Tension in rod AB using principle of moments

[OnlinePhysicsTutor]

Homework Statement

Please see attached picture

Homework Equations

principal of moments

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.

 

[kuruman]

Are you assuming that the force at BC is along the line BA? That would give you 53.1^o, but is not necessarily the case.

 

[OnlinePhysicsTutor]

Are you assuming that the force at BC is along the line BA? That would give you 53.1^o, 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.

 

[kuruman]

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.

 

[kuruman]

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.

 

[OnlinePhysicsTutor]

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.