PHY 105 Lab - Newton's 2nd Law

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Onondaga Community College *

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105

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Mechanical Engineering

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Dec 6, 2023

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PHY 105L – Onondaga Community College 3/2/2015 Page 1 of 6 PHY 105 Lab: Experiment on Newton’s 2nd Law General Instructions This lab report is provided as a Word file and you will need Microsoft Word and Excel to complete it. In case you do not have Microsoft Office, go to the OCC website at https://www2.sunyocc.edu/students, and click the link for Office 365. You will complete the report in this file. While doing so, make sure that you • enter answers in red, and insert graphs in this file where prompted • respect the order of the questions • Do not change the font of the text After you complete the report in Word, save it as PDF and check if it looks nice/professional (so you will not get it back for re-formatting!). Then submit the PDF in Blackboard . (Of course, you should save both the Word and PDF versions for your records.) Maximum number of points: 50 Experiment on Newton’s 2nd Law Introduction: Newton’s 2 nd Law says that the acceleration of a body is directly proportional to the net force acting on that body: F NET = m a . The goal of today’s activity is to check experimentally this fundamental equation of mechanical motion. Experimental set-up: A low friction cart has a force sensor attached on top and can move along a horizontal track under the pull of a hanging object. The force sensor samples the tension in the string at a (controllable) rate of 40 measurements per second. Simultaneously, a motion sensor placed at one end of the track records the position of the cart-sensor system at a (controllable) rate of 40 measurements per second. The Pasco Capstone software uses the position-time measurements to generate velocity-time data which will then be used to determine the acceleration of the cart.

PHY 105L – Onondaga Community College 3/2/2015 Page 2 of 6 Theoretical Analysis Make a sketch of the experimental set-up. Draw free-body diagrams for the cart-force sensor system (of total mass m 1 ) and the hanging object of mass m 2 . The 2 nd Law for each body gives the following equations: T = m 1 a and m 2 g – T = m 2 a; In these equations, T is the tension in the string and a is the acceleration of the system. If the cart is held at rest, the acceleration is zero and the tension equals the weight of the total hanging mass, . Exercise (4 points): Solve for T and a in terms of m 1 , m 2 and g. Show the steps and the result at the end of the report. Measurements (22 points): 1. With no mass pulling on the string, reset the force sensor by pressing the tare side- button. Attach a total mass of XX grams (note that the hanger has 50 g) to the end of the string and position the cart 15-20 cm away from the motion sensor. 2. Click the start icon, wait 5 seconds ( thus allowing the force sensor to record the tension T REST ) and then release the cart free. Catch the cart before it hits the pulley and click the stop icon. 3. Use the software to determine: a) the acceleration a EXP from the slope of velocity-time data, b) the tension T REST when the cart is being held at rest, and c) the tension T MOT when the cart is in motion. 4. Repeat steps 1 through 3 for several increasing masses m 2 , as indicated by the instructor. Record your results in the table below (15 points): m 2 (kg) a EXP (m/s 2 ) T MOT (N) T REST (N) 0.1 0.484 0.95 1.04 0.13 0.626 1.20 1.35 0.16 0.770 1.44 1.62

PHY 105L – Onondaga Community College 3/2/2015 Page 3 of 6 0.19 0.864 1.63 1.82 0.21 0.989 1.84 2.10 0.24 1.090 2.07 2.41 5. Use the balance to measure m 1 , which is the mass of the cart-force sensor system (3 points) m 1 = 1.8275 kg Analysis and Discussion (28 points) 1. Plot T MOT vs a EXP . Add a trendline as suggested by the trend of the data points. Show the trendline equation on chart. (5 points) 2. If Newton’s 2 nd Law is true, how should the graph of T MOT versus a EXP be like? Does the trend of the data points agree with the expectation? (4 points) If Newton’s 2 nd Law is true, our data should fall along a linear line. Tension is proportional to acceleration and the equation is T = m 1 x a. This explanation agrees with the trend of our data because our data falls closely on a linear line. This also explains why the slope of our trend line is relatively close to the mass of the cart-force sensor system.

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PHY 105L – Onondaga Community College 3/2/2015 Page 4 of 6 3. Use the trendline equation to extract the experimental value of m 1 (the mass of the cart). Compare it percentage-wise with the value obtained using the scale. (3 points) Percentage difference = 0.126% (1.8252/1.8275) X 100 = 99.874% 100% - 99.874% = 0.126% 4. Plot T REST vs m 2 . Add a trendline as suggested by the trend of the data points. Show the trendline equation on chart. (5 points) 5. T REST should theoretically equal to the weight m 2 g of the hanging mass. That is, the graph T REST vs m 2 should be linear and its slope should give an experimental value for g. Does your graph agree with the theoretical expectation? Compute the percentage difference of your experimental g to the reference value 9.80 m/s 2 . (3 points) We have a small percent difference of 2.5% between our experimental value and the theoretical value. Therefore, our graph agrees with the theoretical expectation. Percentage difference = 2.5% (9.5575/9.80) x 100 = 97.5% 100% - 97.5% = 2.5%

PHY 105L – Onondaga Community College 3/2/2015 Page 5 of 6 6. Write an overall conclusion for this activity. Summarize how your results checked Newton’s 2 nd Law. Discuss any plausible reasons for the difference between the experimental results and predictions from Newton’s 2 nd Law. (8 points) Newton’s 2 nd Law states that a force acting on a body is equal to the acceleration of that body times it’s mass. Newton’s 2 nd Law is used to prove that Net Force = Tension = mass x acceleration. This explains why our Tension in Motion was proportional to our acceleration from the first graph and why the slope of the line was very close to the mass of our cart-force sensor system. Our data points fell on a linear line and our experimental mass only deviated 0.126% from our actual mass, proving Newton’s 2 nd Law of motion. Tension at Rest = m 2 x gravity. This is because gravity acts on the hanging mass and tension acts in the opposite direction of this force. This explains why Tension at Rest was proportional to the hanging mass with a slope close to the gravitational force. Our data points fell on a linear line and our experimental g only deviated from our reference value (9.80 m/s^2) by 2.5%. This also proves Newton’s 2 nd Law. Overall, both of our graphs prove that Tension = mass x acceleration, which falls in line with Newton’s 2 nd Law of Motion. Possible errors in our experiment could be caused by small air resistance and drag acting on the string connecting the hanging mass to the cart-force sensor system as the hanging mass falls to the floor. The cart did not move straight on the track; therefore, the acceleration could have been affected by the cart hitting the sides of the track. The pulley did not spin loosely, therefore, the acceleration could’ve also been slowed down by this tension. Our intent was to increase the hanging mass by an interval of .03 kg. But we did 0.21 kg instead of 0.22 kg. Therefore, there is a data point screwed from our trendlines in our graphs. This may have affected the equation of our trendline, throwing our results off. Exercise: