Atwood’s Machine Lab

Atwood’s Machine Lab Online Background Newton’s 2nd Law (NSL) states that the acceleration a mass experiences is proportional to the net force applied to it, and inversely proportional to its inertial mass (𝑎 = 𝐹𝑛𝑒𝑡 ). An Atwood’s Machine is a simple 𝑚 device consisting of a pulley, with two masses connected by a string that runs over the pulley. For an ‘ideal Atwood’s Machine’ we assume the pulley is massless, and frictionless, that the string is unstretchable, therefore a constant length, and also massless. Consider the following diagram of an ideal Atwood’s machine. One of the standard ways to apply NSL is to draw Free Body Diagrams for the masses in the system, then write Force Summation Equations for each Free Body Diagram. We will use the standard practice of labeling masses from smallest to largest, therefore m2 > m1. For an Atwood’s Machine there are only forces acting on the masses in the vertical direction so we will only need to write Force Summation Equations for the y-direction. We obtain the following Free Body Diagrams for the two masses. Each of the masses have two forces acting on it. Each has own weight (m1g, or m2g) pointing downwards, and each has the tension in the string pointing upwards. By the assumption of an ideal string the tension is the same throughout the string. Using the standard convention that upwards is the positive direction, and downwards is the negative direction, we can now write the Force Summation Equation for each mass. its (T) 𝑇 − 𝑚1 𝑔 = 𝑚1 𝑎 𝑇 − 𝑚2 𝑔 = −𝑚2 𝑎 In the Force Summation Equations, as they are written here, the letters 𝑇, 𝑔, and 𝑎 only represent the magnitudes of the forces acting on the masses, or the accelerations of the masses. The directions of these vectors are indicated by the +/- signs in front of each term. In these equations the + signs are not actually written out, but they should be understood to be there. Understanding this we can see that m1 is being accelerated upwards at the exact same magnitude that m2 is being accelerated downwards. The reason m2 is being accelerated downwards is due to m2 having a larger weight than m1, and therefore there is a greater downwards acting force on m2 than m1. To solve for the magnitude of the acceleration that both masses will experience, we can simply use the substitution method by solving one equation for the tension T, then substituting that into the other equation. Let’s use the question for mass 1 to solve for the tension, then insert that into the equation for mass 2, then solve for the magnitude of the acceleration. 𝑇 = 𝑚1 𝑎 + 𝑚1 𝑔 𝑇 − 𝑚2 𝑔 = −𝑚2 𝑎 1 (𝑚1 𝑎 + 𝑚1 𝑔) − 𝑚2 𝑔 = −𝑚2 𝑎 𝑚2 𝑎 + 𝑚1 𝑎 = 𝑚2 𝑔 − 𝑚1 𝑔 𝑎(𝑚2 + 𝑚1 ) = 𝑔(𝑚2 − 𝑚2 ) 𝑎= 𝑔(𝑚2 − 𝑚1 ) (𝑚2 + 𝑚1 ) Here we see that the magnitude of the acceleration the two masses experience is given by the ratio of the difference of the two masses and the sum of the two masses all times gravitational acceleration. Since that ratio will always be less than 1, the acceleration will always be less than gravitational acceleration. As the ratio gets closer to 1, then the value of the acceleration of the masses approaches the value of gravitational acceleration. However, as the value of this ratio gets closer to zero, then the value of the acceleration approaches zero as well. Also, comparing the second to last line of the steps to determine the acceleration to Newton’s Second Law we get. 𝐹𝑛𝑒𝑡 = 𝑔(𝑚2 − 𝑚1 ) Here we see that the net force acting on each mass is equal to gravitational acceleration times the difference of the two masses. From the above algebra we can clearly see that 𝐹𝑛𝑒𝑡 = 𝑎(𝑚1 + 𝑚2 ) as well. Setup 1. Go to the following website 2. You should now see the following 2 Procedure: Constant Total Mass 1. Near the bottom center of your screen set Mass of block 1 to 1.1 kg, and record this value in the Constant Total Mass Table for Run 1 in your work sheet. 2. Near the bottom center of your screen set the Mass of block 2 to 0.9 kg, and record this value in the Constant Total Mass Table for Run 1 in your work sheet. 3. Click on the play button which is a bit to the left of the bottom center of the yellow box the Atwood machine is located in. a. The value for acceleration is given near the top left of the yellow box. Record this value in the Constant Total Mass Table for Run 1 in your work sheet. b. Click the reset button right below the bottom right of the yellow box. 4. Repeat this procedure increasing the of Mass of block 1 by 0.1 kg, and decreasing the Mass of block 2 by 0.1 kg for each run and record the new values for the next run in the Constant Total Mass Table in your work sheet till all the rows in the table are filled out. a. Note, the total mass (m1 + m2) for each run should equal 2.0 kg. b. For the last run m1 = 2.0 kg, and m2 = 0.0 kg. c. The software is using g = 10.0 m/s2. Procedure: Constant Net Force 1. Near the bottom center of your screen set Mass of block 1 to 1.1 kg, and record this value in the Constant Net Force Table for Run 1 in your work sheet. 