two variables. If you would like to see the algebraic steps used in this solution click the button to the left below. 6-Step Summary Step 6Step 4 Algebra

13. Algebra for the Equations’ Solution Starting with these two equations we solve the second one for T Now we plug this value for T into the top equation Now we solve this equation for a We solve this for a which gives us a = 3.33 m/s 2. Plugging this value for a back into either original equation gives a value for T. Back

14. Step 6: Examining the Answer This step which is often neglected by beginning problem solvers is very important. Often, if you have made a mistake in the calculations you will know it right away if your answer does not make any physical sense. Let’s check our answers this way. Recall the value of the acceleration, a = 3.33 m/s 2. Is this number reasonable? The largest it could ever be is 9.8 m/s 2 (or 10m/s 2 ) since that is the acceleration caused by gravity in free fall. And the smallest it should ever be is 0 which would be the case if both masses were the same mass. Our value of 3.33 m/s 2 is somewhere in the middle which is a good thing. Now recall the value for the tension force: T= 26.67 N. This value is between the two values of weight for the two masses. Think of it this way: 26.67 N is less than 40 N and so the mass on the left accelerates down. And 26.67 N is more than 20 N so the mass on the right accelerates up. Now that makes sense! We conclude that the solution is probably correct since our answers fall within an expected range. If we had gotten an acceleration greater than 9.8 m/s 2 or a tension force greater then 40N (or less than 20N) we would know that we had made a mistake somewhere. 4kg2kg T T 40 N 20 N 6-Step Summary Practice Problem

15. Now it’s time for you to try to apply these steps. Try this problem. A 12 kg mass and a 3 kg mass hang on opposite sides of a pulley. Determine the acceleration of the system and the force of tension in the cable connecting the masses. Remember the six steps are: Step 1: Draw a Picture of the Situation Step 2: Draw and Label all Force Vectors Step 3: Decide on a Direction for the Acceleration of Each Mass Step 4: Write Newton’s Law for Each Mass Step 5: Solve the System of Equations that Results from Step 4 Step 6: Examine the Answer to See that it Makes Sense Physically Click for a correct drawing Click for correct equations Click for the solution

16. Drawing With Force Vectors 12kg3kg T T 120 N 30 N a a Back

17. Equations Back

18. Final Solution Instead of providing the solution here I have provided another link to the applet web site that you visited at the beginning of this tutorial. Go there and enter the settings for this problem (be sure to make mass #3 – the pulley mass – zero, and make the system frictionless). Check your answer by observing the simulation. Conclusion Applet

19. Conclusion Congratulations on successfully completing this tutorial. At this point if you would like, you may go back to the applet on the web site and experiment further by allowing the pulley to have mass and by allowing the system to have friction. Such complications are beyond the scope of what we will be covering in this course but why not experiment? Thank you for viewing this lesson. Return to Objectives The End