Equilibruim Problem

1. Sep 6, 2009

talaroue

1. The problem statement, all variables and given/known data
A frame ABC is supported in part by cable DBE. Knowing that the tension in the cable is 385 N, determine the components of the force exerted by the cable support at D.

2. Relevant equations
Fx=Fhcos(theata)
Fy=Fcos(theata)

3. The attempt at a solution

I can't seem to get the right angles for Theatay, Theatax, thetaz.

I tried solving for Fh but I still need theatay.

2. Sep 7, 2009

any ideas?

3. Sep 7, 2009

kuruman

All these problems are solved the same way. Draw a free body diagram of the frame, put in all the forces and write three equations (1) The sum of all the x-components of the forces is zero, (2) the sum of all the y-components of the forcces is zero and (3) the sum of all the torques is zero.

The definition of the angles you are looking for is unclear because the picture is very blurry. However, to find angles write the force vectors in unit vector notation and take the appropriate dot products.

Last edited: Sep 7, 2009
4. Sep 7, 2009

talaroue

Let me work on it and see if I can get back to you with the 3 equations

5. Sep 7, 2009

talaroue

the 2 bars going across don't matter right it is just the ropes that matter right? or do the bars exert a force?

6. Sep 7, 2009

kuruman

I cannot say. Can you post a clearer picture? Are the bars segments AB and BC of the frame? Do they touch two perpendicular walls? If they are, then they most assuredly exert forces.

7. Sep 7, 2009

8. Sep 7, 2009

talaroue

They all touch on the axis.....think of them as braces on a wall. A and D are touching the Z-Y plane, while E and C are touching the Y-X plane.

9. Sep 7, 2009

talaroue

I can't come up with the right angles to use for the forces

10. Sep 7, 2009

talaroue

The FBD i came up with has the two tensions going towards point B, while the bars are going away from point B.

11. Sep 7, 2009

kuruman

You have the coordinates of each of the end points of the strings. From these you can write a vector with its tail on the frame and its tip where the string is anchored to the wall. One such vector is
$$\vec{V}=(x_{2}-x_{1})\hat{i}+(y_{2}-y_{1})\hat{i}+(z_{2}-z_{1})\hat{k}$$

The tension is in the same direction. Can you find a unit vector in the direction of vector V? Once you have that, you can find the cosine of any angle by taking dot products with other unit vectors.

12. Sep 7, 2009

talaroue

I need 4 equations like that are needed, 2 for the different tensions and then 2 for the bars across the bottom?

13. Sep 7, 2009

kuruman

Don't you know that one of the tensions is 385 N from the statement of the problem?

14. Sep 7, 2009

talaroue

and how can I tell which angle is the angle for X,Y,Z

15. Sep 7, 2009

talaroue

Oh right, but it says that the tension in the cable is 385 N so is that for the entire cable D+E=385 N?

16. Sep 7, 2009

kuruman

If I understand the problem correctly, both sections of the cable are under the same tension.

17. Sep 7, 2009

talaroue

So if all I need is D which is asked can't I use the equations

Fx=Fcos(theata x)=Fhcos(theata x) I understand that the theatas are different

I thought i could find the angles by just using sin, cos, tan since I know a lot of the dimensions but I visually can't see which angles to use, you know what i mean?

18. Sep 7, 2009

talaroue

Wait i think I got it here is my logic about getting the anwser would this work for all of the problems let me know......(next post)

19. Sep 7, 2009

talaroue

Since it tells me that BC=600 and Dto the y axis is 280 600-280=320 that gives me the distance along the z axis for A to D on the z axis (makes since).

Then using a^2+b^2=c^2 I came up with the hypo. for the right triangle at BA and D on the z axis is 577.

So now for the theata y i can use tan(570/510)......then use that angle in Fcos(theata y)?

I came up with the right anwser but I dont know if my logic is right, and how often that will work.

20. Sep 7, 2009

kuruman

I don't really understands what you have done, but yes you can get the angles from the distances and some trigonometry.

There is an additional equation I forgot to mention in my earlier posting, that the sum of all the z-components of the forces are zero. This brings the total number of equations to four, three from the force components and one from the torques. You have four contact force components to calculate, so you should be OK.