Solving Part B of a Homework Problem Involving Principal of Moments

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Homework Help Overview

The discussion revolves around a homework problem related to the principle of moments, specifically focusing on the forces acting on a system involving rods and angles. The original poster is attempting to solve part B of the problem, which involves calculating forces and angles based on given conditions.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster discusses taking moments about a point and considers the geometry of the triangle involved. Some participants question the assumptions regarding the direction of forces and suggest examining the components of forces acting on the rods. There is also mention of drawing free body diagrams (FBD) to analyze equilibrium.

Discussion Status

The discussion is active, with participants exploring different interpretations of the forces involved. Some guidance has been offered regarding the need to consider both vertical and horizontal components of forces, and the original poster expresses understanding after further analysis.

Contextual Notes

There is a reference to a discrepancy between the original poster's calculations and the solution provided in the textbook, which raises questions about the assumptions made regarding lateral forces and equilibrium conditions.

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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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