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7 Pages

### Lab02--ForceTable

Course: PHYS 152, Fall 2008
School: Purdue
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Word Count: 613

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2: Lab The Force Table R. Napora Today you will be experimenting with a force table to attain a better understanding of vectors. The force table is shown in Fig.[1]. It is a round table with a center pin and angle divisions marked around its outside edge. A ring is placed around the center pin that has strings attached to it. Those strings are run over pulleys at dierent locations around the table and various...

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2: Lab The Force Table R. Napora Today you will be experimenting with a force table to attain a better understanding of vectors. The force table is shown in Fig.[1]. It is a round table with a center pin and angle divisions marked around its outside edge. A ring is placed around the center pin that has strings attached to it. Those strings are run over pulleys at dierent locations around the table and various masses are hung from their ends. Fig.[1]: Schematic diagram of a force table. Each string will exert a force on the ring that is proportional to the mass hung on it1 and in a direction along the line of the string. Notice that force is a vector quantity. Since it has direction as well as magnitude, it is possible for two or more non-zero forces to balance out to a zero net force. This can be seen in the example below. 105 900 0 120 0 75 0 1 The force here is the mass times the acceleration due to gravity at the Earths surface, i.e. F orce = M ass 9.81 m/s2 . The SI unit for force is called the Newton, N . 1 N = 1 kg m/s2 . Phys 152 Lab 2 (v1.0) -1- 150 0 210 165 0 0 195 1800 0 Example: Say you have three forces, one of magnitude 0.20N pointing at 00 , one of magnitude 0.40N pointing at 900 , and one of magnitude 0.45N pointing at 2430 . These forces are represented by arrows in the diagram. We can show that these three vectors add up to a zero net force. We will do this by considering the x and y components separately. 5 13 0 60 0 45 0 30 0 15 0 00 345 0 330 0 22 50 31 5 0 240 0 300 255 0 0 285 2700 0 First considering the x-components: Fnet,x = (0.20 N ) cos(00 ) + (0.40 N ) cos(900 ) + (0.45 N ) cos(2430 ) Fnet,x = 0.20N + 0N + (0.20N ) Fnet,x = 0 Now considering the y-components: Fnet,y = (0.20N ) sin(00 ) + (0.40N ) sin(900 ) + (0.45N ) sin(2430 ) Fnet,y = 0N + 0.40N + (0.40N ) Fnet,y = 0 We see that these three vectors sum to zero in both the x and y directions. If you were to set these three forces up on the force table, the ring would remain centered around the pin without being moved in any direction. If the net force on the ring were ever not zero, there would be an imbalance and the ring would be pulled in some direction (if the center were pin not there to stop it). For the remainder of this lab, you will be given sets of forces and be asked to solve for the force(s) needed to cancel them out. You will then test your predictions on the force table. I would also like you to sketch all of the force vectors on the diagrams provided. Neatness counts! When sketching vectors be sure to draw them such that their lengths are proportional to their magnitudes. A few warnings: (1) Be sure that your calculator is not in radians mode. (2) When using inverse trigonometric functions, be sure that the angle you nd is in the correct quadrant. A neat, clear picture helps tremendously with this. Phys 152 Lab 2 (v1.0) -2- 105 900 0 120 0 75 0 60 0 45 0 5 13 0 30 0 Case I. vector A B C A+B+C =0 mass, kg 0.200 0.200 magnitude, N direction 300 1200 x-component y-component veried: Phys 152 Lab 2 (v1.0) 150 0 210 165 0 0 15 0 195 22 31 50 5 0 1800 0 00 345 0 330 0 240 0 300 255 0 0 285 -3- 2700 0 105 900 0 120 0 75 0 60 0 45 0 5 13 0 30 0 Case II. vector A B C A+B+C =0 mass, kg 0.200 0.150 magnitude, N direction 200 800 x-component y-component veried: Phys 152 Lab 2 (v1.0) 150 0 210 165 0 0 15 0 195 22 31 50 5 0 1800 0 00 345 0 330 0 240 0 300 255 0 0 285 -4- 2700 0 105 900 0 120 0 75 0 60 0 45 0 5 13 0 30 0 Case III. vector A B C mass, kg 0.200 0.150 veried: Phys 152 Lab 2 (v1.0) 150 0 210 165 0 0 15 0 195 A+B+C =0 magnitude, N direction 00 900 x-component 22 31 50 5 0 1800 0 00 345 0 330 0 240 0 300 255 0 0 285 -5- 2700 0 y-component 105 900 0 120 0 75 0 60 0 45 0 5 13 0 30 0 Case IV. vector A B C mass, kg 0.300 veried: Phys 152 Lab 2 (v1.0) 150 0 210 165 0 0 15 0 195 A+B+C =0 magnitude, N direction 00 900 2400 x-component 22 31 50 5 0 1800 0 00 345 0 330 0 240 0 300 255 0 0 285 -6- 2700 0 y-component 105 900 0 120 0 75 0 60 0 45 0 5 13 0 30 0 Case V. vector A B C D A+B+C +D =0 mass, kg 0.100 0.200 0.300 magnitude, N direction 300 900 2250 x-component y-component veried: Phys 152 Lab 2 (v1.0) 150 0 210 165 0 0 15 0 195 22 31 50 5 0 1800 0 00 345 0 330 0 240 0 300 255 0 0 285 -7- 2700 0
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