Having a hard time with visual of FBD

In summary, the conversation revolved around writing a position vector for point D, with the initial assumption that the z component was 0. However, upon further examination, it was determined that this assumption was incorrect and that the rope must be sagging. The conversation then shifted to discussing the correct x and y components for point D and the possibility of redrawing the diagram from different perspectives. The conversation also touched on the frustration of doing calculations, as well as the use of online solvers and an old solution to the problem. Finally, there was a mention of difficulties with posting attachments.
  • #1
Saladsamurai
3,020
7
I am trying to write a position vector for point D

Picture3.png


I originally thought that the z component was 0...but now common sense is telling me that it can't be...the rope must be sagging.

Are my first to components correct or am I visualizing this incorrectly? :
8i+7j-(something)k <---or are my x and y's wrong too?
 
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  • #2
look more carefully at the diagram... (for D) x and y are both smaller than you have stated, and z is not zero.
It may help to try to redraw the diagram looking from other perspectives, starting with looking downwards.
Here's a clue... x=3 for point D.
 
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  • #3
mda said:
look more carefully at the diagram... (for D) x and y are both smaller than you have stated, and z is not zero.
It may help to try to redraw the diagram looking from other perspectives, starting with looking downwards.
Here's a clue... x=3 for point D.
And it is y=2 correct And z=-2?

I need to look at the fact that the ropes AD and DC are not parellel to the x and y axes so I can't use the 7 and 8 foot measurements that correspond.

Maybe I should redraw the 3-D case in two 2-D cases instead?
 
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  • #4
I am so goddamn sick of doing all of these calculations just to get the wrong answer! WTF WTF WTF WTF!
 
  • #5
Here is my work. This is my third attempt in 3 hours. Am I F'ing stupid? It's okay to tell me. Disregard the attempt to solve the equations. It is a retarded substitution. I am not even sure it would work. But, I now see the easy elimination. But the equations themselves are somehow incorrect.

When I enter them into an online solver, only T_A comes out correct. I don't know if that helps me figure out where my error is...but this is killing me.

The equations are the ones marked (1) (2) and (3)

Picture4.png
 
  • #6
Assume the position vector of B as (0,0,0). Wright down the position vectors of A, D, C. and then find vector DA, DB and DC. Find the unit vectors of them. Let p, q and r be the magnitudes of the tensions along DA, DB and DC. consider the points A', B' and C' are the points on the plane containing D which is parallel to the plane ABC. Now take the projection of the three tensions on this plane. For equilibrium two conditions should be satisfied. i) Z- components of the tensions should be equal to 20lb, and projections of the tensions should satisfy the Lami's theorem. You can find out the required angles by noting down the proper distances.
 
  • #7
0 5546 Tc is wrong. It should be 0.3244Tc. Try this value.
 
  • #8
rl.bhat said:
0 5546 Tc is wrong. It should be 0.3244Tc. Try this value.

Why? 2/3.606=.5546
 
  • #9
I have been vindicated! My instructor posted an old solution to the problem in which it was set up differently. My position vectors should be correct.
 
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  • #10
rc should be 0i + 7j + 0k and rdc should be -3i + 5j + 2k
 
  • #11
rl.bhat said:
rc should be 0i + 7j + 0k and rdc should be -3i + 5j + 2k

Thank you sir!

Casey
 
  • #12
Say, Saladsamurai, what are you using to post your attachments? I'm just getting "red x's" at your illustrations...
 
  • #13
Saladsamurai said:
I am so goddamn sick of doing all of these calculations just to get the wrong answer! WTF WTF WTF WTF!

heh yes I also hate position vectors but there are much worse problems for that then the type you posted. Mid term exam had a equivalent systems question and there were dimensions everywhere, at least 20 dimensions markings - in millimeters as well of course. :cry:
 
  • #14
dynamicsolo said:
Say, Saladsamurai, what are you using to post your attachments? I'm just getting "red x's" at your illustrations...

Photobucket...is anyone else seeing red x's??

Casey
 

1. What is a free-body diagram (FBD)?

A free-body diagram is a visual representation used in physics and engineering to illustrate the forces acting on a single object. It shows all the external forces acting on the object and their directions, but does not include any internal forces.

2. Why is it important to use an FBD?

An FBD helps to simplify complex problems by breaking down all the forces acting on an object into individual components. It also helps to identify any unknown forces and determine their magnitude and direction, making solving physics problems easier and more accurate.

3. How do I create an FBD?

To create an FBD, start by identifying the object you want to analyze and draw a simple sketch of it. Then, identify all the external forces acting on the object, such as weight, normal force, friction, and applied forces. Draw arrows to represent the direction and magnitude of each force, and label them accordingly.

4. What are some common mistakes to avoid when drawing an FBD?

One common mistake is to mix up the direction of the forces. Make sure to draw the arrows in the correct direction according to the given problem. Another mistake is to include internal forces, such as tension in a rope, in the FBD. Only external forces should be included in an FBD.

5. Can an FBD be used for objects in motion?

Yes, an FBD can be used for objects in motion as it helps to analyze the forces that affect an object's motion. In such cases, the FBD will also include the acceleration vector, which represents the direction and magnitude of the object's acceleration. This can help in solving problems involving Newton's laws of motion.

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