Ekins P7A Lab 3

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Apr 3, 2024

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Physics 7A Dynamics Lab, v.5.0 p-1 Lab 3: Dynamics Introduction This lab has two purposes. First, starting in the prelab, you'll address increasingly difficult versions of a standard exam-level force problem about a cart and pulley. Then in lab you'll test your answers experimentally. In addition, you'll do experiments and answer questions designed to highlight some of the subtle conceptual aspects of Newton’s laws, and how these relate to common-sense force intuitions. Prelab Questions: An exam-level force problem [Complete these questions before coming to lab! Your GSI will discuss them at the beginning of the lab.] Consider the lab set-up below. A cart of mass m_,,, is connected to a vertically hanging mass, of mass my,,.. The cart moves with negligible friction. Inlab, you will give the cart a brief push away from the pulley, and then let it go. —— 1 —_— adjustable end stop Figure 7.1 Equipment Setup A. Draw separate free-body diagrams (FBDs) for the cart and for the hanging mass. Your diagrams should describe those objects after you finish pushing the cart, but while it's still traveling away from the pulley. Label each force acting on each object. T F(n) F(9) Flg) B. After you finish pushing the cart, what force (if any) is making the cart slide away from the pulley? Explain. After the initial push, the acceleration caused by the applied force makes the cart continue to slide away from the pulley. There is no applied force currently present though. Even though the applied force is no longer acting upon the object, the object will remain moving in the direction of the acceleration until the point at which the tension of the pulley overcomes the initially applied force.

Physics 7A Dynamics Lab, v.5.0 p.2 C. Interms of m,, My, and g, find the cart’s acceleration. Show your work here. (During the lab, you will plug in the relevant numbers, to make a numerical prediction.) Horizontal Mass: F =m2a =T - m2g Cart: F =m1a =T (left is positive in the horizontal direction) m2a = (m1a) - m2g a(m2-mil) =-m2g a=-m2g/(m2-m1) D. Consider the time after the cart reverses direction and starts moving back towards the pulley. (i) Inwhat ways, if any, must we modify the free-body diagrams from question A? Explain, and draw the correct FBDs here. The free-body diagrams will be changed to have a tension that is smaller than the force of gravity for the hanging mass FBD. The cart's FBD will need to have no applied force, making the tension the sole horizontal force. T T F(n) Fo F@ (ii) Is the new acceleration bigger than, smaller than, or the same as the cart’s acceleration when it was traveling away from the pulley? Relate your answer to (i). The new acceleration is in the opposite direction, so the new acceleration will be less than the original. In (Di), the tension of the cart FBD is equal to the net force as the acceleration will only be in the negative horizontal direction at this interval of time. End of pre-lab questions. The lab starts on the next page GSI's Initials:

Physics 7A Dynamics Lab, v.5.0 p.3 Cart & hanging mass 1. In this first experiment, you'll test your prelab prediction about the acceleration. ¢ Remove the brake from the cart, if it's attached. ¢ Usea scale to measure m,,,, and .. Don't forget to include the mass of the “hook” in . With these values, predict the cart’s acceleration, using your answer to prelab question C. My =_5509 Mygng =59 Predicted acceleration = _0-089 m/s"2 4 If the track is currently set up as a ramp, lay it flat on the lab bench, and make sure it’s level. 4 Using the computer software "7A_MotionDetector.mbl" from our earlier labs, set up the screen to show an acceleration graph—and no other graphs. 4 Set up the motion detector, and run the experiment. To determine the value of the acceleration as accurately as possible, adjust the acceleration axis. A Measured acceleration = 01 mis"2 2. By what percentage does the measured acceleration differ from the predicted acceleration? Show your work. Measured value - expected value / expected value x 100% (0.1-0.089) /0.089 = 0.1235 = 12.35% 3. List some reasons why your predicted and measured acceleration disagree. How could the experiment be improved? The predicted acceleration differs from the measured acceleration because there are other forces that we considered negligible in our calculations. For example, the track had friction and the string moving over the pulley had friction so these forces were unaccounted for in our expected value. Additionally, the string has a mass that was not taken into consideration. The experiment could be improved by utilizing materials with smaller amounts of friction and mass (in regards to the string).

