Unformatted text preview: Part 1 Holder & Weight (grams) 550 550 550 550 550 Angle Between cords (degrees) 7 25 50 80 100 Conclusion: When the angle is very small between the two cords holding up the weight, the combined weight readings on each scale should be very close to what it is holding up. this case it is 550 grams and at an angle of 7o the sum of the scales is 540 grams. As angle increases so does the sum of the two scales, even to almost double the actual weight being held. When the angle between the two cords is 100o the sum of the scale 840 grams. Part 2 Force #1 Force #2 Force #3 Diagonal #1 Diagonal #2 Diagonal #3 Scale Reading (grams) 480 400 460 Length (cm) 4.8 4.1 4.5 Scale: 100 grams = 1 centimeter Conclusion: Transforming the scale readings from grams into centimeters then drawing these at the direction given, any two lines can be used to find the length that the third line should be. In my diagram the angle between lines A and B is 120o. If you draw a lin off of A towards line B and then a 120o line off of B towards line A, the intersection o two lines should, if drawn closely enough, be exactly the same length away from the of the chart as the third line, line C, is in length. Scale #1 (grams) 260 270 300 360 430 Scale #2 (grams) 280 290 300 340 410 cords holding up the weight, the e very close to what it is holding up. In sum of the scales is 540 grams. As the , even to almost double the actual wo cords is 100o the sum of the scales is Direction (degrees) 0 120 232 Direction (degrees) 180 from A 180 from B 180 from C into centimeters then drawing these lines d to find the length that the third line es A and B is 120o. If you draw a line 120o f B towards line A, the intersection of these actly the same length away from the center Part 3 Horizontal Components: HA = 30 g cos( 30 ) HB = 90 g cos( 150 ) HC = 60 g cos( 270 ) HR Vertical Components: VA = 30 g sin( VB = VC = R= 90 g 60 g 30 ) = = = = = = = = 25.98 77.94 0 51.96 15 45 60 0 51.96 0 51.96 HR2+VR2= R2 R= R in Radians = Convert Degrees into Radia 30 = 150 270 0 = = = sin( 150 ) sin( 270 ) VR = 51.96 g tan θ R θR = θ anti  R = Conclusion: Here I was able to calculate the direction and amount of force needed to bring the three random forces back to equilibrium. Since the Vertical force was already in equilibrium it is safe to say that the HR value will be the weight needed to bring the forces to equilibrium a angle of 0o. On the last page of the handout I drew a to scale model of my exact observatio rt Degrees into Radians 0.52 2.62 4.71 0 51.96 1.63 d to bring the three y in equilibrium it is ces to equilibrium at an my exact observations. ...
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This note was uploaded on 04/21/2010 for the course PHYS 1004 taught by Professor Dr.good during the Spring '09 term at Ouachita Baptist.
 Spring '09
 DR.GOOD
 Physics, Force

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