8a-lab2-forces-and-accelerations

8A Lab2 Forces And Accelerations Introductory Physics (University of California, Berkeley) Scan to open on Studocu Studocu is not sponsored or endorsed by any college or university 8A Lab2 Forces And Accelerations Introductory Physics (University of California, Berkeley) Scan to open on Studocu Studocu is not sponsored or endorsed by any college or university Downloaded by Kunduz Ablimit (dmablimit@gmail.com) lOMoARcPSD|16589059

LAB 2 Forces And Accelerations In this laboratory, we will be analyzing a spring scale and a cart and track system with a various assortment of masses. The goal of this lab is to better familiarize you with Newton’s Laws as well as forces and accelerations in general. Given Quantities • Track Length = 0.8m • Mass of Weight Hook = 5g • Mass of Cart = 520g • Mass of Pulley = 30g • Mass of String ≈ 0.5g • Mass of Rectangular Metal Block = 500g 2.1 Analyzing Net Force And Accelerations In this part of the lab, we will try to better understand the relationship between forces and accelerations by using a mass and a spring scale. As background, a spring scale works in a similar fashion to an ordinary household scale. When a weight is attached to it, its spring stretches and the scale displays the weight of the attached object in units of Newtons. Figure 2.1: 500 Gram Mass on Spring 105 Kunduz Ab limit Downloaded by Kunduz Ablimit (dmablimit@gmail.com) lOMoARcPSD|16589059

2.1. Analyzing Net Force And Accelerations 8. Draw a Free Body Diagram for the mass and scale system as it is moved upward quickly from the floor. Indicate the relationship between the “Force of the Scale” and “Weight of the Mass” and indicate the direction of acceleration if applicable. 9. Before we proceed any farther, let’s do another thought experiment. What would happen to the reading of the scale if you were to move it from shoulder level to the floor rather quickly? Would the reading on the scale increase, decrease, or stay the same? Briefly explain your reasoning. 10. With the 500g mass attached, have one teammate hold the scale and mass at shoulder level then have them move it toward the ground rather quickly while the rest of the team observes the reading of the scale. What happens to the reading of the scale as the mass is moved quickly downward? Was your group’s hypothesis correct? If not, explain why. 11. Draw a Free Body Diagram for the mass and scale system as it is moved downward quickly. Indicate the rela- tionship between the “Force of the Scale” and “Weight of the Mass” and indicate the direction of acceleration if applicable. 12. Using the principles you learned above explain how someone would feel in an elevator as it initially moves upward, as it is traveling upward, and as it comes to a stop. Specifically, explain whether someone would feel lighter, heavier, or the same weight at these three points and briefly explain why using a combination of Free Body Diagrams and brief explanations. 107 7 . After watching this being done in the video . I would say that my hypothesis was correct because in the video , it appears that the reading on the scale increases quite rapidly and by a large margin as the demonstrator quickly / suddenly moves the scale upwards . Again , this is likely due to the upwards acceleration causing the net upwards force to be larger than the downwards force , thus causing the reading to be larger than the actual weight of the mass . I * acceleration is present in the upwards direction FBD : ^ Force of the scale Rnefantinnnnhinnbhwntonrnen : since acceleration is present in the upwards • direction , the upwards force of " the Force of the scale " is of a larger ✓ weight of the Mass magnitude than the downwards force of " weight of the Mass " . I predict that the reading on the scale would decrease . I predict this because the net force would no longer be zero because the downward force , which used to be just the weight of the mass cm ✗ g) . would now include the downward acceleration , which in turn would cause the reading on the scale to decrease . After watching this being done in the video , I would say that my hypothesis is correct because in the video , it appears that the reading on the scale decreases quite rapidly and by a large margin as the demonstrator quickly / suddenly moves the scale downwards . Again , this is likely due to the downward 's acceleration causing the net force to be negative , thus causing the reading to be smaller than the actual weight of the mass . I * acceleration is present in the downwards direction FBD : Relationshipblwforcesimnm.si rice acceleration is present in the •^ Force of the scale downwards direction , the downwards force of " The weight of the mass " is of a greater magnitude than the upwards force of " the weight of the Mass Force Of the scale " . v Point 1 : Initially Moves Upward : Point 2 : Traveling Upward : Point 3 : coming to a stop : ^ ^ •^ Force Of Elevator @ force of elevator force of elevator • y µ agg mass > force of elevator y m g g g mass = force of elevator y m g g g force Of elevator > mass At this 1St point , someone would feel As the elevator is traveling Upward , As the elevator comes to a stop . someone would lighter because the force in the someone would likely just feel the same feel heavier since while its coming to a stop , downwards direction is greater than weight because the movement would the acceleration is in the upwards direction , the upwards force , thus making the net be at a constant velocity , meaning there thus meaning the upwards force is greater than force negative , which ultimately makes would be no acceleration present in either the downwards force , causing you to feel direction and that both present forces would YOU feel lighter than normal . be of equal magnitude , causing you to feel the same . heavier than normal . Downloaded by Kunduz Ablimit (dmablimit@gmail.com) lOMoARcPSD|16589059