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Physics 7A Dynamics Lab, v.5.0 p-4 4.* After you let the cart go, as it moves back and forth across the track, is the tension in the string more than my,, g, equal to my,. g, less than m,, ¢? Explain, both mathematically and conceptually. As the cart moves back and forth, the tension of the string changes and oscillates among being less than, equal to, and greater than m(hang) g. As the cart moves away from the pulley, the hanging mass moves upwards. The only forces acting on the hanging mass are tension in the upwards direction and the force of gravity, which is constant. In order for the hanging mass to move upwards as the cart moves away from the pulley, the tension must be greater than m2g. When the cart moves towards the pulley, the hanging mass moves in the downward direction, so the tension must be less than the force of gravity. There is a point in time when the tension value is changing that it will equal m2g because it cannot achieve a value of + - m2g without being equal to m2g at a point. Horizontal Mass: F =m2a =T - m2g Cart: F =m1a =T (left is positive in the horizontal direction) T =m2g + m2a when cart is moving in the positive horizontal direction T>m2g T =m2g - m2a when cart is moving in the negative horizontal direction T<m2g 5. (Prediction) What will the position, velocity, and acceleration graphs of the cart look like if you repeat the above experiment, but with the following twist: after the cart has reversed direction and traveled about half way back towards the pulley, you'll catch the hanging mass, allowing the string to go slack. In other words, you'll “turn off” the tension force. As usual, draw your predictions before doing the experiment. Let “towards the pulley” be the positive direction, and neglect friction. The actual graphs were very close to the predictions N T art hanging @t hanging ant hanging direction caught direction caught direction caught ¢ Before running the experiment, pull down the View menu, select Graph Layout and choose “Three panes”. Make sure one of the graphs shows position, another shows velocity, and a third shows acceleration. As usual, you can change graphs by double clicking them. Make sure they all have the same time scale. Set the time scale onall three graphs equal to 4 sec. (You can change this setting if needed; but all three graphs must have the same time scale.) ¢ Make sure the motion detector is at the other end of the track from the pulley. ¢ Run the experiment. Change the scale of the graphs, if needed, to see the results more clearly. # If the actual graphs disagree with your predictions, sketch them on the above axes, using a different color or a dashed line. Omit the beginning segment of the graph when your hand was still pushing the cart.

Physics 7A Dynamics Lab, v.5.0 p.5 Newton’s laws and force intuitions The following series of questions is designed to help you develop a deeper intuitive understanding of some subtle aspects of Newton's laws. 6. (Quick Prediction) On this question, write down your immediate gut reaction; don’t calculate. A car cruises down the highway at constant velocity 50 mph. The backwards force of wind resistance and friction have a combined strength of 5000 newtons. The car’s engine causes a forward force to be exerted on the car. Is this forward force less than 5000 newtons, equal to 5000 newtons, or greater than 5000 newtons? Answer immediately, and don’t continue until you have done so. Equal to 5000 N Most people initially guess that the forward force must be greater than the 5000-newton backward force, or else the car wouldn't go forward. (Even if you personally didn’t have that intuition, you can probably see why most people would say that) Deeper reflection reveals, however, that the forward force must equal the backward force. Explain why. Because the velocity is explicitly constant, there must be no acceleration. Because of this, the forward and backward forces must be equal so that the net force is equal to 0, meaning there is no acceleration. Instead of abandoning the commonplace intuition from question 6, it's better to reconcile that intuition with Newton’s 2nd law. Here's one way to do it. (a) Before reaching cruising speed, when the car was speeding up from 0 to 50 mph, was the forward force greater than, equal to, or less than the backwards force? Explain. The car was accelerating during this time, so there must be a nonzero net force. Because the net force is nonzero, the forward and backward forces must not cancel and because the car is accelerating in the forward direction, the total acceleration will be in the forward direction, making the forward force greater than the backward force during this interval of time. (b) So, when someone has the intuition that a “net forward force is needed to make an object move forward,” in what sense is that intuition right, and in what sense is it wrong? Although a net force is needed to make an object accelerate, according to Newton’s second law of motion, a Force is directly proportional to acceleration. This does not include any information on the velocity of an object, and when a velocity is constant, there is no acceleration. According to Newton'’s first law of motion, an object in motion will stay in motion so long as there are no external forces acting to change this state. So a net forward force is needed to change the state of motion (at rest to in motion or in motion to at rest), but a net force is not needed to keep an object in motion.

