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Lesson - Fletcher's Trolley Name: _____________________________________ The applet simulates the motion of two blocks connected by a string that is guided over a pulley. Prerequisites Students should understand the vector properties of magnitude and direction and be familiar with adding vectors graphically by the Tip-to-Tail method. They should also have a working knowledge of basic trigonometry. Learning Outcomes This lesson will give insight into the physics concepts of force (especially tension), free-body diagrams, and Newton's second law of motion. The system studied involves objects moving with constant accelerations, and the lesson will review concepts related to motion with constant acceleration. Instructions Students should know how the applet functions, as described in Help and ShowMe. Many of the step-by-step instructions in the following text are to be done in the applet. Some physical assumptions have been made about the system simulated by the applet. These are described in Assumptions. Contents Acceleration Observed Acceleration Calculated from the Masses of the Blocks Tension in the String When the Blocks are Moving Appendix A - Equations for Motion with Constant Acceleration Acceleration Observed Imagine two blocks connected by a rope. Now, imagine that one of the blocks is hung over the edge of a table via a pulley. What would happen to the blocks if they were free to move? Would both blocks begin to accelerate, and if so would they accelerate at the same rate if the rope does not stretch? The apparatus used to answer such questions is known as Fletcher's Trolley. We will investigate these questions using a simulated Fletcher's Trolley. Fletcher’s Trolley – Lesson 1 The applet will be used to observe the acceleration of two blocks on Fletcher's Trolley. a. Click the reset button ( ). The lower edge of block 2 will then be at elevation h = 1.30 m, and the yellow platform at h = 0 m, as in figure 1. The masses of the blocks will be: m1 = 600 g m2 = 400 g The pulley's mass will be zero. These mass values will be assumed in all of the following explanations, except where otherwise noted. Figure 1 b. Click on block 1 and drag it until block 2 is at elevation h = 3.00 m. Maximize ( ) the applet window if you are not able to drag block 1 far enough to the left. c. Click play ( ) and observe the motion of the blocks. Describe the motion of the blocks using the terms speed, time and acceleration. Fletcher’s Trolley – Lesson 2 The acceleration of the blocks can be determined by calculating the slope of a velocity Vs time graph. Click replay ( ) and Step ( ) through the motion. Record the velocity at each of the following times in the table below. a. The time and velocity data is found on the left side of the applet display panel. Using this data, construct a velocity Vs time graph. Velocity Vs. Time time (t) s velocity (v) m/s 0.00 0.0 0.04 _________ 0.08 _________ 0.12 _________ 0.16 _________ 0.20 _________ 0.24 _________ 0.28 _________ 0.32 _________ 0.36 _________ b. Calculate the slope (acceleration) of the best-fit line on the velocity Vs time graph. Show your work below. c. The constant slope indicates that the acceleration is constant. Compare your value with that shown by the applet (a = 3.92 m/s2). d. Why would one expect the accelerations of the blocks in Fletcher's Trolley to be constant? The force of ____________ acting on block 2, which drives the system, is constant. Newton's second law of motion says that the net force acting on an object is proportional to the object's acceleration. Thus, if the forces acting on an object are constant, then the same must be true for the acceleration of the object. Fletcher’s Trolley – Lesson 3 Acceleration Calculated from the Masses of the Blocks An engineer who wants to design a system with a specific acceleration must be able to calculate the acceleration from the parameters defining the system. In the case of Fletcher's Trolley, these parameters are the masses of the two blocks and the magnitude of the acceleration due to gravity (g). Given the parameters illustrated on Figure 2, how can one calculate the magnitude of the acceleration (a) of the two blocks? You know that gravity is providing an acceleration on one of the blocks and that the two blocks have to accelerate at the same rate because they are attached by a string that can not stretch. Applying Newton's second law to the two blocks separately will give two equations for two unknowns. The two unknowns are acceleration and tension. Figure 2 • acceleration (a) Both blocks have accelerations of the same magnitude (a) because the string is assumed not to stretch. • string tension (T) Note that for a massless and frictionless pulley and a massless string, the tension (T) in the string is the same on both sides of the pulley and everywhere along the string and equal to the magnitudes T1 and T2 of the forces exerted by the string on blocks 1 and 2, respectively: T = T1 = T2 The acceleration of the blocks can be determined by setting up two basic equations and