Fall22 Atwood's Machine Lab Online 8.16.22

Atwood’s Machine Lab Online Background Newton’s 2 nd 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 = F net m ). An Atwood’s Machine is a simple 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 m 2 > m 1 . 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 its own weight ( m 1 g, or m 2 g ) pointing downwards, and each has the tension ( T ) 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. T − m 1 g = m 1 a T − m 2 g =− m 2 a In the Force Summation Equations, as they are written here, the letters T , g, ∧ 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 m 1 is being accelerated upwards at the exact same magnitude that m 2 is being accelerated downwards. The reason m 2 is being accelerated downwards is due to m 2 having a larger weight than m 1 , and therefore there is a greater downwards acting force on m 2 than m 1 . 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. 1

T = m 1 a + m 1 g T − m 2 g =− m 2 a m ( ¿¿ 1 a + m 1 g )− m 2 g =− m 2 a ¿ m 2 a + m 1 a = m 2 g − m 1 g a ( m 2 + m 1 ) = g ( m 2 − m 2 ) a = g ( m 2 − m 1 ) ( m 2 + m 1 ) 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. F net = g ( m 2 − m 1 ) 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 F net = a ( m 1 + m 2 ) as well. 2

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. 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 (m 1 - m 2 ) for each run should equal 0.7 kg. d. For the last run m 1 = 2.0 kg, and m 2 = 1.3 kg. e. The software is using g = 10.0 m/s 2 . 4

Analysis of Atwood’s Machine Lab Name_____________________________________ Course/Section____________PHY 1951___________________ Instructor________________________________ Constant Total Mass Table (20 points) Run m 1 (kg) m 2 (kg) m 1 +m 2 (kg) a(m/s 2 ) F net (N) 1 1.1 0.9 2.0 1.00 2.0 2 1.2 0.8 2.0 2.00 4.0 3 1.3 0.7 2.0 3.00 6.0 4 1.4 0.6 2.0 4.00 8.0 5 1.5 0.5 2.0 5.00 10. 6 1.6 0.4 2.0 6.00 12 7 1.7 0.3 2.0 7.00 14 8 1.8 0.2 2.0 8.00 16 9 1.9 0.1 2.0 9.00 18 10 2.0 0 2.0 10.00 20 Complete the above chart. Use the acceleration and total mass to calculate F net = a ( m 1 + m 2 ) . Show some calculations to receive credit. Examples of Runs 1-3: Run 1: F net = 1.00 m / s 2 ( 1.1 kg + 0.9 kg )= 2.0 N Run 2: F net = 2.00 m / s 2 ( 1.2 kg + 0.8 kg )= 4.0 N Run 3: F net = 3.00 m / s 2 ( 1.3 kg + 0.7 kg )= 6.0 N 1. What is a real-world application of an Atwood's Machine? (4 points) The first thing that comes to mind is a pulley system used on wealthy old houses for laundry and 5

Constant Net Force Table (20 points) Run m 1 (kg) m 2 (kg) m 1 +m 2 (kg) a(m/s 2 ) F net (N) 1 1.1 0.4 1.5 4.67 7.0 2 1.2 0.5 1.7 4.12 7.0 3 1.3 0.6 1.9 3.68 7.0 4 1.4 0.7 2.1 3.33 7.0 5 1.5 0.8 2.3 3.04 7.0 6 1.6 0.9 2.5 2.80 7.0 7 1.7 1.0 2.7 2.59 7.0 8 1.8 1.1 2.9 2.41 7.0 9 1.9 1.2 3.1 2.26 7.0 10 2.0 1.3 3.3 2.12 7.0 Complete the above chart. Use the acceleration and total mass to calculate F net = a ( m 1 + m 2 ) . Show some calculations to receive credit. Examples of Runs 1-3: Run 1: F net = 4.67 m / s 2 ( 1.1 kg + 0.4 kg )= 7.0 N Run 2: F net = 4.12 m / s 2 ( 1.2 kg + 0.5 kg )= 7.0 N Run 3: F net = 3.68 m / s 2 ( 1.3 kg + 0.6 kg )= 7.0 N 5. For the Constant Net Force data (Table 2), using Excel, or some other graphing software, plot a graph of, a vs 1/M tot , with the trendline displayed on the graph. Make sure to turn the graph in with your lab worksheets. (15 points) 7

0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 2 2.5 3 3.5 4 4.5 5 f(x) = 7.02 x − 0.01 Acceleration vs. 1/Total Mass 1/Total Mass (1/kg) Acceleration (m/s2) 6. (a) What are the units of the slope? (4 points) The units of the slope are Newtons. (b) What physical quantity does the slope of the best-fit line represent? (4 points) It represents the net force, which is consistently 7.0 N. 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) Newton’s third law of motion states that for every force, there’s an equal and opposite reactive force. Due to the nature of the pulley system (and the way how the two masses are attached), this law can be applied to the tension between the masses. 8. Above, we derived an equation for the acceleration: a = g ( m 2 − m 1 ) ( m 2 + m 1 ) . Briefly explain what the numerator and denominator are in a physical sense. (5 points) The denominator is just the total mass, as in the sum of both masses (in kg). The numerator is 8