Equilibrium Particles in Space

Objective: To determine the tension of cables both theoretically and experimentally. We will hang three cables at different locations on the ceiling of the force table. We will then tie about a 6kg mass to the end of the cables and measure the length of the cables. We will then calculate the tension using this information. Afterwards we will find the actual force using the spring scales at the top of the cable. The two answers will be compared and the percent error calculated. An explanation must be given for any percent difference.


Equipment Needed:
- 3 Cables -3 Hooks
- Load -Meter stick
- 3 Scales -Force table

Experimental Procedure: We get access to a force table and make sure it is up to working conditions. We then screw 3 hooks at 3 separate locations on the ceiling in the forces table, we then tighten them with nuts and washers. Then a spring scale is attached to each hook. Next cut a piece of string and tie one end to a hook. Do the preceding 2 more times for each hook and scale. Next tie all three strings together and place a 6.05kg mass on the point the strings meet. Then find a point on the ceiling that is perpendicular to the point where all the strings meet. This will be your zero point origin on the x, y, and z axes. Also find the distance from the origin on the ceiling to all three hooks along the x, y, and z axes. Then measure the distance of the string make sure to include the spring scale. Record all the measurements.


Theory: Equilibrium of particles in space is the theory that when the particle O is in equilibrium the sum of all forces is zero.
Fx = 0 Fy = 0Fz = 0


Results: In centimeters and kilograms and Newton.
A (-49, 73.5, 4.5) AO=87.5cm TOA = 24.5N
B (-10, 73.5 -45.5) BO=87cm TOB =14.7N
C (40, 73.5, 19) CO=86.5cm TOC =33.3N O (0, 0, 0) W=6.05kg
Theoretical Calculations: Only using x, y, and z position not the cable lengths or
tension found in experiment.

I. W= -6.05kg (9.8m/s2) = -59.35N

II. = A-O = (-49cm) î + (73.5cm) ĵ+ (4.5cm) k
____________________
OA=√ (-49)2+ (73.5)2+ (4.5) 2 = 88.45cm

λ OA = = -.55 î +.83 ĵ +.05 k
88.45

TOA λOA = -.55 TOA î +.83TOA ĵ +.05TOA k

III.  =B-O = (-10cm) î + (73.5cm) ĵ + (-45.5cm) k
______________________
OB=√ (-10) 2 + (73.5) 2 + (-45.5) 2 = 87.02cm

λ OB = = -.11 î +.84 ĵ -.52 k
87.02

TOB λOB = -.11 TOB î +.84TOB ĵ +.52TOB k

IV.OC = C-O = (40cm) î + (73.5cm) ĵ + (19cm) k
__________________
OC=√ (40) 2+ (73.5) 2+ (19) 2 =85.81cm

λ OC = OC= -.47 î +.86 ĵ +.22 k
85.81
TOC λOC = -.47 TOC î +.86TOC ĵ +.22TOC k

V. ∑Fx = 0
-.55 TOA î -.11 TOB î -.47 TOC î = 0
∑Fy = 0
83TOA ĵ +.84TOB ĵ +.86TOC ĵ – 59.35N = 0
∑Fz = 0
.05TOA k+.52TOB k+.22TOC k = 0


VI. Solve for the three unknowns with the three equations with these results:

TOA = 22.6N TOB = 15.4N TOC = 31.3N





Percent Error: E = theoretical value (estimated)
A = measured value (actual)
| E – A | × 100
A
Since I only used the position in space to calculate the cables tension and length I will have to find the percent error for OA, OB, OC, TOA, TOB and TOC.

OAE = 88.45 OBE = 87.02 OCE = 85.81
OAA = 87.50 OBA = 87.00 OCA = 86.50
Percent error = 1.09% = 0.02% = 0.80%

TOA E = 22.6N TOB E = 15.4N TOC E = 31.3N
TOA A = 24.5N TOB A = 14.7N TOC A = 33.3N
Percent error = 7.76% = 4.76% = 6.01%


Conclusion: The most likely reason for the error in the cable lengths is imprecise measurements by the people who did the measurements. When the meter stick was placed and the line plotted a small missed reading was put down. This gave the small percent errors for the lengths. This lead to a larger error when the figures were used in the more complicated tension formulas. The percent errors in tension then became much more magnified.

Repeat the experiment with out OB:
Given: A (-48, 74.5, -5.8) W = 6.05kg
C (41.5, 74.5, 6) TOA = 32.34N TOB = 35.28N


Theoretical Calculations: Only using x, y, and z position not the cable lengths or
tension found in experiment.


1. = A-O = (-48cm) î + (74.5cm) ĵ+ (-5.8cm) k
____________________
OA=√ (-48)2+ (74.5)2+ (-5.8) 2 = 88.81cm

λ OA = = -.54 î +.84 ĵ -.07 k
88.81

TOA λOA = -.54 TOA î +.84TOA ĵ +.07TOA k





2. OC =C-O = (-41.5cm) î + (74.5cm) ĵ + (6cm) k
______________________
OC=√ (-41.5) 2 + (74.5) 2 + (6) 2 = 85.49cm

λ OC = OC= -.49 î +.87 ĵ -.07 k
85.49

TOB λOB = -.49 TOB î +.87TOB ĵ +.07TOB k

3. ∑Fx = 0
-.54 TOA î-.49 TOB î = 0
∑Fy = 0
.84TOA ĵ+.87TOB ĵ -59.35N= 0
∑Fz = 0
.07TOA k+.07TOB k = 0

4. Solve for the two unknowns with the three equations with these results:

TOA = 32.99N TOB = 36.36N

5. Percent Error: E = theoretical value (estimated)
A = measured value (actual)
| E – A | × 100
A

OAE = 88.81 OCE = 85.49
OAA = 87.50 OCA = 86.50
Percent error = 1.50% =1.16%

TOA E = 32.99N TOC E = 36.36N
TOA A = 32.34N TOC A = 35.28N
Percent error = 2.01% =3.06 %

Conclusion: The tension on the cables increased as expected with the removal of cable OB. The percent error was closer to the actual measurements mostly because there were fewer things to plot.