Solving Vector Measurement Concerns in Lab

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SUMMARY

The discussion focuses on resolving discrepancies in vector measurements in a laboratory setting. The user initially calculates the resultant vector (R) using trigonometric functions, yielding a value of 26 cm, which is significantly lower than the measured value of 29.4 cm. The correct approach involves recalculating the X and Y components without relying solely on trigonometric functions, leading to more accurate results. The user’s alternative method, which sums the components directly, provides a more accurate resultant vector measurement.

PREREQUISITES
  • Understanding of vector components and resultant vectors
  • Basic knowledge of trigonometric functions (sine and cosine)
  • Familiarity with the Pythagorean theorem for calculating resultant vectors
  • Experience with laboratory measurement techniques
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  • Review vector addition and subtraction techniques in physics
  • Study the application of trigonometry in non-right triangles
  • Learn about error analysis in experimental physics
  • Explore advanced vector analysis tools and software
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Physics students, laboratory technicians, and educators involved in teaching vector measurement and analysis techniques.

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http://img366.imageshack.us/img366/5171/phwq5.jpg Hello there. In lab, we measured vectors. I'm having a little concern on how to get X and Y. If I follow the books method, R_calc comes out too small compare to measure R. (Big % error).

X:
15.5+12.4cos37 = 25.4
Y:
0+9.3sin37 = 5.4
R:
square root of (25.4^2+5.4^2) = 26cmIf I do it my way, R is a lot closer to measure.

X:
15.3+12.4= 27.9cm
Y:
0 + 9.3 = 9.3 cm
R:
square root of (27.9^2+9.3^2) = 29.4 cmThe measurements (x1/2; y1/2; R m) are pretty accurate. What am I doing wrong? Thanks for any help.
 
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OK, I will try to point out a couple of things:

Agent said:
http://img366.imageshack.us/img366/5171/phwq5.jpg


Hello there. In lab, we measured vectors. I'm having a little concern on how to get X and Y. If I follow the books method, R_calc comes out too small compare to measure R. (Big % error).

X:
15.5+12.4cos37 = 25.4
I'm not sure I understand what you are attempting, but as I see it your x value is not correct. R is made up of x1 and b in this case. You need to find b, to solve it this way. So use trig to find b instead of x2. But it isn't a right triangle, so the resultant is a little more difficult to solve for.Y:
0+9.3sin37 = 5.4
R:
square root of (25.4^2+5.4^2) = 26cm

This is correct:If I do it my way, R is a lot closer to measure.

X:
15.3+12.4= 27.9cm
Y:
0 + 9.3 = 9.3 cm
R:
square root of (27.9^2+9.3^2) = 29.4 cm


The measurements (x1/2; y1/2; R m) are pretty accurate. What am I doing wrong? Thanks for any help.
 
Last edited by a moderator:

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