2. F ORCES A ND A CCELERATIONS 2.2 Newton’s First Law / Analyzing Simple Accelerations Take your track and place it flat against the table, and then place your cart on the track. Ensure that the track is flat and level, and that the cart can remain at rest while on the track. Adjust the track as necessary. Ensure that the “Bubble Level” is securely attached to the cart using the piece of putty, and that the bubble is centered between the lines when the cart is at rest i.e. that the level is parallel to the table. Assume that all surfaces are perfectly smooth, which means the cart does not slow down due to friction. Refer to the picture below. Figure 2.3: Cart on Level Track 1. Briefly explain Newton’s First Law and provide an every-day example below. 2. Before we begin the next part of the lab, let us briefly think about what is inside of a bubble level. Consider that a bubble level contains some sort of fluid, and some sort of gas. For our purposes, we can assume that the level contains a combination of water and air. Which of these components likely has more mass, the water or the air bubble? Now we are going to try and make some predictions. For each of the motions described below, predict where the bubble will be in the level and sketch the location of the bubble within each box. For each case, try and briefly explain your prediction. It may also be helpful to think about what happens to the fluid that is in the level during these time periods. Case 1: Cart moves to the right and is speeding up. Case 2: Cart moves to the right with constant velocity. Case 3: Cart moves to the right and is slowing down. Explanation: Explanation: Explanation: Now you will try it out. Take your cart and place it at the end of the track farthest away from the bumper. With your hand, give the cart a gentle but quick tap toward the end of the track with the bumper. 1. While your hand is pushing the cart, sketch the position of the bubble. What can you say about the cart’s 108 Newton 's First Law essentially states that objects at rest tend to stay at rest unless an external force acts on it , thus causing it to move . An everyday example of this law would be a book that just sits on a table at rest and remains at rest unless an external force acts on it . I believe that the water likely has more mass than the air bubble . . ① ° . ° . I feel that in this scene rio , the bubble I feel that as the cart is moving at a 1 feel that the bubble would shift to would move to the right because there constant velocity , there would be no the left because the cart would be is acceleration present in the right acceleration present in either left or de accelerating in the right direction , direction and the fluid , which has more right direction , thus resulting in the causing the fluid , which has more mass , mass than the bubble , would shift back - bubble not shifting and just staying to shift forwards , in turn causing the bubble wards cleft ) causing the bubble to shift in the middle . forwards ( right ) . to shift backward ( left ) . Downloaded by Kunduz Ablimit (dmablimit@gmail.com) lOMoARcPSD|16589059

2. F ORCES A ND A CCELERATIONS Case 1: The cart moves up the incline as someone is pushing the cart. Case 2: After being released, the cart travels up the incline Case 3: The cart reaches the top of the incline and travels back down the ramp. Explanation: Explanation: Explanation: Now we will try it out. With your hand, give your cart a gentle but quick tap up the incline. 1. While your hand is pushing the cart, sketch the position of the bubble. What can you say about the cart’s acceleration during this time, is it zero or nonzero? Draw a Free Body Diagram of the cart at this time and if applicable, draw an arrow that points in the direction of the cart’s acceleration. 2. After your hand is no longer touching the cart, sketch the position of the bubble. What can you say about the cart’s acceleration during this time, is it zero or nonzero? Draw a Free Body Diagram of the cart at this time and if applicable, draw an arrow that points in the direction of the cart’s acceleration. 3. What happens to the position of the bubble when the cart starts to roll back down the incline? Sketch the position of the bubble. Draw a Free Body Diagram of the cart at this time and if applicable, draw an arrow that points in the direction of the cart’s acceleration. 4. Using Newton’s 1st law, try to explain what is happening to both the bubble and the fluid during the different periods of motion. Were your predictions correct? 110 ° . Mlk : Mbs BANK : I feel that when the cart is being I feel that the bubble would shift Pushed up the incline , the fluid is back to the middle and remain I feel that the bubble would shift going to shift backwards , causing at the middle as it travels up the to the right because the fluid , the bubble to shift forwards , which incline because there would be no which has a greater mass , would would be the left . external forces acting upon the also be shifting to the right , but since cart , thus causing the bubble to it is heavier , it would push the bubble remain at rest in the middle . to the right . FBD : o . * the cart 's acceleration is nonzero ( positive ) r F hand 7 Fn ( y ) r ✗ final ⑥ ✓ mg ✓ ✗ initial - direction of acceleration FBD : Mh o . Mfg * the cart 's acceleration is zero ^ T Fn ly ) ✗ final Be > ✗ initial vmg MMMM • . * the cart 's acceleration is non - zero c negative , FBD : r ^ Fncy ) ✗ final ⑥ ✓ ✗ initial ✓ mg J direction of acceleration in the first period of motion , the external force of the hand causes the fluid to shift to the right / Which would technically be backwards on the incline , thus causing the bubble to shift to the left . So , Newton 's 1st law applies here because the external force of the hand causes the bubble and fluid to shift from their rest position . In the second period of motion , the fluid and bubble both remain at rest ( middle ) since there is no external force acting on the cart . In the third period of motion , the fluid shifts in the downwards direction of the incline , and this external force causes the bubble to shift to the right because water is heavier than gas . So , again we see Newtons first law applying here because the external force ( downward acceleration ) causes a shift of the bubble and fluid from rest . What I said would happen did end up happening in the video so my predictions were correct . Downloaded by Kunduz Ablimit (dmablimit@gmail.com) lOMoARcPSD|16589059

