Cartesian Vector Form - Door with 2 Chains

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Homework Help Overview

The problem involves analyzing the forces acting on a door held open by two chains, with specific tensions given for each chain. The objective is to express these forces in Cartesian vector form, considering both 2D and 3D components.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the breakdown of forces into 2D and 3D components, with some expressing confusion about the transition from 2D calculations to 3D. There are attempts to clarify the relationships between the forces and the geometry involved, particularly regarding angles and distances.

Discussion Status

Some participants have provided guidance on how to approach the problem, suggesting methods to visualize the forces and break them down into components. There is acknowledgment of differing interpretations and calculations, with ongoing questions about specific values and methods.

Contextual Notes

Participants note constraints such as missing information about distances and angles, which complicate the calculations. There is also mention of homework rules that may limit the use of certain methods or assumptions.

BuckBee
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Homework Statement


[/B]
The door is held open by the means of 2 chains. If the tension in AB and CD is Fa = 300 N and Fc = 250 N, respectively, express each of these in Cartesian Vector Form

Homework Equations



Sin / cos / tan

The Attempt at a Solution



The angle of FA at B is atan(1.5sin30/(1+1.5cos30 )) then +y direction force A is 300cos that angle and -z direction force is 300sin that angle

There are 2 forces in 2D but I don't understand the 3D. There are 3 forces x, y and z

See Image Attached
 

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BuckBee said:
There are 2 forces in 2D but I don't understand the 3D.
Take it in stages. Drop a vertical from C to a point G below it on the ground. The tension has a vertical component and a horizontal one from G to D.
 
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Thanks but I am still a bit confused with the (x) force. I am really not sure about the 3D calculations. The horizontal length for C to D is 2 meters with a Fc of 200N

+y direction force is tan(1.5sin30/(1+1.5cos30) = X then 300 CosX

-z direction force is tan(1.5sin30/(1+1.5cos30) = X then 300 SinX
 
BuckBee said:
The horizontal length for C to D is 2 meters
No, that's just the X component. What is the whole horizontal distance? Hence what is tan(∠CDG)?
 
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ok I admit I have no idea how to do the 3D x, y, z forces for CD. I have no idea what formula to use or where to start

The 2D y, z forces for AB I understand that, see the calculations in the image attached below, I am sure that's correct

17311394_10211780872818684_1088220014_o.jpg
 
p3.jpg
 
BuckBee said:
how to do the 3D x, y, z forces for CD.
C, D, G lie in a vertical plane. That's just 2D.
Use Pythagoras to find the distance DG, hence find ∠CDG.
 
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cool thanks but we don't know the distance DC to find out the length DG using Pythagoras
p6.jpg
 
ok got it!

p7.jpg
 
  • #10
So does that look correct? If so I can then work out the x, y, z forces. Also how do I express the answer in cartesian vector form?
 
  • #11
BuckBee said:
So does that look correct? If so I can then work out the x, y, z forces. Also how do I express the answer in cartesian vector form?
Yes, that's it. You might want to keep one more digit of precision during the calculation... I have 3.14 where you have 3.1.
Cartesian vector form probably means as three numbers in parentheses: (Fx, Fy, Fz).
 
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  • #12
Great, thanks for the help, I think I've got it now. Can you please double check the calculations and answers and let me know if correct, Cheers

cvf1.jpg
 
  • #13
cvf2.jpg
 
  • #14
cvf3.jpg
 
  • #15
I'm getting rather different numbers. Where does the 2.23 come from? 1+0.75(√3)/2=2.3.
But you've basically cracked it, so I'll show you an easier way. You do not need to calculate the angles in degrees.
In 2D, the components of a unit force in the direction (x,y) are x/h, y/h where h=√(x2+y2), right?
In 3D, that extends simply to x/h, y/h, z/h where h=√(x2+y2+z2).
 

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