Lab4 Newton's Laws1

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Physics: Newton’s Laws Objectives :  Experiment 1: perform a series of motions and analyze the results to explain Newton’s 1 st Law of Motion.  Experiment 2: investigate Newton’s 3 rd Law of Motion by observing forces exerted on objects.  Experiment 3: demonstrate the mechanical laws of motion using a simple assembly named the Atwood machine that is similar to that used by Rev. George Atwood in 1784 to verify Newton’s 2 nd Law. Introduction : Newton’s Laws govern the motion of objects and we will be studying how these laws apply to object close to and on Earth’s surface. Newton’s 1 st Law (N1L) talks about the law of inertia, where an object at rest will remain at rest or an object moving in constant straight-line motion will continue in that motion unless acted on by a net external force. Inertia is dependent on an object’s mass. Newton’s 2 nd Law (N2L) provides us with a mathematical representation of how to calculate net force using the product of an object’s mass and acceleration. Newton’s 3 rd Law (N3L) discusses force pairs between two interacting objects. Force pairs must occur between at least two interacting object and will have the same magnitude in size, but point in opposite directions. Pre-Lab Questions: (Be sure to write using complete sentences) Use the free body diagram of the pulley (Figure 5) to answer the Pre-Lab Questions. Figure 5: Free Body Diagram: 2 objects with mass hanging on a pulley by string.

1. Draw a free body diagram for M 1 . Use a draw tool of your choice, or simply insert an image of a hand- drawn figure. a. 2. Draw a free body diagram for M 2 . Use a draw tool of your choice, or simply insert an image of a hand- drawn figure. a.

3. Apply Newton’s 2nd Law to write the equations for M 1 and M 2 . You should get two equations with Tension in the string, weight for each mass and accelerations for each mass (a 1 and a 2 ), in variable form. a. M 1 a 1 = M 1 g – T b. M 2 a 2 = M 2 g – T c. T = M 1 g – M 1 a 1 = M 2 g – M 2 a 2 4. Pre-Lab Question 3 results in two equations with three unknowns. A third equation is required to solve the system. Formulate the third equation. (Hint: think about how the two accelerations relate to one another). a. M 1 g – M 1 a 1 = M 2 g – M 2 a 2, b. a 1 = -a 2, c. M 1 g – M 1 a = M 2 g + M 2 a, d. (M 1 + M 2 )a = M 1 g – M 2 g, e. a = (M 1 g – M 2 g)/( M 1 + M 2 ) EXPERIMENT 1: Newton’s 1 st Law of Motion

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Procedure Part 1 1. Fill the container with about 4 inches of water. 2. Find an open space outside to walk around in with the container of water in your hands. 3. Perform the following activities and record your observations of each motion in Table 1. a. Start with the water at rest (on top of a table). Grab the container and quickly accelerate it. b. Walk with constant speed in a straight line for 15 feet. c. After walking a straight line at constant speed, make an abrupt right-hand turn. Repeat with a left-hand turn. d. After walking a straight line at constant speed, stop abruptly. Part 2 4. Place a 3x5 notecard on top of a Styrofoam  cup. 5. Place a washer on the middle of the 3x5 notecard. 6. Hold the Styrofoam  cup with your non-dominant hand and flick the notecard with your dominant hand so it moves off of the Styrofoam  . Record your observations in Table 2. 7. Repeat steps 1-3 four times for a total of five trials. Data Tables Table 1. Motion of Water Observations Motion Observations a The water within the container tried to stay at rest and oscillated/splashed from front to back of the bowl in the opposite direction of acceleration. b The water didn’t break its surface tension and remained still. c Similar to accelerating forward the water wanted to continue in the direction it was traveling in and splashed out in the opposite direction of each turn from the carriers (my) perspective d The water wanted to stay in motion and oscillated from back to front and then splashed in front of the bowl. Table 2. Observations after Flicking Notecard Off of Cup Trial Observations 1 The card flew off and the washer dropped right into the middle of the bucket 2 The card flew off a little slower and the washer didn’t fall exactly in the middle, it looked like it traveled with the card a bit

