Drawing FBD: Cable Passes Over a Smooth Peg

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xxxyzzz
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Homework Statement
Determine the resultant internal loadings on the cross-section at point E. The cable passes over a smooth peg at C.
Relevant Equations
N/A
Structural diagram of a frame: member A to D to right end, with a vertical post rising from it to pinned joint C; cable from wall point B to C, and a second cable from C to the right end over a pulley; points A, D, E labeled; 0.5 m dimensions and 1 m spacings; uniform load 3 kN/m downward along the bottom member.

Attempting to draw the FBD for this problem but was wondering what to do about the cable, especially when I determine the internal loadings at E. At C, do I only draw one force arrow (CB), or draw both on the left and right side?
 

Discussion

xxxyzzz said:
At C, do I only draw one force arrow (CB), or draw both on the left and right side?
You draw ALL of the forces extant at that point. How many are there?
 
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xxxyzzz said:
The cable passes over a smooth peg at C.
...
At C, do I only draw one force arrow (CB), or draw both on the left and right side?
Welcome!
The cable, just like it happens to any rope or any chain, can only pull, it can't push anything.
In order to pull an object, it needs to be solidly attached to it.
Please, note that the problem states that the cable is not attached to the point C, it can only freely slide over it.
 
Hello, thank you for the welcome and responses. I am unsure if the notifications will be received if I do not click on the reply button, but I drew my idea of the FBD below and what I think all the forces are. I did not think that the cable is only pulling. Is this a correct assumption of the forces?
Hand-drawn free-body diagram of a horizontal beam with a pin reaction Ax pointing right and Ay pointing up at the left end, a 12 kN downward load at the middle, a 45° cable force FBC at the right end, and point E marked on a vertical member above the beam. Dimensions 1.5 m and 1.5 m are labelled.

And to find the internal loadings at E, do I proceed with this?
Hand-drawn free-body diagram of a vertical bar with points C at the top and E at the bottom, showing a red force arrow labeled FBC at 45 degrees at C, and red arrows labeled VE, NE and ME at E.

Thanks all.
 
I believe that the equivalent point load to the distributed one should not be 12 KN.
What Ve stands for?
 
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Lnewqban said:
I believe that the equivalent point load to the distributed one should not be 12 KN.
What Ve stands for?
My apologies, I screenshotted the problem from the textbook where w=3 kN/m, while my homework problem had w=4 kN/m. It should be 9 kN in my FBD. Ve is meant to be the internal shear force at E.
 
No apologies needed.
Force at C is incomplete.
If the structure is in equilibrium, the summation of forces and moment acting on each of its elements must be zero.
 
Draw the ENTIRE FBD all at one time with ALL of the forces. Always do that when solving this kind of problem
 
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https://www.physicsforums.com/attachments/313781Are these all the forces? I did not think that C had any reaction forces because it is smooth.
 
phinds said:
First, your link is broken and second PUT IT RIGHT HERE. This is like pulling teeth.
I was editing my post since I was unsure about my diagram, which was probably why the link was broken (might have accidentally deleted it). I believe I forgot some basic concepts from Statics since it has been a while, so I am having trouble understanding what forces I am missing. I do not mean to make you feel like you are "pulling teeth." Would like to ask you to be patient with me. Here is the reupload.
Free-body diagram of a bent beam with pin support A at the left end and a 12 kN downward load at a point 1.5 m from A, with point D marked. A vertical member carries points C and E and has a 1.5 m horizontal dimension to its right, with reaction labels Ax, Ay, Cx, Cy and force FBC drawn at 45 degrees.
 
xxxyzzz said:
I was editing my post since I was unsure about my diagram, which was probably why the link was broken (might have accidentally deleted it). I believe I forgot some basic concepts from Statics since it has been a while, so I am having trouble understanding what forces I am missing. I do not mean to make you feel like you are "pulling teeth." Would like to ask you to be patient with me. Here is the reupload.
View attachment 313784
Check your math on the distributed load. The reaction force at C, doesn’t exist.
 
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xxxyzzz said:
Here is the reupload.
View attachment 313784
This diagram is much better.
Time to compare forces and moments to determine the value of the cable tension Fbc.
With that value on hand, a vectorial addition of the tensions (same value, different directions) acting on C should let you know what types and what magnitudes of loads the cross-section E of the vertical element is withstanding.