Forces: A joint which is part of a loaded framework carries a vertically downward load of 14kN.

1

12345!!

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Hey guys,
I don’t suppose any of you are able to help me out on this one?

If you could provide a breakdown on how this is accurately calculated it would be very much appreciated. I can calculate to the degree so know I’m doing something not quite right as the following 2dp are incorrect.
03D0505D-994F-4A6E-AC95-5A3CF89F98F3.jpegDBA11135-CECF-423D-9368-63FCA453DF63.jpeg
 
I'm assuming the system is in equilibrium. If so, then [imath]\displaystyle \sum F_x = 0[/imath] and [imath]\displaystyle \sum F_y = 0[/imath].

Force_1.jpg
 
I’m not sure
skeeter said:
I'm assuming the system is in equilibrium. If so, then [imath]\displaystyle \sum F_x = 0[/imath] and [imath]\displaystyle \sum F_y = 0[/imath].

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Hi there,

I am not sure on this one. If you don’t mind could you please provide the workings out in order to come up with the final answers. That would be very much appreciated
 
12345!! said:
I am not sure on this one. If you don’t mind could you please provide the workings out in order to come up with the final answers. That would be very much appreciated
Click to expand...

Actually, the way for you to learn is not to have yet-another fully-working example (like you already have in your textbook, your class notes, and whatever online resources from which you've tried to study), but is to have put your own effort in.

So please reply with a clear listing of your thoughts and efforts so far, so we can see where you're getting stuck. (Read Before Posting)

Thank you!
 
I fully understand what you are saying however I cannot seem to figure out this particular question. Here is a similar working out.
Many thanks.IMG_3936.jpeg
 
Last edited by a moderator:
12345!! said:
I fully understand what you are saying however I cannot seem to figure out this particular question. Here is a similar working out.
Many thanks.View attachment 36629
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Okay, so you've done similar things. which means you should be able at least to get started on this one. So please reply showing your start and how far you've been able to get. Thank you!
 
In the horizontal direction, force components to the right = force components to the left

[imath]R\cos(35) = 11 + Q\cos(45)[/imath]

In the vertical direction, force components up = force components down

[imath]R\sin(35) + Q\sin(45) = 14[/imath]

solve the system of equations for R and Q in kN.
 
12345!! said:
Hey guys,
I don’t suppose any of you are able to help me out on this one?

If you could provide a breakdown on how this is accurately calculated it would be very much appreciated. I can calculate to the degree so know I’m doing something not quite right as the following 2dp are incorrect.
View attachment 36625View attachment 36626
Click to expand...

  1. I am going to assume that all the forces and the lines OR & OQ are on the plane of the paper
  2. I'll draw a free-body-diagram depicting the applied forces (along the line OP and vertical) and reaction forces (along the lines OQ and OR)
  3. Then following response #8 (using assumption of static equilibrium) I'll obtain 2 linearly-independent equations involving forces along OQ & OR (response #8)
  4. Solve for those two (reaction) forces using algebra.
 
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