Solving Part B of a Homework Problem Involving Principal of Moments

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SUMMARY

The discussion focuses on solving part B of a homework problem related to the principle of moments, specifically involving forces acting on rods in a triangular configuration. The user initially calculated a force of 250N at an angle of 53.1 degrees but found a discrepancy with the book's solution of 290N at 43.6 degrees. Through further analysis, including the consideration of lateral forces and equilibrium conditions, the user determined that a lateral force of 210N on rod AB was necessary to reconcile the differences and arrive at the book's answer.

PREREQUISITES
  • Understanding of the principle of moments
  • Knowledge of Free Body Diagrams (FBD)
  • Familiarity with Newton's 3rd Law of Motion
  • Basic trigonometry for calculating angles and forces
NEXT STEPS
  • Study the application of the principle of moments in static equilibrium problems
  • Learn how to construct and analyze Free Body Diagrams (FBD)
  • Explore the effects of lateral forces in structural mechanics
  • Review trigonometric functions in the context of force resolution
USEFUL FOR

This discussion is beneficial for students studying mechanics, particularly those tackling problems involving static equilibrium and the principle of moments. It is also useful for educators looking for examples of common misconceptions in force analysis.

OnlinePhysicsTutor
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Homework Statement


Please see attached picture

Homework Equations


principal of moments
IMG_3477.jpg


The Attempt at a Solution


part a) is fine. For part b, I would Normally take moments about C. The force acting on BC at B must be the tension in AB. The geometry of the triangle gives and angle of 53.1 to the horizontal, and moments gives a force of 250N. However the solution in my book says 290N at and angle of 43.6 degree to the horizontal, so I'm not sure if I have missed something, taking into account compressive tension along BC doesn't seem to work. Any pointers would be appreciated.
 
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Are you assuming that the force at BC is along the line BA? That would give you 53.1o, but is not necessarily the case.
 
kuruman said:
Are you assuming that the force at BC is along the line BA? That would give you 53.1o, but is not necessarily the case.
edit; I assume the book answer is taking into account some lateral force also on AB, but I don't see how I would calculate that.
 
Look at rod BC. What would the forces acting on it look like? How big a vertical component do you have at B acting on rod BC? Rod BA would have an equal amount acting on it at B but in the opposite direction (Newton's 3rd law). Now, if you can find the horizontal component on BA at point B, you can find the tangent of the angle that you are seeking without any assumptions about its value.
 
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OnlinePhysicsTutor said:
edit; I assume the book answer is taking into account some lateral force also on AB, but I don't see how I would calculate that.

Draw a FBD of rod BA and demand that it be in equilibrium, just like you did for the entire triangle. You know (or should be able to figure out) all but the lateral force.

On edit: Don't forget there is a force with a vertical and horizontal component at point A as well.
 
I get it now, thank you. I did moments on AB, gives a lateral force of 210N, which then works out to give the answer in the book.
 

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