Torque/moment of beam in equilibrium

[Hyperfluxe]

Homework Statement

http://i.imgur.com/5plXa.png

Homework Equations

ƩF_x = 0, ƩF_y = 0, ƩM = 0

The Attempt at a Solution

I realize that this is an equilibrium problem and that the sum of the forces in the x and y directions is 0, and the sum of the moments is 0. I draw two free body diagrams, one at point A (the contact point between the left wall and beam), and one at point B (the edge). Each FBD has the forces exerted by the bar on the wall - I call it F_a and F_b respectively. Also, both points have normal forces which are perpendicular to the surface. F_Na is perpendicular to the wall while F_Nb is perpendicular to the rod/edge. Also incorporating the weights (W_a and W_b), I get four equations.

What I can't figure out is how to formulate an equation for the sum of the moments, and subsequently solving the problem for theta. Any help would be tremendously appreciated.

 

[tiny-tim]

Hi Hyperfluxe!

I realize that this is an equilibrium problem and that the sum of the forces in the x and y directions is 0, and the sum of the moments is 0. I draw two free body diagrams, one at point A (the contact point between the left wall and beam), and one at point B (the edge). Each FBD has the forces exerted by the bar on the wall - I call it F_a and F_b respectively. Also, both points have normal forces which are perpendicular to the surface. F_Na is perpendicular to the wall while F_Nb is perpendicular to the rod/edge. Also incorporating the weights (W_a and W_b), I get four equations.

(you don't need to separate the weight, one W will do )

You should be able to find two equations for F_Na and θ.

What is your moments equation (about B)?

 

[Hyperfluxe]

Ok, what I did instead is making a sum of the moments equation about point A. So, from my FBD, I get 3 equations. F1 is the force between the left contact point and the wall, and F2 is the force from the edge contact point. W is the weight (assuming it acts down at the center of the bar), I get:

1 - F1 = F2sin(theta) ----> I didn't need this equation

2 - F2cos(theta) = W

3 - (F2cos(theta))(0.325)-(W)(x) + (F2sin(theta))(y)
where x=0.5cos(theta) and y=0.325tan(theta) from geometry.

I equate W, cancel out F2, and solve for theta using a some algebra and a trig identity to get theta = 29.98degrees.

Is my method and answer correct? Thanks!

 

[tiny-tim]

Hi Hyperfluxe!

2 - F2cos(theta) = W

3 - (F2cos(theta))(0.325)-(W)(x) + (F2sin(theta))(y)
where x=0.5cos(theta) and y=0.325tan(theta) from geometry.

I equate W, cancel out F2, and solve for theta using a some algebra and a trig identity to get theta = 29.98degrees.

Is my method and answer correct? Thanks!

Yes, your method is fine.

(though you could just have said F₂0.5secθ, instead of (cosθ + sinθtanθ) )

(i haven't chekced your numerical result)

 

[Hyperfluxe]

Thank you very much! Turns out I didn't have enough knowledge on moments on Friday, so I wasted 2 hours trying to solve that question. It feels really satisfying right now though!

 

[tiny-tim]

(You obviously needed practice on moments, so I didn't suggest this before …)

There are only three external forces, and you know the directions of all of them …

so can you see a way to find θ just by drawing some lines on the diagram? ​