# Homework Help: Determine the magnitudes and nature of forces in space truss

1. Apr 8, 2017

### DevonZA

1. The problem statement, all variables and given/known data

2. Relevant equations

equivalent tension coefficient tDE= applied force/distance between points D and E
= 17/SQRT(1[^2+2^2+4^2) = 3.7097

The component of the 17kN in the x-direction = Δx X tDE = -1 X 3.7097
= -3.7097
The component of the 17kN in the y-direction = Δy X tDE = -2 X 3.7097
= -7.4194
The component of the 17kN in the z-direction = Δz X tDE = 4 X 3.7097
= 14.8388

There are three unknown forces, therefore we need three equations.

3. The attempt at a solution

I then need to multiply or divide one of these equations to get the unknown values the same so that I can add or divide to get rid of unknown term(s) to solve.
I've tried various options and cannot seem to get to the correct answers.
Working backwards from the given answers I get:

FBD/tBD = SQRT[2^2+3^2=0^2]
tBD = 2.3796

FCD/tCD = SQRT[3^2=2^2=1^2]
tCD = 1.5528

FBD= 8.58kN (S)
FCD= 5.81kN (S)

2. Apr 8, 2017

### haruspex

You seem to be treating the applied force as being along the line DE, but the diagram makes it look as though it is vertical and the text does not mention E. Is there a later part to the question which mentions E?

3. Apr 8, 2017

### haruspex

Would that be up or down?

4. Apr 8, 2017

### haruspex

Check thecoefficients on tAD and tCD, and why no tBD?

5. Apr 9, 2017

### DevonZA

Hi Haruspex

Firstly there is no mention made of E.
How will I get equivalent tension coefficient without using point E?

Component of 17kN would be down? Therefore -14.8388.
No tBD because the Z co ordinate at B is 0.

6. Apr 9, 2017

### haruspex

Then none of your calculations related to E have any value. Why is it in the picture? Just to confuse, maybe?
If it is straight down, as pictured, then the z component is ...?
Your equation c) is inconsistent with the other two. E.g. in the x direction for tAD you had a coefficient -3. I can see that this comes from subtracting the x coordinate of point D from that of point A. All the coefficients in a) and b) follow that pattern, but not those in c).
The z coordinate of D is ... and the z coordinate of A is .... so the coefficient of tAD in the z equation should be ...?

7. Apr 9, 2017

### DevonZA

I believe E is in the picture to confuse. Unfortunately there are no similar examples in my textbook which makes this question rather difficult to attempt.

If straight down then the z component is negative?

Z coordinate of D is 4.
Z coordinate of A is 1.
Z coordinate of B is 0.
Z coordinate of C is 1.

Does this look correct?

8. Apr 9, 2017

### haruspex

How are you getting 14.8388? Seems to me you are still using E to get that.
The coefficients look like the right magnitudes now, but not sure about the signs. It depends how you are defining tAD etc.
What equations do you have now for the x and y directions?

Regarding point E, it looks to me that in the examples on which you based your equations E would have been a point along the line of action of the applied force. I.e. the force was applied along the line DE, not vertically. If you wish to follow those examples then you need to change E to be a point in the line of action of the force in this case. The origin would be an easy choice.

9. Apr 9, 2017

### DevonZA

The same question appears in the textbook but only the solutions are given. E is not shown here so I am going to ignore it especially as they do not refer to it anywhere in the question.

I need to find the equivalent tension coefficient:
teq = the applied force / distance between points D and ?

To find the x,y and z component of the 17kN force: teq is then multiplied by Δ x, y and z (again I assume I need to calculate the difference in coordinates between point D and ?)

10. Apr 9, 2017

### haruspex

not sure I can do much more than reiterate my last post. Please try to understand it.
Judging from your attempt to use E, I assume that such a point existed in some earlier example, and that E was in the line of the applied force. That is the only way your method would make sense. In the present case, the shown point E is not in the line of the applied force, so if you want to use exactly the same method then you must replace it with a point E which is in that line. The origin would look like an easy choice. So just repeat what you did before but using (0, 0, 0) for the coordinates of E.

11. Apr 9, 2017

### DevonZA

I will be back with an updated attempt :)

12. Apr 9, 2017