F(applied Physics 7A Dynamics Lab, v.5.0 p.6 A good summary of the previous page is to say that a net forward force is needed to get an object moving but not to keep it moving. Newton’s 2nd law, F,,, = ma, requires us to make this distinction between initiating motion and maintaining motion. But once that distinction is made, the intuition from question 6 is still useful. ¢ To “feel” this distinction, attach the brake to your cart, and place it at rest on the floor. Then, using a spring, pull the cart at constant velocity. The length of the spring indicates how hard you're pulling. Notice that a bigger force is needed to get the cart moving than is needed to keep it moving. Coefficient of friction In this section, with the brake still attached to the cart, you'll return to the first experiment (i.e., cart moves along track while connected to hanging mass). After thinking about how friction changes the graphs, you'll figure out the coefficient of friction between the cart and the track. 10.* (Prediction) If you give the cart a brief push away from the pulley, what will the position, 11. velocity, and acceleration graphs look like? This is just like question 5, except there’s friction, and you never catch the hanging mass. cart t art art reverses reverses reverses direction direction direction 4 Run the experiment. Change the scale of the graphs, if needed, to see the results clearly. ¢ If the actual graphs disagree with your predictions, sketch them on the above axes, using a different color or a dashed line. Please omit the beginning segment of the graph when your hand was still pushing the cart. Explain why the acceleration becomes smaller in magnitude after the cart reverses direction. Hint: Draw a free-body diagram (FBD) of the cart when it's moving away from the pulley (after your hand stops pushing it), and a separate FBD of the cart when it's moving toward the pulley. The force of friction is causing the cart to decelerate, so the acceleration is non-constant and will become smaller in magnitude after reversing directions. Fo F) F(friction) F(friction) 4— Tension Tension E Fo) (9)

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Physics 7A Dynamics Lab, v.5.0 p.7 12.* Using the data on the computer screen, determine the coefficient of kinetic friction between the cart and the track. The coefficient of kinetic friction (,) is related to the kinetic friction force (£) by the equation f, = #,N. Show your work below. [Hint: When we take friction into account, the tension in the string is different as the cart moves in the two directions. Can you see why?] m1 = cart mass m2 = hanging mass Fnet=T-f=mla f=T-mla f=uN (T-mia) =u (mig) Fnet(hanging) = T - m2g = m2a T =m2a+m2g (m2a + m2g) - m1a =u (m1g) u=(m2a+m2g-mia)/(mig) u = [ (0.005 kg)(0.06 m/s"2) + (0.005 kg)(9.81 m/sA2) - (0.5 kg)(0.06 m/ $12) /(0.5 kg)(9.81 m/sh2) u=39x1043

Physics 7A Dynamics Lab, v.5.0 p.8 13. (Do this one if you have time; make sure you complete #14 even if you don’t do this.) Design and carry out an experiment to find the coefficient of static friction (1) between the cart and the track (with the brake in place). Write your procedure below (nothing formal—just a brief description of what you did), and your calculations and results. Did not have time

Physics 7A Dynamics Lab, v.5.0 p-9 14. Anairplane of mass 5000 kg flies in a straight line at steady speed of 70m/s. Wind resistance pushes backwards on the airplane with a force of 2000 newtons. Is the forward force that the engine exerts on the airplane greater than, less than, or equal to 2000 N? Explain. The forward force is equal to 2000 N because the velocity is constant, therefore there is no acceleration, meaning there is no net force, so the forward force must equal the backward force.

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Ekins P7A Lab 3

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