Force balance for angled supports with 60 kN load

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rihitz
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Homework Statement


Hand-drawn free-body diagram of a truss joint B with members to supports A and C, a downward load P = 60 kN at B, angle 45 degrees at A and 20 degrees at C.
[/B]
P=60kN
angle alpha = 45 degrees
angle beta = 20 degrees

The task is to calculate the force on the base A and C, and to draw vectors that represent those forces.

Homework Equations

The Attempt at a Solution


I did similar examples at the class, but when I had to do this one on the test, I just could not get it done.

If I could get the balanced equations down correctly, I would manage to do the rest. (balanced force equations on x and y axis)

Sorry if there are wrong terms used, I am not a native speaker. Ill try to explain if there is anything unclear.

Thanks!
 

Discussion

What have you tried so far?

I'm no physicist, but I would think that some there might exist some r, such that r*(sin 20 + sin 45 ) = 60. This might help you balance out the y direction.
 
RUber said:
What have you tried so far?

I'm no physicist, but I would think that some there might exist some r, such that r*(sin 20 + sin 45 ) = 60. This might help you balance out the y direction.
Well the R should be right above the P in that case, which cancels the P itself. But that didnt help me out. Or I am drawing something incorectly
 
I was thinking that r would be a scalar...rsin(20) is the y component of the vector acting at point C, r sin 45 would be the y component of the vector acting at point A.
 
Or perhaps that is too simple. Let's take another look at this...
You have two force vectors W and V, each one has magnitude |W| and |V|, respectively.
You want to find those magnitudes such that the following equations are satisfied:
## |W| \sin 45 + |V| \sin 20 = 60##
##|W| \cos 45 = |V| \cos 20##
Does that seem more reasonable?
 
Handwritten sketch of a truss with joints A, B and C: member Fw runs from A up to B at 45 degrees, member Fv runs from B down to C at 20 degrees, and a downward load P = 60 kN acts at B. Below it, the equation sum Fy = 0 gives Fw + Fv - P = 0, rewritten as P/sin45 + P/sin20 = 60, then 84.85 + 175.42.

I am a bit confused.

Is this equation for the projection of all forces on y-axis wrong?
 
I also don't get why on the second equation both magnitudes have to be equal? Id say they won't be equal, since the angles are different
 
If F and G are the compression forces in rods AB and BC, what are the components of these forces in the x- and y directions?
 
Chestermiller said:
If F and G are the compression forces in rods AB and BC, what are the components of these forces in the x- and y directions?
In the picture F and G is Fw and Fv. Is that right? In that case components are. P/sin45 and P/sin20 for the y axis. And P*cos45 and P*cos20 for the x axis. Atleast if I got it right.
 
rihitz said:
In the picture F and G is Fw and Fv. Is that right? In that case components are. P/sin45 and P/sin20 for the y axis. And P*cos45 and P*cos20 for the x axis. Atleast if I got it right.
You didn't get it right.

Step 1 - Forget about P.

Step 2 - From the trigonometry, what are the components of F and G in terms of F, G, and the angles.
 
Chestermiller said:
You didn't get it right.

Step 1 - Forget about P.

Step 2 - From the trigonometry, what are the components of F and G in terms of F, G, and the angles.
Handwritten diagram of a truss joint B with members to supports A and C, member forces F and G, a 45 degree angle at A, a 20 degree angle at C, and a downward load P = 60 kN at B. Below are component equations: Fx = cos45° · F, Fy = sin45° · F, Gx = cos45° · G (crossed out and corrected from 90°), Gy = sin20° · G.


Like this?
 
Yes. Very nice.

The signs on the G components should be - in the x direction and + in the y direction. Now, using the upper junction point as your free body, do a force balance in the x direction and a force balance in the y direction on the junction point. What do you get?
 
Chestermiller said:
Yes. Very nice.

The signs on the G components should be - in the x direction and + in the y direction. Now, using the upper junction point as your free body, do a force balance in the x direction and a force balance in the y direction on the junction point. What do you get?
I hope I understood correctly what you asked due to the language barier.

But I guess this is what you meant:
Handwritten free-body diagram of a truss joint B with members to supports A and C at 45 and 20 degrees, a downward 60 kN load P, and force equations for F and G with sum of forces equal to zero.
 
Chestermiller said:
Excellent. Now solve for F and G.

Also, now compare with Ruber's post #5.
Is this correct?
Handwritten force-balance working on squared paper: a truss sketch with members F and G meeting at a peak, a 60 kN downward load P, and 45 and 20 degree angles, followed by equilibrium equations solved for F = 62.21 kN and G = 46.83 kN.
 
Your methodology is correct. I didn't check your algebra. If you want to do that, just substitute your results into the two force balance equations and see if the equations are satisfied.
 
Chestermiller said:
Your methodology is correct. I didn't check your algebra. If you want to do that, just substitute your results into the two force balance equations and see if the equations are satisfied.
Great! Thanks! To both of you.
So my mistake was that I tried to make components of F and G from P? While I actually had to make them from the F and G itself.
 
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Chestermiller said:
Your methodology is correct. I didn't check your algebra. If you want to do that, just substitute your results into the two force balance equations and see if the equations are satisfied.
Im sorry. Could you please tell me where I am wrong again? I am using the same methodology, but I have a feeling its incorrect, cause I can't find the second force momentum equation.

The A, Ax, Ay, C , Cy and Cx is added by me. Everything else was given.
 
Why am I getting a negative Cy, when it should be positive?
 
rihitz said:
Im sorry. Could you please tell me where I am wrong again? I am using the same methodology, but I have a feeling its incorrect, cause I can't find the second force momentum equation.

The A, Ax, Ay, C , Cy and Cx is added by me. Everything else was given.

I do not understand the connection between this post and the rest of this thread. What are Ax etc.?
 
haruspex said:
I do not understand the connection between this post and the rest of this thread. What are Ax etc.?
Oh sorry.. the image was not uploaded.
Here:
Handwritten statics sketch of a beam with pin support A, roller support C and a vertical load P = 60 kN at end Z, with labelled lengths 10 m and 8 m and an angle of 35 degrees, followed by force equilibrium equations and their worked solutions.