Is My Tension Calculation Correct in Mechanics Problem?

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SUMMARY

The tension calculation for the mechanics problem is confirmed to be correct at 14123N. The solution involves using the equilibrium equations, specifically \(\sum F_x = 0\) and \(\sum M_A = 0\), to derive the tension in the cable. The moment at point C about point B is calculated as 49050N, leading to a tension of 7007N when considering the angle of 29.74 degrees. The method of using the intersection of lines of action for the forces provides a visual verification of the calculated tension.

PREREQUISITES
  • Understanding of static equilibrium principles in mechanics
  • Familiarity with trigonometric functions and their applications in physics
  • Knowledge of moment calculations and their significance in beam analysis
  • Ability to interpret and solve problems involving forces and tension
NEXT STEPS
  • Study the principles of static equilibrium in greater depth
  • Learn about moment calculations in mechanics, focusing on beam problems
  • Explore trigonometric applications in physics, particularly in force resolution
  • Investigate graphical methods for solving equilibrium problems, such as force triangles
USEFUL FOR

Students studying mechanics, particularly those tackling problems involving tension and equilibrium in beams, as well as educators looking for effective teaching methods in physics problem-solving.

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Homework Statement



Please see the attachment, Q2

Homework Equations



[tex]\sum[/tex]Fx = 0
[tex]\sum[/tex]MA = 0

The Attempt at a Solution



for part A, i get 14123N as the tension, moment at C about B, 9.8x1000x5 = 49050
49050/7 = 7007N
tan-1 (4/7) = 29.74o
TSin(29.74) - 7007 = 0
TSin(29.74) = 7007
T=7007/sin(29.74) = 14123N

Is this correct?

Also I had a go at part b, however, went round in circles, so can anyone help me?

Cheers

Andrew
 

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There are 3 forces that matter to the beam. First the tension in the cable; second the 1T weight; third the reaction at C. The lines of action of these three must meet at one point if the object is to be in equilibrium from a moment point of view. If you can identify that point of intersection, that will give you the direction of the reaction at C. If you now move those three lines apart to form a triangle, the sides will be in proportion to the forces. You should be able to get it from there. That will incidentally check your first answer for yourself. I accept that there are more algebraic ways of doing this, but with triangles etc, you can see what is going on.
 

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