vf_dundooah_218

# vf_dundooah_218 - /35 Winter Quarter 2008 Winter Quarter...

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Unformatted text preview: /35 Winter Quarter 2008 Winter Quarter 2008 Experiment: Vector Forces Your name: Anish Dundoo Partner: Eric Stetz Performed on: February 4 th , 2008 TA: Hyundeok Song Section number: 218 Promptness % (100 means it was on time): 100 Additional Scores (+)/Penalties (-): Abstract (4) The Vector forces in this experiment were made using a triple beam balance with the weight stationed on a pulley system at different angles. Uncertainties such as magnitude of the vector u{f}=9g and the direction of the vector u{ θ }=1 ° which is the minimum amount of weight and the angle needed respectively to get the system out of equilibrium were taken into account. The aim for the experiment is to verify Newton’s Second Law of motion that states that the net force acting on a body in equilibrium is zero. While carrying out the experiment, the angles that we picked were 27 ° , 141 ° , 156 ° 301 °. In order to form a balanced system with the ring in the middle of the peg, we tried different weights that would balance the forces out. The weights obtained were 140g , , 150g 158g and finally 265g . Finally we added the vectors to obtain the resultant vector which theoretically should have equaled zero. On comparing, the calculated value is 4.9±23grams. This does not compare closely to the measured value of 0. Sample Calculations (5) Calculating the x component of each magnitude force vector(x displacement): = × ( ) x1 r1 cos θ1 Where again r is the force vector = × x1 140g cos27 = x1 124.7g Calculating the sum of the x components: = × + × + + Fx f1 cosθ1 f2 cosθ2 f3cosθ3 f4cosθ4 = × +( × +( × +( × ( ) Fx 140g cos27 150g cos141 158g cos156 256g cos 301 =- . Fx 4 32g Calculating the y component of each magnitude force vector(y displacement): = × ( ) y1 r1 sin θ1 Where r is the force vector Error: Reference source not found, 09/17/04, Error: Reference source not found...
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vf_dundooah_218 - /35 Winter Quarter 2008 Winter Quarter...

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