2.2. Newton’s First Law / Analyzing Simple Accelerations 2.2.2 Circular Motion Using the same principles as earlier, we will now make some predictions and observations regarding circular motion. In the front of the room, there is a rotating chair with a bubble level attached to it. Figure 2.5: Chair Like before, we are going to try and make some predictions. For each of the motions described below, predict where the bubble will be in the level and sketch the location of the bubble within each box and indicate which direction the center of the circle is relative to the level by drawing an arrow. For each case, try and briefly explain your prediction. It may also be helpful to think about what happens to the fluid that is in the level during these time periods. Case 1: The chair is being slowly spun clockwise with a constant speed. Case 2: The chair is released and continues rotating clockwise. Case 3: The chair is being slowly spun counter clockwise with a constant speed. Explanation: Explanation: Explanation: Now we will try it out. Go to the front of the room and make sure that the bubble is in the center of the level (i.e. the level is level) while the chair is at rest. Take your hand and gently spin the chair in the clockwise direction. Stop it and then slowly spin it again in the counter clockwise direction. 1. Using Newton’s first law, try to explain what is happening to both the bubble and the fluid during the different periods of motion. Were your predictions correct? 2. Briefly describe what happens to the speed and velocity of the bubble level during its circular motion. In particular, are they constant or changing? How does this influence the position of the bubble? 111 centers Yfg• . µ ffg centers • - YANK MY . . > center I predict that as the chair is released I predict that as the chair is being spun I think that it actually does not matter and continues to spin clockwise , the clockwise slowly , the bubble will begin in bubble is just going to shift more towards whether the stool is being spun in a clockwise the middle and then slowly shift towards the the middle since the centrifugal force will Or counterclockwise direction because either center of the rotation because the centrifugal push the fluid away from the center , which way / the centrifugal force will Push the fluid force would shift the fluid away from the center , again causes the bubble to shift towards the away from the center , thus shifting the bubble which shifts the bubble towards the center . center of rotation . towards the center of rotation . in the first period of motion , there is an external force present : centrifugal force . This external force causes the fluid to shift away from the center of rotation which then shifts the bubble towards the center , as seen in the video . In the second point of motion , the centrifugal external force is still present which , as we know from Newton 's 1st law . Prevents the fluid and bubble from staying at a rest position ; in the video , we see the fluid still shifted away from the center and the bubble shifted towards the center . In the third period of motion , though the turn is counterclockwise , the external centrifugal force is still present thus shifting the fluid out of rest and away from the center and also the bubble out of rest and towards the center . in the first and third case I believe that the speed is constant because it is stated in the descriptions themselves that the spinning is done at a constant speed ; however , in the first and third cases , the velocity is changing because velocity accounts for direction and in case one , the direction is clockwise , whereas the direction is counterclockwise in the third case . The constant speed and changing velocity in cases one and three influence the bubble by causing it to be towards the center , regardless of the direction of the rotation . In the second case , the speed and velocity are both changing because the stool gradually slows down to a stop . This causes the bubble to meter . Shift towards the center for the main duration of the spin and then as the spin slows down , the bubble shifts back to the center of the bubble Downloaded by Kunduz Ablimit (dmablimit@gmail.com) lOMoARcPSD|16589059