solving them in terms of acceleration. Setting Up the Basic Equations Step 1: define the directions. The coordinate axes (directions) are defined in the applet. • Horizontal: positive to the right • Vertical: positive downward Fletcher’s Trolley – Lesson 4 Step 2: view the following free body diagrams (FBD) by clicking the FBD button ( ) that will illustrate the forces acting on each block. FBD 1: Equation for Block 1: FBD 2: Equation for Block 2: Applying Newton's second law Applying Newton's second law (Fnet = ma) to block 1 yields the (Fnet = ma) to block 2 yields the following equation: following equation: (2) (1) N1 and W1 cancel Equations (1) and (2) are the fundamental equations governing the motion of the system. Each of these equations can be used independently to determine the tension in the string. a. Using equation (1) calculate the tension in the string if m1 = 0.600 kg and a = 3.92 m/s2. Show your work here. (Verify your answer using the applet.) b. Using equation (2) calculate the tension in the string if m2 = 0.400 kg and a = 3.92 m/s2. Show your work here. (Verify your answer using the applet - your answer should be identical to part (a).) Fletcher’s Trolley – Lesson 5 Solving the Equations for Acceleration Step 3: combine equations (1) and (2) to generate an equation to solve for the acceleration of the system. Adding equations (1) and (2) causes T to cancel. Manipulating the equation in terms of acceleration gives: Solving for the acceleration gives: Verify that this is the value for the acceleration as displayed by the applet. Given the following masses, calculate the acceleration of the system. Verify your answer on the applet by adjusting the mass values to match those given below. The mass values on the applet are controlled by the mass slider. Show your work below. m1 = 0.600 kg m2 = 0.500 kg Given the following masses, calculate the acceleration of the system. Verify your answer on the applet by adjusting the mass values to match those given below. The mass values on the applet are controlled by the mass slider. Show your work below. m1 = 0.600 kg m2 = 0.800 kg Fletcher’s Trolley – Lesson 6 Tension in the String When the Blocks are Moving How can one calculate the tension present in the string when the blocks are moving? The tension is not measured as easily as the acceleration. One would have to insert a strain gauge into the string, which would not be massless and would have an effect on the conditions in the system including the tension. Therefore, it is particularly important that one knows how to calculate the tension. For example, an engineer who needs to design a system like Fletcher's Trolley will want to know how strong to make the string so that it will not break under the tension that it will need to sustain. If the acceleration has already been measured or calculated, the easiest way to find the tension is to substitute the value for acceleration (a) into either equation (1) or (2). For example, Using Equation 1 Using Equation 2 The applet displays a value of 2.4 N, which is consistent with these calculations. Notice that the tension value of 2.35 N is less than the weight of block 2. The weight of the block 2 is calculated by W2 = m2g = (0.400 kg)(9.81 m/s2) = 3.92 N. What would the acceleration of the system be if the tension were equal to the weight of block 2? (Hint: study the free body diagram of block 2 to answer this.) Fletcher’s Trolley – Lesson 7 Given the following masses, calculate the tension in the string. Verify your answer on the applet by adjusting the mass values to match those given below. The mass values on the applet are controlled by the mass slider. Show your work below. m1 = 0.600 kg m2 = 1.100 kg Given the following masses, calculate the tension in the string. Verify your answer on the applet by adjusting the mass values to match those given below. The mass values on the applet are controlled by the mass slider. Show your work below. m1 = 0.600 kg m2 = 1.400 kg Fletcher’s Trolley – Lesson 8 Appendix A - Equations for Motions with Constant Acceleration In the equations in the table below, it is assumed that an object is moving along an x-axis with constant acceleration (ax). For a motion along the y-axis with constant acceleration, the equations are analogous. You would only replace the subscript x with y. Notation Both velocity, vx, and acceleration, ax, will be written with the subscript x. Quantities that carry the subscript x can be either positive or negative. The subscript notation distinguishes these quantities from others that can take on only positive values (or zero), like the speed, v, and the magnitude a of the acceleration. Quantities without subscript are equal to the absolute values of the corresponding quantities with subscript. For example: v = |vx| and a = |ax| Equations for Acceleration, Velocity, Displacement and Speed Acceleration (A1) (constant) Velocity (A2) (A3) Displacement (A4) Speed (A5) Fletcher’s Trolley – Lesson 9

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posted: | 2/6/2010 |

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