# Truss mechanics question

1. Dec 3, 2016

### garr6120

Given the diagram in the file. I am trying to find the force for the member AB and I do not understand why they use the ratio 40/41 and why they are using 6.3 as the y distance because we have 9ft given and we use the ratio 4/5 instead of 40/41.

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2. Dec 3, 2016

### Mapes

You haven't shown any of the calculations, but it's immediately clear that the length of AD must by 41 ft because of the Pythagorean Theorem. Therefore, the tangent of the angle is 40/41. Presumably this is relevant.

3. Dec 3, 2016

### garr6120

Sorry I should of shown them my bad. I calculated: the sum of moments around G. which is equal to -1.8kips(14ft)-0.9kips(28ft+(9/15(Fab))(9ft)
solving for the Fab I get a force of 9.33kN however the answer is 8.20kN and they use a ratio of 40/41 times 6.3 ft. I was wondering how to approach this problem and what am I doing wrong. Isn't the moment of Fab 9ft above G?

4. Dec 3, 2016

### Mapes

Ah, got it. Good identification of the correct free-body diagram to use! But the moment arm of AB is the perpendicular distance to G. The way I'd calculate it is to say that the cosine of the far right angle is d / 28 ft (where d is the moment arm) and that the cosine is also 9 ft / 41 ft from the larger triangle. (Note that the first hypotenuse is on the bottom; the second is on the top.) This gives the same answer. There are probably a few ways to work it out; it would be good practice to try to find another way.

EDIT: Sorry, earlier I should have written that the sine, not the tangent, is 40/41.

5. Dec 3, 2016

### garr6120

I have a question though I understand that AB is the perpendicular distance above G therefore you cannot find a moment using the length of AB in the x direction. However, I am still confused about the length in x being 40 from G shouldn't the length be 12ft. Am I thinking about Moments wrong?

6. Dec 4, 2016

### Mapes

I'm not sure what you mean when you say "the length in x being 40 from G". You may be referring to a calculation that I don't see. The moment arm of AB around point G is shown below. There are multiple ways to find this distance; the approach I took was to first find angle CDH, whose tangent is 9 ft / 40 ft. Then I used the fact that the sine of the angle is d / 28 ft, where d is the perpendicular distance from AB to G.