3D Equilibrium with cables (statics)

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SUMMARY

The discussion focuses on solving a statics problem involving a mast supported by three cables, specifically determining the tensions in cables AC and AD, as well as the vertical force exerted by the mast. The user initially calculated tensions using vector equations and the sum of forces in the x, y, and z directions, but received feedback indicating errors in their results. Key vectors involved include AB, AC, and AD, with tensions calculated as 910.2 N for Fad and -694.2 N for Fac, both of which were marked incorrect. The solution emphasizes the importance of accurate arithmetic and the use of fractions for precision.

PREREQUISITES
  • Understanding of vector mathematics in three dimensions
  • Knowledge of static equilibrium and force summation
  • Familiarity with unit vectors and their application in physics
  • Basic skills in algebra for solving equations
NEXT STEPS
  • Review the principles of static equilibrium in three-dimensional systems
  • Practice calculating tensions in cable systems using vector analysis
  • Learn about the significance of using fractions in calculations for improved accuracy
  • Explore common mistakes in vector arithmetic and how to avoid them
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Students studying statics, engineers working with cable systems, and anyone involved in structural analysis or mechanics.

kirab
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Homework Statement



Diagram is attached.

The mast OA is supported by 3 cables. If cable AB is subjected to a tension of 500N, determine the tension in cables AC and AD and the vertical force F which the mast exerts along its axis on the collar at A.

a = 6m
b = 3m
c = 6m
d = 3m
e = 2m
f = 1.5m
g = 2m

Homework Equations



For this assignment, I'm pretty sure the prof just wants us to consider sum of Fx, Fy and Fz, and to take no moments.

The Attempt at a Solution



I have a complete solution, but apparently it's not the right answer according to the teacher assistant (I don't know what the right answer is).

so vector AB = 2i + 3j - 6k
AC = -1.5i + 2j - 6k
AD = -3i -6j - 6k

and I get that vector Fab = (142.85i + 214.285j - 428.55k)N

Fac = (-0.2308(Fac)i + 0.30769(Fac)j - 0.923076(Fac)k)N (where the Fac's on the right side of the equation represent the magnitude of the force *vector* Fac)
and

Fad = (-0.333(Fad)i - 0.666(Fad)j - 0.666(Fad)k)N

and of course, F = (Fk)N

where i, j and k are all unit vectors in the direction of the x, y and z axis, respectively.

From that, I have that:

sum of Fx = 0

Equation 1: 142.85i - 0.2308(Fac)i - 0.333(Fad)i = 0

sum of Fy = 0

Equation 2: 214.285j + 0.30769(Fac)j - 0.666(Fad)j = 0

sum of Fz = 0

Equation 3: -428.55k - 0.923076(Fac)k - 0.666(Fad)k + F(k)= 0

That's 3 equations in 3 unknowns. I then rearranged equation 1 to get Fac in terms of Fad. I plugged this Fac equality into equation 2, to get Fad alone.

My answer for Fad is 910.2 N. Apparently that's wrong. Plugging that in elsewhere, I get that Fac = -694.2 N and that F = 393.9 N, which are both wrong according to the marker.

Any suggestions/corrections? Did I not approach the question properly or assume too much?

Thanks
 

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This problem is a simple concurrent forces on point A. The sum of this force will equal a resultant force. You tackled the problem correctly, first find the position vectors, then the normalize (unit vectors). Check for mistakes in your numbers. I checked your position vectors and they are right. I also checked on unit vector and it was right.

By the way, I recommend using fractions instead of decimals whenever possible, so your answers will tend to be more accurate (your rounding is a bit off sometimes).
 
Aha, it appears that it was just a simple arithmetic mistake!

Thanks for the tip.
 

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