Lab #2 - Vectors Name Katie Ellis Groups Names Eden Emily Felicia Introduction\/Purpose Using a vector force table with a center post and ring this

# Lab #2 - Vectors Name Katie Ellis Groups Names Eden Emily...

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8/30/18 Vectors Name: Katie Ellis Groups Names: Eden, Emily, Felicia Introduction/Purpose: Using a vector force table with a center post and ring, this experiment aimed to exemplify the three different ways to add vectors and how those vectors look graphically. By experimentally guessing the third or fourth vector that balances out the existing force vectors, we got to find the equilibrant in the system of forces, theoretically. Throughout this lab, the data should reflect the error in the groups guess of the equilibrant and exemplify the three ways to add vectors (experimentally, graphically, and mathematically). Data/Calculations: Task #1 (given vectors: F 1 = 1.47N at 0° and F 2 = 1.47N at 90°) Experimental Results: Equilibrant Vector: F 3 = 2.10N at 225° Mass measured: 214g (1x10 -3 ) = 0.214kg x 9.8(m/s 2 )= 2.10N How it was done: By placing the two given vectors where directed with the proper amount of weight on each string, the third vector was found by pulling a third string around the vector force table until the ring was centered around the centered pin. The goal was to find the direction of the third vector and then the magnitude after the string was secured with the attachable pulley. Once the ring became very close to center at the 225°, we secured the string with the attachable pulley and began adding masses (g) to the end of the string. Adding 50g at a time, then 20g, then 10g, we kept decreasing until all the vectors balanced out with the ring centered around the centered pin on the vector force table. This gave us how many grams of mass needed to balance the system, which becomes the magnitude of the equilibrant vector. To convert those grams into newtons, you have to convert grams to kilograms and multiply by the force of gravity to finally get the magnitude of the new vector. Graphical Results:
* Adding graphically F 1 = (1.47,0) and F 2 = (0,1.47) to get F 3 = (-1.47,-1.47) *Graphical representation of F 1 = (1.47,0) and F 2 = (0,1.47) to produce the experimental equilibrant F 3 = (-1.48,-1.48) Mathematical Results: General equations needed: F 3 = -F 1 -F 2 or F 1 +F 2 =-F 3 *F 3 is the resultant F 1 = 1.47N at 0° *convert to cartesian* (1.47 x cos 0°, 1.47 x sin 0°)= (1.47,0) F 2 = 1.47N at 90° *convert to cartesian* (1.47 x cos 90°, 1.47 x sin 0°)= (0,1.47) F 3 = (-1.47,-1.47) *calculations* (1.47,0)+(0,1.47)= (1.47,1.47) x (-1) =F 3 *convert to polar* (
(-1.47 2 + -1.47 2 )=2.10N, tan -1 -1.47/-1/47= 45°+180°= 225°) = F 3 = (2.08, 225) Comparison/Analysis: Equilibrant Vector : F 3 = 2.10N at 225° *convert to cartesian* (2.10 x cos 225°, 2.10 x sin 225°)= F 3 = (-1.48,-1.48) Graphical Vector : F 3 = (-1.47,-1.47) *convert to polar* ( (-1.47 2 + -1.47 2 )=2.10N, tan -1 -1.47/- 1/47= 45°+180°= 225°) = F 3 = (2.08, 225°)

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• Spring '15
• Farris

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