2. Near the bottom center of your screen set the Mass of block 2 to 0.4 kg, and record this value in the Constant New Force Table for Run 1 in your work sheet. 3. Click on the play button which is a bit to the left of the bottom center of the yellow box the Atwood machine is located in. a. The value for acceleration is given near the top left of the yellow box. Record this value in the Constant Total Mass Table for Run 1 in your work sheet. 3 b. Click the reset button right below the bottom right of the yellow box. 4. Repeat this procedure increasing the of Mass for both blocks by 0.1 kg, and recording their new values for in the Constant Net Force Table for the next run until all the rows in the Constant Net Force Table are filled out. c. Note, the difference bet the two masses (m1 - m2) for each run should equal 0.7 kg. d. For the last run m1 = 2.0 kg, and m2 = 1.3 kg. e. The software is using g = 10.0 m/s2. 4 Analysis of Atwood’s Machine Lab Name Stefany Pascua Course/Section PHY-1951-005 Physics for Scientist and Engineers 1 Laboratory Instructor Stephen Flowers Constant Total Mass Table (20 points) Run 1 2 3 4 5 6 7 8 9 10 m1(kg) 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 m2(kg) 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0 m1+m2 (kg) 2.0 2.0 2.0 2.0 2.0 2.0 2.0 2.0 2.0 2.0 a(m/s2) 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 Fnet(N) 2 4 6 8 10 12 14 16 18 20 Complete the above chart. Use the acceleration and total mass to calculate 𝐹𝑛𝑒𝑡 = 𝑎(𝑚1 + 𝑚2 ). Show some calculations to receive credit. Fnet = a (m1 + m2) 1.0 (2.0) = 2 N 2.0 (2.0) = 4 N 3.0 (2.0) = 6 N 4.0 (2.0) = 8 N 5.0 (2.0) = 10 N 6.0 (2.0) = 12 N 7.0 (2.0) = 14 N 8.0 (2.0) = 16 N 9.0 (2.0) = 18 N 10.0 (2.0) = 20 N 1. What is a real-world application of an Atwood's Machine? (4 points) The real-world application of an Atwood’s Machine is an elevator. 2. For the Constant Total Mass data (Table 1), using Excel, or some other graphing software, plot a graph of Fnet vs. a, with the trendline displayed on the graph. Make sure to turn this graph in with your lab worksheets. (15 points) 5 3. (a) What are the units of the slope? (4 points) Units of the slope are kilograms. (b) What physical quantity does the slope of the best-fit line represent? (4 points) The slope of the best-fit line represents mass. Constant Net Force Table (20 points) Run 1 2 3 4 5 6 7 8 9 10 m1(kg) 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 m2(kg) 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 m1+m2 (kg) 1.5 1.7 1.9 2.1 2.3 2.5 2.7 2.9 3.1 3.3 a(m/s2) 4.67 4.12 3.68 3.33 3.04 2.80 2.59 2.41 2.26 2.12 Fnet(N) 7.005 7.004 6.992 6.993 6.992 7 6.993 6.989 7.006 6.996 Complete the above chart. Use the acceleration and total mass to calculate 𝐹𝑛𝑒𝑡 = 𝑎(𝑚1 + 𝑚2 ). Show some calculations to receive credit. Fnet = a (m1 + m2) 4.67 (1.5) = 7.005 N 6 4.12 (1.7) = 7.004 N 3.68 (1.9) = 6.992 N 3.33 (2.1) = 6.993 N 3.04 (2.3) = 6.992 N 2.80 (2.5) = 7 N 2.59 (2.7) = 6.993 N 2.41 (2.9) = 6.989 N 2.26 (3.1) = 7.006 N 2.12 (3.3) = 6.996 N 5. For the Constant Net Force data (Table 2), using Excel, or some other graphing software, plot a graph of, a vs 1/Mtot, with the trendline displayed on the graph. Make sure to turn the graph in with your lab worksheets. (15 points) 6. (a) What are the units of the slope? (4 points) The units of the slope are m/s2*kg. (b) What physical quantity does the slope of the best-fit line represent? (4 points) The slope of the best-fit line represents net force. 7 7. In this experiment, we made the assumption that the tension and the acceleration experienced by the two subsystems, the two different masses, were exactly the same. Why are these good and/or valid assumptions? (5 points) There are good and/or valid assumptions because it is possible for a system to have the same values of both quantities, although tension and acceleration are two completely different physical quantities. 8. Above, we derived an equation for the acceleration: 𝑎 = 𝑔(𝑚2 −𝑚1 ) . (𝑚2+𝑚1 ) Briefly explain what the numerator and denominator are in a physical sense. (5 points) In a physical sense, the denominator is the total mass of the system while the numerator is the gravitational force acting on the difference of masses. 8