3 The card flew off and the washer dropped right into the middle of the bucket 4 The card flew off and the washer dropped right into the middle of the bucket 5 The card flew off and the washer dropped right into the middle of the bucket Post-Lab Questions 1. Explain how your observations of the water and washer demonstrate Newton’s law of Inertia. a. Newton’s first law of motion states that an object will not change its motion unless acted on by an unbalanced force. Both the water in the bowl and the washer on the notecard experiments demonstrate this phenomenon by putting a force on an exterior object, the notecard or bowl. Since the objects within/on the container were at rest, according to Newton, they should stay at rest or in their straight line of motion since there was no direct unbalanced force acting on either. The water in the bowl demonstrates this phenomenon very easily visually by splashing out of the bowl when there is a force/acceleration applied to the bowl. The washer experiment demonstrates Newton’s first law by having the cup identify the area directly below the washer and that it falls directly in with no force applied directly. 2. Draw a free-body diagram of your containers of water from the situation in Part 1, motion d. Draw arrows for the force of gravity, the normal force (your hand pushing up on the container), and the stopping force (your hand accelerating the container as your stop). What is the direction of the water’s acceleration? a. The water doesn’t have any acceleration as prior to stopping it was moving at a constant velocity so it’s acceleration does not have a direction as it stayed in constant motion in accordance with Newton’s first law of motion. 3. Describe two instances where you feel forces in a car, in terms of inertia, that apply to this experiment. Start your response by stating the appropriate Newton’s Law and continue by explaining how it applies to the scenario. a. Just like the water in the bowl experiment, one can think of oneself as the water inside the bowl while driving a car, the car as the bowl. While making a left or right turn at high speeds a

passenger or drivers body obeys Newton’s first law of motion/inertia and will sway in the opposite direction of the turn because it wants to stay in its original straight line of travel. In order to stay upright drivers and passengers must use their core body strength to fight the force of the turn and Newton’s first law. Another instance while driving where one feels the effects of Newton’s first law is accelerating fastly or coming to an abrupt stop. In either instance the car changes speed quickly and drastically, by either speeding up or slowing down, the driver/passengers inside do not have an unbalanced force acting on the initial force so their bodies/heads will swing forwards or back depending on whether the car is stopping or starting; this is the exact same as the water splashing out of the bowl in motions A and D. EXPERIMENT 2: Newton’s 3 rd Law & Force Pairs Procedure Part 1 1. Make sure the spring scales are calibrated using the standard masses. 2. Hook the handle of the 5 N spring scale to the hook of the 10 N spring scale. 3. Holding the 10 N spring scale stationary, pull the hook of the 5 N spring scale until the force reads 5 N on it. Record the force on the 10 N spring scale in Table 3. You should be using 1 decimal place. 4. Repeat Steps 2 and 3 with the 10 N spring scale hanging from the 5 N spring scale. Record the force on the 5 N spring scale in Table 3. Part 2 5. Suspend the 0.5 kg mass in the air using the 10 N spring scale. Record the force on the 10 N spring scale in Table 4. 6. Tie one end of one of the pieces of string to the 0.5 kg mass and the other end to the hook of the 10 N spring scale. 7. Suspend the mass in the air by lifting the 10 N spring scale. Record the force on the 10 N spring scale in Table 4. 8. Untie the end of the string attached to the 0.5 kg mass and tie it to the hook of the 5 N spring scale. 9. Hook the 0.5 kg mass to the handle of the 5 N spring scale. Suspend the mass, scales, and string by holding the handle of the 10 N spring scale. Record the values of the spring scales in Table 4. 10. Secure the pulley on a tabletop by tying string to one of the hooks. Then, use masking tape to secure the string to a tabletop so that the hook on the top of the pulley lays flat on the side of the tabletop (Figure 6). 11. Using the mass setup from Step 5, place the string over the pulley by unhooking one of the spring scales, feeding the string through the pulley and reattaching the string to the hook of the spring scale (Figure 7).

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12. Hold the 10 N spring scale in place so that the scales and mass are stationary. Record the values for both spring scales in Table 4. Data Tables Table 3. Forces on a Stationary Spring Force on Stationary 10 N Spring Scale (N) 4.9 Force on Stationary 5N Spring Scale (N) 4.9 Table 4. Spring Scale Force Data Suspension Set Up Force (N) on 10 N Spring Scale Force (N) on 5 N Spring Scale 0.5 kg Mass with short string on 10 N Spring Scale 4.9 0.5 kg Mass with String on 10 N Spring Scale 4.9 0.5 kg mass, string and 5 N Spring Scale on 10 N spring scale 5.3 Immeasurable… > 5N 0.5 kg mass, string and 5 N Spring Scale on 10 N spring scale on Pulley 4.9 4.9 Post-Lab Questions 1. Compare the magnitude of the forces on both spring scales after you pulled the 5 N spring scale suspended on the 10 N spring scale. Be sure to use a %difference when asked to compare two numerical values.

