Solving for the internal forces and reactions

[chunchunmaru]

Homework Statement

I'm having a hard time finding the reactions at the supports and the force exerted by the spring.

Given:
20 lbs
25 lbs
35 lbs
k = 200 lb/ft
lo = 2.5 ft
members are 4 ft long

Homework Equations

ΣM
∑Fy = 0
∑Fx = 0

The Attempt at a Solution

I tried solving for the reactions first. I know that at pin A it would react with two forces upward and downward. At D, it would only react upward since it is a roller.

I first solved for the x-direction reaction at A and got 35 lbs to the left because it would resist the 35 lbs force to the right at D.

I'm kind off confused on what is the direction of the force exerted by the spring. I imagine the spring being stretched when the downward forces are in effect. I think they would just exert a left and right forces but I don't know how to get their values. Any idea how?

Then I tried solving for the moment at A using the unstretched length of the spring as the leverage. Or can it be solved without using the moment equation? After that I'm stuck.

 

[Orodruin]

Hint: The internal forces (such as the forces exerted by the spring) are irrelevant for the force and torque equations of the entire contraption.

 

[chunchunmaru]

So when solving for the reactions we neglect the force exerted by the spring? Then when do we use it?

If so, when I tried solving for the reactions. I got the RAx = 35 lbs to the left. Then I tried summing the moment at point A using the length of the unstretched spring length for the leverage and I got RDy = 25 lbs upward. After that, I all the forces in the y-direction and got RAy = 20lbs.

I thought it was fine until I solved for the internal forces on the members. I tried solving the force at the member BD but using the joint method, it didn't make any sense because it's direction should be opposing the RDy upward and the 35 lbs force to the right. There is no such direction that would cancel the forces out.

 

[Orodruin]

Then when do we use it?

The free body diagram of the entire object is not the only free body diagram you can draw. All parts of the object must be in equilibrium so you will need to consider more than one fbd.

 

[chunchunmaru]

Is it the one when the spring is stretched?

 

[Orodruin]

The spring is always stretched. Your object consists of two different rods. Both must be in equilibrium for the system to be.

 

[chunchunmaru]

Hmm so I would need to dismember the two members? but that is after I solve for the reactions right?

 

[Orodruin]

You can do it in any order you want. The important thing is to find enough constraints to solve for your unknowns. Note that you only need the fbd of one of the members. If the entire system and one member are in equilibrium, the second member must also be.

 

[chunchunmaru]

I tried to dismember one of the members to get the force exerted by the spring, but I have a question about the pin at the midpoint of the members.

For example I will dismember member ABC, such that it has reactions forces at A, 25 lbs force downward at C, and a force exerted by the spring on C but I don't know what to do with the forces coming from EB and BD would they be represented as a one force or should they be separate forces?

 

[Orodruin]

Does it matter? The only relevant issue for the fbd is where there are forces and what the total force at each point is.