2.3. Analyzing Forces And Accelerations Now we will take our cart (with the metal mass bar and an additional 30 grams of mass on top of it), and our hook (with a total mass of 30 grams) and position them so we can conduct experiments to observe the motion of the cart. Place the cart so that the front is 80cm away from the bumper (i.e. approximately at position 20cm ). Have one person hold the cart still, keeping it at rest. Add 25g of mass to the end of the string with the hook and place it over the top of the pulley. Have someone in your group take the stopwatch and reset it so that it reads a value of zero. When your teammate is ready, let go of the cart and time how long it takes from the moment the cart is released, to the moment the cart hits the Styrofoam bumper. 5. Repeat this process three times, record the times below and then calculate the average time: Mass Attached to the Hook 25g (Total: 30g) Time (s) Trial 1 Trial 2 Trial 3 Average Time 6. Using the kinematic equation for position and the average time you just recorded, calculate the acceleration of the cart and the hook. Mass Attached to the Hook Acceleration of the Cart / Hook ( m / s 2 ) 25 grams (30 grams total) Now we will take the 30 grams of additional mass that is on top of the cart, and place it on the hook. Just like before, have someone in your group take the stopwatch and reset it so that it reads a value of zero. When your teammate is ready, let go of the cart and time how long it takes from the moment the cart is released, to the moment the cart hits the bumper. 7. Repeat this process three times, record the times below and then calculate the average time: Mass Attached to the Hook 55g (Total: 60g) Time (s) Trial 1 Trial 2 Trial 3 Average Time 8. Using the kinematic equation for position and the average time you just recorded, calculate the acceleration of the cart and the hook. Mass Attached to the Hook Acceleration of the Cart / Hook ( m / s 2 ) 25 grams (30 grams total) 9. What is the ratio of the acceleration you calculated from part (8) over the rate of acceleration you calculated in part (6)? Is this ratio what you would expect? 113 1. 765 1. 85s 1. 865 1. 82s ✗ = that 2 trot + Xo 0=1-2 all . 82 ) -104.82 ) -1 0.8 a = - 0.879m IS 2 0 = 0.91 a + 0+80 ÷i=:Y ¥ a = - 0.879 1. 13s 1. 40s 1. 20s 1. 245 ✗ = Eat 2 + rot + Xo 0=1-29 ( 1.24 ) -1 0 ( 1.24 ) + 0.8 - 1 . 290 M IS 2 ÷i=% ¥ - a = - 1.290 - 1. zqom , gz since the mass on the hook was doubled , I was expecting more of a z :| ratio where the value I would want is 2 , but what I ended up with was about 1.5 which is not extremity - 0.879mHz = 1. 468 off , but still is a tad surprising and not what I expected . Downloaded by Kunduz Ablimit (dmablimit@gmail.com) lOMoARcPSD|16589059

2. F ORCES A ND A CCELERATIONS 10. How do your observed rates of acceleration compare to the theoretical values of acceleration you calculated in parts (3) and (4)? Are they higher than, lower than, or about the same as your predictions? 11. What things might have affected the acceleration of the cart in real life that we did not account for in our simplified calculations? 2.3.1 Statics And Motion On An Inclined Ramp Remove the 500g mass bar from the top of your cart. Remove the extra circular masses form your hook. Take your track and place it on the metal bar so that it is inclined again. Make sure that the black bumpers on the bottom of the track are behind the metal bar, to ensure that the track will not slide off during the next part the laboratory. Again take the string and attach it to the cart; take the other end of the string with the hook attached and place it over the pulley. Figure 2.7: Cart and Pulley on a Sloped Track 1. Draw a Free Body Diagram for the Cart and the Hook separately for this positioning. 2. Theoretically, it is possible to place the cart on the middle of the ramp and have the system be at equilibrium (at rest). How could this be done? 3. Calculate what angle you would need to incline the ramp at, to have the system of the cart (mass = 520 grams) and the hook with 100 grams of added mass (105 gram total mass) to remain at rest. What angle would be needed? 114 My observed rates of acceleration were both greater than the theoretical rates of acceleration I calculated , though they were both in the negative direction . This is likely just due to human error in the experiment either by the demonstrator in the video or , more likely , due to me making mistakes in my timing . Things in real life that may not have been accounted for could be experimental error made by humans , wind resistance , or even the force of friction . FBD for the hook : FBD for the cart : ^ Tension Tension r n Fn • • Um , g mg cost ✓ mzgmgsino yes this is possible actually . If a system is at equilibrium , that means the total net force on the cart and the hook will be zero . I think the only way this state could be reached at rest would be if we physically changed the incline angle and made it less steep . 1- = mg = 0.105 × 9.8 = 1.029 > The angle that should be needed for the system to remain at rest is 11.650 1. 029 = sin 0 0 = 11.65° Downloaded by Kunduz Ablimit (dmablimit@gmail.com) lOMoARcPSD|16589059

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