a. Comparing the values of the 0.5 kg mass and 5N spring scale attached to the 10N spring scale, between when the pulley was involved – 4.9N – and when it wasn’t – 5.3N – provides a percent difference of 7.84% b. 200*(5.3 – 4.9)/(5.3 + 4.9) = 7.84% c. Comparing the values between when the 0.5kg mass was placed on the 5N spring and when it was placed on the 10N spring, without the pulley, is impossible to calculate as a percent difference since the value was over 5N which is immeasurable on a 5N spring force scale. 2. Compare the magnitude of the forces on both spring scales after you pulled the 10 N spring scale suspended on the 5 N spring scale. a. Comparing the values of the 0.5 kg mass and 10N spring scale attached to the 5N spring scale, between when the pulley was involved and when it wasn’t is also impossible as the first measurement, without using the pulley, was >5N which is immeasurable on a 5N spring scale as previously stated. b. On the other hand, comparing the values between when the 0.5kg mass was placed on the 5N spring and when it was placed on the 10N spring, with the pulley, is possible. Since both springs measured 4.9N when hanging the other spring and mass over the pulley, the have a percent difference of 0.0%. 3. Use Newton’s 3rd Law to explain your observations in Post-Lab Questions 1 and 2. Students should state N3L first before explaining how it applies to the scenario. a. Newton’s Third Law of Motion states that for every action, there is an equal and opposite reaction. This means that any force exerted on a body will create a force of equal magnitude but in the opposite direction on the object that exerted the first force. This explains our observations in questions 1 and 2 because when the weight and spring scale are attached to the other spring scale without a pulley, the force exerted (being measured) is equal to (0.5kg + m spring force scale )*g. When the pulley is added to the scenario, the total amount of force in the system is the same but it changes the distribution and direction of the force so that the force being measured is just equal to 0.5kg*9.81m/s 2 . 4. Compare the force on the 10 N spring scale when it was directly attached to the 0.5 kg mass and when there was a string between them. a. Considering there was no percent difference between the 0.5kg mass being measured on the spring force scale using a long vs short string, and the tension in the string is just equal to m*g anyway; theoretically, if the masses were able to be attached directly to the spring force scales there would be absolutely no difference in measurement because without the string, as long as it has negligible mass, the only force being measured is still m*g. 5. Compare the force on the two spring scales in Steps 5 and 6. What can you conclude about the tension in a string? a. Tension in the string is equal to m*g as long as there are no other external forces (or masses like question 3) in the system.

EXPERIMENT 3: Newton’s 2 nd Law & the Atwood Machine Procedure Part 1 1. Support the pulley so that objects hanging from it can descend to the floor. Do this by tying a short piece of string to one of the pulley hooks. Use a piece of masking tape to secure the string to a tabletop or door frame so that the pulley hangs plumb (Figure 6). Find an open space outside to walk around in with the container of water in your hands. Note : A higher pulley support will produce longer time intervals, which are easier to measure. 2. Thread a piece of string through the pulley so that you can attach washers to both ends of the string. The string should be long enough for one set of washers to touch the ground with the other set near the pulley. (You may attach the washers using a paperclip or by tying them on). 3. Use the spring scale to weigh the set of fifteen washers. Divide the total mass by fifteen to find the average mass of a washer. Record the total mass of the washers and average mass of one washer in Table 5. 4. Attach seven washers to each end of the string. 5. Observe how the washers on one side behave when you pull on the washers on the other side. Answer Post Lab Question 1 based on your observations. 6. Add the remaining washer to one end of the string so one side of the string has seven washers (M1), and the other has eight washers attached to it (M2). 7. Determine the approximate mass of M1 and M2. Record these masses in Table 5. 8. Place M1 on the floor. Use the tape measure to measure the height that M2 is suspended while M1 is on the floor. Measure the distance M2 will fall if you release the light set when it is in contact with the floor. Record the distance in Table 5. 9. Time how long it takes for M2 to reach the floor. Repeat Steps 7-8 four more times (five times total), recording the values in Table 5. Calculate and record the average time in Table 5. 10. Calculate the acceleration (assuming it is constant) from the average time and the distance the washers moved. Use an appropriate kinematic equation from chapter 2. Part 2 1. Transfer one washer, so that there are six on one end of the string (M1) and nine on the other (M2). 2. Determine the approximate mass on each end of the string. Record the mass in the Part 2 section of Table 5. 3. Repeat steps 8-10 of Part 1. Record the data in the Part 2 section of Table 5. Data Tables :

