# "Statics" Question - Not for School
**URL:** https://forums.speedlife.net/t/statics-question-not-for-school/44817
**Category:** NYSpeed Automotive
**Created:** [March 11, 2008, 11:26am UTC](https://forums.speedlife.net/t/statics-question-not-for-school/44817 "2008-03-11T11:26:56Z")
**Posts on this page:** 17
**Page:** 1
### Author:  [@LUKE\_L](https://forums.speedlife.net/u/LUKE_L)
#### Post date: [March 11, 2008, 11:26am UTC](https://forums.speedlife.net/t/statics-question-not-for-school/44817/1 "2008-03-11T11:26:56Z")
I’m too rusty to remember how to do simple force equations at the moment and don’t feel like diggin out my Statics book right now.
Anyone want to give this a go for me?
AB = 6 inches
BC = 19 inches
AD = 40 inches
Weight of AD = 40 lbs.
I need to know the Force of Vector BC to support AD.
EDIT: Angle A is 50 degrees…Sorry
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### Author:  [@sureshot007](https://forums.speedlife.net/u/sureshot007)
#### Post date: [March 11, 2008, 11:27am UTC](https://forums.speedlife.net/t/statics-question-not-for-school/44817/2 "2008-03-11T11:27:53Z")
no.
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### Author:  [@Bacon](https://forums.speedlife.net/u/Bacon)
#### Post date: [March 11, 2008, 11:32am UTC](https://forums.speedlife.net/t/statics-question-not-for-school/44817/3 "2008-03-11T11:32:38Z")
42
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### Author:  [@Fry](https://forums.speedlife.net/u/Fry)
#### Post date: [March 11, 2008, 11:35am UTC](https://forums.speedlife.net/t/statics-question-not-for-school/44817/4 "2008-03-11T11:35:54Z")
What’s the length of AC?
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### Author:  [@Fry](https://forums.speedlife.net/u/Fry)
#### Post date: [March 11, 2008, 11:36am UTC](https://forums.speedlife.net/t/statics-question-not-for-school/44817/5 "2008-03-11T11:36:53Z")
Or any of the angles?
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### Author:  [@Fry](https://forums.speedlife.net/u/Fry)
#### Post date: [March 11, 2008, 11:40am UTC](https://forums.speedlife.net/t/statics-question-not-for-school/44817/6 "2008-03-11T11:40:20Z")
Angles muthafucka! Angles!
Can’t solve. Need more info.
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### Author:  [@jrod0187](https://forums.speedlife.net/u/jrod0187)
#### Post date: [March 11, 2008, 11:50am UTC](https://forums.speedlife.net/t/statics-question-not-for-school/44817/7 "2008-03-11T11:50:30Z")
What does this have to do with general automotive? Fuzzy is correct the answer is 42
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### Author:  [@Fry](https://forums.speedlife.net/u/Fry)
#### Post date: [March 11, 2008, 11:53am UTC](https://forums.speedlife.net/t/statics-question-not-for-school/44817/8 "2008-03-11T11:53:02Z")
> [@jrod0187](#):
>
> What does this have to do with general automotive? Fuzzy is correct the answer is 42
> [@Fry](#):
>
> Can’t solve. Need more info.
As for the Gen Auto thing, maybe he’s designing something custom for the Grand Am? We’ll give him time to explain…
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### Author:  [@LAFENGAS](https://forums.speedlife.net/u/LAFENGAS)
#### Post date: [March 11, 2008, 11:55am UTC](https://forums.speedlife.net/t/statics-question-not-for-school/44817/9 "2008-03-11T11:55:08Z")
not enought info… unless you want the answer as a variable
angles? fixed points? pivot points? what are the lenghts of each/every segment? The only force applied is uniformly along the length of AD?
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### Author:  [@LUKE\_L](https://forums.speedlife.net/u/LUKE_L)
#### Post date: [March 11, 2008, 12:36pm UTC](https://forums.speedlife.net/t/statics-question-not-for-school/44817/10 "2008-03-11T12:36:47Z")
EDIT: Angle A is 50 degrees. Yeah that would have helped. Sorry everyone. I feel like a db.
Length BC is Adjustable. Pivot points at A B and C. Length AC is unknown really but if you made it variable to give options that would be fine.
