Optimizing Cable Direction for Minimum Force in Vector Addition of Forces

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SUMMARY

The discussion focuses on optimizing the direction of a third cable to minimize the force it exerts in a vector addition scenario involving three cables. The known forces are 1200 N at 45 degrees and 800 N at 30 degrees, resulting in a calculated resultant force of 1605.28 N at an angle of 16.22 degrees. The objective is to determine the angle θ for the third cable that minimizes its force while maintaining equilibrium in the x-y plane. The solution involves vector analysis and trigonometric calculations to find the optimal direction.

PREREQUISITES
  • Vector addition in physics
  • Trigonometric functions and their applications
  • Understanding of equilibrium in two-dimensional forces
  • Basic knowledge of force components in the x-y plane
NEXT STEPS
  • Study vector resolution techniques in physics
  • Learn about equilibrium conditions for static systems
  • Explore optimization methods in force analysis
  • Review trigonometric identities and their applications in force calculations
USEFUL FOR

This discussion is beneficial for physics students, engineers involved in structural analysis, and anyone interested in optimizing force applications in mechanical systems.

Suy
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Homework Statement


Three cables pull on the pipe such that they create a resultant force having magnitude FR. If two
of the cables are subjected to known forces, as shown in the figure, determine the direction θ of
the third cable so that the magnitude of force F in this cable is a minimum. All forces lie in the
x–y plane.What is the magnitude of F? Hint: First find the resultant of the two known forces.
Prob._2-31.jpg

Homework Equations





The Attempt at a Solution


I have the solution
but i don't understand "minimum" in this question?
"the third cable so that the magnitude of force F in this cable is a minimum. "
 
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Suy said:
...determine the direction θ of the third cable so that the magnitude of force F in this cable is a minimum.

[tex]F_{R}=\left[1200sin45\textdegree+800cos30\textdegree\right]\hat{i}+\left[1200cos45\textdegree-800sin30\textdegree\right]\hat{j}[/tex]

[tex]F_{R}=1541.35\hat{i}+448.53\hat{j}[/tex]

[tex]\left|F_{R}\right|=1605.28[/tex]

[tex]tan^{-1}\left (\frac{448.53}{1541.35}\right)=16.22\textdegree[/tex]

Cheers!
 
wat is the solution...can u give me?
 

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