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Table 5. Motion Data Procedure 1 Motion Data Mass of M 1 : 20 g calibration mass = .02 kg Mass of M 2 : 50 g calibration mass = .05 kg Height (m): 0.76 Trial Time (s) 1 .58 2 .59 3 .60 4 .59 5 .57 Average .59 Average Acceleration (m/s 2 ) 4.37 Procedure 2 Motion Data Mass of M 1 : 10g calibration mass = .02 kg Mass of M 2 : 20 g calibration mass = .07 kg Height (m): 0.66 Trial Time (s) 1 .50 2 .52 3 .53 4 .48 5 .49 Average .50 Average Acceleration (m/s 2 ) 5.28

Post-Lab Questions 1. What do you observe about the motion of the washers when you give one set a downward push compared to the other set? Does it stop before it reaches the floor? Explain this behavior. a. When pushing the heavier mass downwards, the masses accelerate faster than when starting from rest and eventually come to a rest on the floor. When pushing the lighter mass however, the mass travels down for some time before traveling back up as the heavier masses force due to gravity is larger and counteracts the smaller mass’s force due to gravity. The smaller mass only touches the floor if the push provides a large enough force, it still rebounds back upwards as the heavier mass will always produce a larger force when there are no external forces acting on the system. 2. Draw a free body diagram for M 1 and M 2 in Procedure 1. Draw force arrows for the force due to gravity acting on both masses (F g1 and F g2 ) and the force of tension (F T ). Also draw arrows indicating the direction of acceleration, a . a.

3. Use Newton’s Second Law to write an equation for each of the free body diagrams you drew in Question 2 (use the correct signs to agree with your drawings). These equations should be similar to those found in Pre-Lab Question 2. For both equations, rearrange them to isolate the force of tension (F T ) variable on the left side. Do not insert any numbers yet. a.  F 1 = m 1 g + m 1 a – T i. T = m 1 g + m 1 a b.  F 2 = m 2 g - m 2 a – T i. T = m 2 g - m 2 a 4. Set the two resulting expressions for the force of tension from Part 1 equal to one another (as long as the string does not stretch, the magnitude of the acceleration in each equation is the same). Replace F g1 and F g2 with M 1 and M 2 , respectively. Rearrange the resulting expression for the acceleration, a , by isolating a to the left side of the equation. You should now have an expression for the acceleration of a in terms of both masses. Then, go back to Question 3 and solve for the F T by substituting your expression for the acceleration a , into one of the expressions for the tension (F T ). This should result in an expression for the tension (F T ) in terms of the masses. Again, no numbers to plug in here yet. a. Setting the two equations equal to each other and solving for a, obtains a = g*(m 2 – m 1 )/(m 1 + m 2 ). Substituting this formula for a in the formula for T, obtains T = m 2 g - m 2 g*(m 2 – m 1 )/(m 1 + m 2 ). i. T = 0.28N ii. .05*9.81 – (.05*9.81*.03)/(.07) = .280286 5. Calculate the acceleration for the two sets of data you recorded using the expression for acceleration you determined in Post-Lab Question 4, and compare these values to those obtained by measuring distance and time using percent difference. The percent difference should be within 15% to receive full credit. Cite two factors that may cause discrepancies between the two values. a. Procedure 1 i. Acceleration (using N2L): (9.81*.03)/.07 = 4.20 m/s 2 ii. Acceleration (using kinematics): (2*.76)/.59 2 = 4.37 m/s 2 iii. Percent difference: 200*(4.37 – 4.20)/(4.37 + 4.20) = 3.87% b. Procedure 2 i. Acceleration (using N2L): (9.81*.05)/.09 = 5.45 m/s 2 ii. Acceleration (using kinematics): (2*.66)/.5 2 = 5.28 m/s 2 iii. Percent difference: 200*(5.45 – 5.28)/(5.45 + 5.28) = 3.17% 6. Calculate the tension in the string for the falling washers for both procedures and for the one where the masses were equal. Use the expression for the tension derived in Post-Lab Question 4. Show all calculations. From these two values, and the one where the masses were equal, what trend do you observe in the tension in the string as the acceleration increases? a. Tension in procedure 1 = .28N as calculated in question 4. b. Using formula from question 4 to solve for tension in procedure 2: i. T = .07*9.81 – .07*9.81*(.07 – .02)/(.07 + .02) = .3052

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c. Using formula from question 4 to solve for tension in equal masses i. Since a = g*(m 2 – m 1 )/(m 1 + m 2 ), and m 2 = m 1 , a must equal 0 ii. T = mg – 0 d. These calculations show that tension increases as acceleration increases. If both masses were the weight of the lighter mass in either procedure 1 or 2, the tension would be equal to m*g and would be less than the tension when the blocks have different mass and create an acceleration/force that adds into the tension.

Lab4 Newton's Laws1

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