Fry…you are correct. I am looking to install some struts under my hood (ie. Adios Senior Prop Rod). I could just buy a “AutoZone” equivalent that would work but where is the fun in that?
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### Author:  [@Fry](https://forums.speedlife.net/u/Fry)
#### Post date: [March 11, 2008, 12:55pm UTC](https://forums.speedlife.net/t/statics-question-not-for-school/44817/11 "2008-03-11T12:55:03Z")
To make the math easier we’ll make ABC a 6-8-12 right triangle.
The verticle Force at the centroid of the beam is 40 pounds.
Fsin(50)=40
F=25.7=perpendicular Force at the centroid.
So that makes torque at A = 25.7 \* 20 = 514 in-lbs
Perpendicular force at C = 514/12 = 43 pounds
43/cos(50) = 62.5 pounds
Maybe. I haven’t done statics in probably 5 years. Might have made a mistake. A vertical load on BC of 62.5 pounds seems reasonable though. :shrug:
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### Author:  [@Nikuk](https://forums.speedlife.net/u/Nikuk)
#### Post date: [March 11, 2008, 1:03pm UTC](https://forums.speedlife.net/t/statics-question-not-for-school/44817/12 "2008-03-11T13:03:05Z")
I just want to know if Fuzzy was right.
Because if he wasn’t then why did jrod jump in to verify it?
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### Author:  [@Fry](https://forums.speedlife.net/u/Fry)
#### Post date: [March 11, 2008, 1:15pm UTC](https://forums.speedlife.net/t/statics-question-not-for-school/44817/13 "2008-03-11T13:15:53Z")
> [@Nikuk](#):
>
> I just want to know if Fuzzy was right.
>
> Because if he wasn’t then why did jrod jump in to verify it?
I think jrod forgot a question mark.
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### Author:  [@ILCisDEAD](https://forums.speedlife.net/u/ILCisDEAD)
#### Post date: [March 11, 2008, 1:20pm UTC](https://forums.speedlife.net/t/statics-question-not-for-school/44817/14 "2008-03-11T13:20:55Z")
Fuzzy is correct with 42
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### Author:  [@fairgentleman\_Z](https://forums.speedlife.net/u/fairgentleman_Z)
#### Post date: [March 11, 2008, 1:32pm UTC](https://forums.speedlife.net/t/statics-question-not-for-school/44817/15 "2008-03-11T13:32:05Z")
[hi-jack]This thread reminded me of a few…
> **[Dark Roasted Blend: Balancing the Forces](http://www.darkroastedblend.com/2007/03/balancing-forces.html)**
>
> Dark Roasted Blend
> **[Dark Roasted Blend: Balancing the Forces, Part 2](http://www.darkroastedblend.com/2007/08/balancing-forces-part-2.html)**
>
> Dark Roasted Blend
[\hi-jack]
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### Author:  [@LUKE\_L](https://forums.speedlife.net/u/LUKE_L)
#### Post date: [March 12, 2008, 3:36am UTC](https://forums.speedlife.net/t/statics-question-not-for-school/44817/16 "2008-03-12T03:36:47Z")
> [@Fry](#):
>
> To make the math easier we’ll make ABC a 6-8-12 right triangle.
>
> The verticle Force at the centroid of the beam is 40 pounds.
> Fsin(50)=40
> F=25.7=perpendicular Force at the centroid.
>
> So that makes torque at A = 25.7 \* 20 = 514 in-lbs
> Perpendicular force at C = 514/12 = 43 pounds
>
> 43/cos(50) = 62.5 pounds
>
> Maybe. I haven’t done statics in probably 5 years. Might have made a mistake. A vertical load on BC of 62.5 pounds seems reasonable though. :shrug:
I really can’t disagree because I too haven’t done this in like 4 years.
Thanks for working this out for me.
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### Author:  [@Bacon](https://forums.speedlife.net/u/Bacon)
#### Post date: [March 12, 2008, 6:06am UTC](https://forums.speedlife.net/t/statics-question-not-for-school/44817/17 "2008-03-12T06:06:36Z")
> [@Nikuk](#):
>
> I just want to know if Fuzzy was right.
>
> Because if he wasn’t then why did jrod jump in to verify it?
It is always the answer, it’s the answer to everything you know and see.