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PHYSICS 182A/195L LAB REPORT - LAB 5: NEWTON’S SECOND LAW Lab 5: Newton’s Second Law San Diego State University Department of Physics Physics 182A/195L TA: Alvin Yassuiae Lab partner 1: Matthew Ying, Olivia Sekimoto Lab partner 2: Tyler Shonnard, Emily Gerken Date: October 3, 2023 Score: Theory Newton's second law of motion explains the relationship between forces and the acceleration of bodies. “If a net external force acts on a body, the body accelerates. The direction of acceleration is the same as the direction of the net force. The mass of the body times the acceleration vector of the body equals the net force vector.” This can be expressed by the vector equation: where is the sum of all forces acting on a mass m, and is the acceleration of that mass. If we are dealing with a one-dimensional system, then Newton’s law simplifies to: Tension Tension (T) represents the net force present on a rope, string, wire, etc. It is constrained by the forces on either end, and remains constant as long as the rope is considered rigid and unchanging in length. By hanging a mass on one end, and attaching the other end to a separate mass, a system is developed. Cart and hanging mass system In this lab we will be analyzing the car and hanging mass system shown in the figure below. 1 Department of Physics
A cart of some mass is tied on one end of the string with some accessory mass attached, and the other end of the string is held over a pulley by a hanging mass . Assuming the only external force in the system occurs from gravity pulling on the hanging mass , a system of equations can be used to find the net acceleration of the system. This system of equations will be entirely one-dimensional. First let’s consider the cart and the hanging mass separately. Using Newton’s second law, we can relate the forces acting on the cart to the tension in the string : . We treat the cart stacked with mass as a single object with mass . When it comes to the hanging mass, there are two forces included in the sum: the force downward from gravity and the force upward from the tension : . Careful thought will reveal that the accelerations and are the same. If the string connecting the masses has a fixed length, then any displacement of results in the same displacement of , and so all three masses have the same velocities and accelerations. A definition will be made to replace these accelerations, . The tension is also the same for both bodies; one can be substituted into the other to form a single equation, . Isolating , this is then rewritten into the form: . We now define to be the force from the hanging mass and to be the total mass of the system: , .
PHYSICS 182A/195L LAB REPORT - LAB 5: NEWTON’S SECOND LAW So that we can write: . Finally, the acceleration of all masses can be solved for by dividing the above equation by the total mass : . Procedure Setup 1. Measure and record the masses of the Smart Cart (scale should be located at the front of the lab). Record the value in SI units in Table 1 and the first row of Table 2 below. 2. Check that the PASCO Universal Interface is turned on and connect the Smart Cart via the Hardware Setup tab (left side of the screen). 3. Level the track with the Torpedo Level. Once the track is level, the Smart Cart should not move in either direction when placed on the track. 4. Check that a gate with elastic band (shown in Figure 2) or a magnetic End-Stop is located on the end of the track between the Smart Cart and the pulley. We do not want the Smart Cart to hit the pulley or fall off the track. 5. Check that the Common Rate is set to 40.00 Hz (located at the bottom of the screen next to Recording Conditions .) 6. Drag over 3 graphs, one for each of the parts below. They should be set to track velocity. 7. Tie a loop in both ends of a braided physics string. Loop one end on the Smart Cart hook, pass the string under the elastic band (or through the End-Stop hole), and hang a mass hanger on the string over the pulley. 8. Make sure that the string is parallel with the track from the Smart Cart hook to the pulley (i.e. the pulley and hook should be at the same height). The string should not touch any other part of the setup between the hook and the pulley. Part A 1. Place 0.040 kg in the Smart Cart in order to set kg. 2. The mass hanger weighs 0.005 kg. Add a 0.015 kg mass from the Mass and Hanger set for a total hanging mass of 0.020 kg. 3. Pull the Smart Cart back as far as possible without allowing the mass hanger to touch the pulley. Allow the pulley to come to rest (we do not want it to be swinging wildly when recording data). 4. Start Recording and release the Smart Cart. 5. Click Stop after the Smart Cart strikes the elastic band (or End-Stop). 6. Click on the Data Summary button (left Toolbar). Right click on the run you just made and rename the data ‘Part A’. 7. Use the data selection tool (yellow pen icon) to select only the linear portion of your data. 3 Department of Physics
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8. Use the linear curve fit tool on the PASCO graph interface to find the slope of your velocity versus time graph. 9. Record the measured acceleration (slope) in Table 2 below. 10. Fill in the appropriate column of Table 2 below. 11. Include a copy of your PASCO graphs in the Analysis section. Right click on the edge of the PASCO graph object and select “Copy Display”. Paste into this document with “Ctrl+v”. Part B Repeat steps 3 - 11 from Part A with the same total mass , but this time with double the hanging mass . To accomplish this: 1. Place 0.020 kg on the Smart Cart in order to set kg. 2. The mass hanger weighs 0.005 kg. Add a 0.035 kg mass from the Mass and Hanger set for a total hanging mass of 0.040 kg. Part C Repeat steps 3 - 11 from Part A with double the total mass and the same hanging mass as in Part B. In order to double the total mass but keep the hanging mass the same, we will have to adjust the mass added to the cart. 1. Set kg, where is the total mass from Part B. 2. The mass hanger weighs 0.005 kg. Add a 0.035 kg mass from the Mass and Hanger set for a total hanging mass of 0.040 kg. Data (Please enter your data in blue!) Table 1: Cart mass (kg) 2.51g x 1kg/1000g= 0.251 kg Table 2: Masses, Accelerations and Forces! Part A Part B Part C (kg) 0.251 0.251 0.251 (kg) 0.040 0.020 0.522 (kg) 0.020 0.040 0.040 (kg) 0.311 0.311 0.833
PHYSICS 182A/195L LAB REPORT - LAB 5: NEWTON’S SECOND LAW ( ) F(hanging) = Mass(hanging) * g 0.196 0.392 0.392 ( ) 0.521 1.11 0.646 ( ) 0.630 1.26 0.471 Percent error 20.9% 13.5% 2.48% Analysis Part A Plot A: Copy your velocity graph for part A into the box below. 1. What happened to the force going from Part A to part B? It increased 2. What happened to the total mass going from Part A to Part B? Stays the same 5 Department of Physics
3. What happened to the acceleration going from Part A to Part B? It increased Part B Plot B: Copy your velocity graph for part B into the box below. 4. What happened to the force going from Part B to part C? Stay the same 5. What happened to the total mass going from Part B to Part C? It doubled
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PHYSICS 182A/195L LAB REPORT - LAB 5: NEWTON’S SECOND LAW 6. What happened to the acceleration going from Part B to Part C? It decreased by factor of ½ Part C Plot C: Copy your velocity graph for part C into the box below. 7. What happened to the force going from Part A to part C? It doubled 8. What happened to the total mass going from Part A to Part C? It doubled 7 Department of Physics
9. What happened to the acceleration going from Part A to Part C? Acceleration increased Questions 1. Let be some mass and be some force acting on that mass (the numerical values are not important). Write in terms of the other two variables given. a1= f1/m1 2. Suppose that we double the mass of the object in question 1 so that , but we maintain the same force so that . Express in terms of . What happened to the acceleration? a2=f2/2m1; acceleration decreased by a half 3. Suppose instead that we double both the mass and the force, so that and . Now what is in terms of ? What happened to the acceleration? a2= 2F1/2M1; There’s no change in acceleration, so acceleration stays the same. 4. The gravitational force acts on all objects in the cart and hanger system, including the cart’s mass . Should we have included in our equations? Why or why not? No because the track counteracts the force of gravity so gravity shouldn’t be taken into account in the equations.