Need help with Statics

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Need help with Statics

2009-08-26 19:52:18

TheunforgivenII: Member; +12|6701

The gusset plate is subjected to the forces of four members. Determine the force in member B and its proper orientation θ for equilibrium. The forces are concurrent at point O. Take F = 12 kN.

https://img137.imageshack.us/img137/9205/problem36.th.jpg

F = 12 kN
F₁ = 8 kN
F₂ = 5 kN
θ₁ = 45°

So then I applied the equilibrium equations:
right positive ΣF₁ = 0; F₁-Tcos(θ)+F₂sin(θ₁) = 0
up positive ΣF₂ = 0; -Tsin(θ)-F₂cos(θ₁) + F = 0

Now I don't know what to do from here. I don't know what numbers to plug in and to solve for what unknown.

2009-08-26 19:54:19

Uzique: dasein.; +2,865|6890

I dunno but your first sentence sounds like something from the script of a niche-genre porn movie.

_{libertarian benefit collector - anti-academic super-intellectual}. _{http://mixlr.com/the-little-phrase/}

2009-08-26 19:55:54

ATG: Banned; +5,233|6948|Global Command

Do your own homework, mmkay?

2009-08-26 19:58:30

TheunforgivenII: Member; +12|6701

ATG wrote:
Do your own homework, mmkay?

I have been doing my own homework. I've been staring at this problem for about 4 hours and still can't figure it out

Last edited by TheunforgivenII (2009-08-26 19:58:52)

2009-08-26 20:00:30

Bevo: Nah; +718|6940|Austin, Texas

Wikipedia?

I have to take statics next semester I think, that or in two semesters..

2009-08-26 20:02:26

TheunforgivenII: Member; +12|6701

Bevo wrote:
Wikipedia?

I have to take statics next semester I think, that or in two semesters..

wikipedia is not really a trustworthy website to get information on.

2009-08-26 20:04:33

Stubbee: Religions Hate Facts, Questions and Doubts; +223|7163|Reality

2 equations and 2 unknowns seems pretty simple. solve for one variable in one equation and plug answer into second. oh and sin and cos of 45 deg are the same so it is even easier.

The US economy is a giant Ponzi scheme. And 'to big to fail' is code speak for 'niahnahniahniahnah 99 percenters'

2009-08-26 20:05:57

Bevo: Nah; +718|6940|Austin, Texas

TheunforgivenII wrote:
Bevo wrote:
Wikipedia?

I have to take statics next semester I think, that or in two semesters..
wikipedia is not really a trustworthy website to get information on.

Their math stuff looks fairly fine to me. I mean, I wouldn't go on there looking for an exact formula, but it may spark inspiration on how to solve it. Most of that core stuff is heavily regulated.

or if you want something more "reliable"

http://www.wolframalpha.com/

2009-08-26 20:09:04

Flaming_Maniac: prince of insufficient light; +2,490|7126|67.222.138.85

Okay I think I am repeating what you already wrote in those equations, but since how you wrote them is confusing a little bit I'm going to say it again for shits and giggles.

So what you need to do is write an equation for all of the vertical forces and set it equal to zero, and write an equation for all the horizontal forces and set it equal to zero. You should get each term by multiplying the force by the sin or cos of the angle to the horizontal for the vertical and horizontal forces respectively. You should have three finite terms and one term with a variable in each equation (one term will be zero in each it looks like).

Now you need to set up three systems of equations. Your variables are horizontal force, vertical force, and theta. Your equations are the two mentioned above which get you a Fsintheta term and a Fcostheta term, and you can write those into a third equation.

I think you need three equations because you don't know the horizontal vector or the vertical vector. You aren't just looking for the magnitude, which is usually how these problems are set up at a lower level.

I am not *completely* sure about this answer, but this is certainly how I would try to work it out with a system of three equations.

2009-08-26 20:31:49

TheunforgivenII: Member; +12|6701

Flaming_Maniac wrote:
Okay I think I am repeating what you already wrote in those equations, but since how you wrote them is confusing a little bit I'm going to say it again for shits and giggles.

So what you need to do is write an equation for all of the vertical forces and set it equal to zero, and write an equation for all the horizontal forces and set it equal to zero. You should get each term by multiplying the force by the sin or cos of the angle to the horizontal for the vertical and horizontal forces respectively. You should have three finite terms and one term with a variable in each equation (one term will be zero in each it looks like).

Now you need to set up three systems of equations. Your variables are horizontal force, vertical force, and theta. Your equations are the two mentioned above which get you a Fsintheta term and a Fcostheta term, and you can write those into a third equation.

I think you need three equations because you don't know the horizontal vector or the vertical vector. You aren't just looking for the magnitude, which is usually how these problems are set up at a lower level.

I am not *completely* sure about this answer, but this is certainly how I would try to work it out with a system of three equations.

what?

2009-08-26 20:43:20

Flaming_Maniac: prince of insufficient light; +2,490|7126|67.222.138.85

...

Three equations, three variables. Horizontal force, vertical force, theta. First two equations are horizontal and vertical force equilibrium, third is setting the discovered horizontal and vertical forces equal in terms of theta. Solve via substitution.

It is extremely annoying to do my best at helping, and then get no more a specific question about which part you didn't understand than "what?" to my entire response.

2009-08-26 20:56:31

mcminty: Moderating your content for the Australian Govt.; +879|7141|Sydney, Australia

Flaming_Maniac wrote:
...
It is extremely annoying to do my best at helping, and then get no more a specific question about which part you didn't understand than "what?" to my entire response.

This. Also.. can you give is a larger diagram? Hand draw it out, if you must..

edit.. eh. now i see.. i knew i disliked imageshack for a reason

It's been a while since I've done statics.. and now I'm on to the BComm stuff, I don't really remember most of it.. But..

TheunforgivenII wrote:
The gusset plate is subjected to the forces of four members. Determine the force in member B and its proper orientation θ for equilibrium. The forces are concurrent at point O. Take F = 12 kN.
http://img137.imageshack.us/img137/9205 … m36.th.jpg

F = 12 kN
F₁ = 8 kN
F₂ = 5 kN
θ₁ = 45°

So then I applied the equilibrium equations:
right positive ΣF₁ = 0; F₁-Tcos(θ)+F₂sin(θ₁) = 0
up positive ΣF₂ = 0; -Tsin(θ)-F₂cos(θ₁) + F = 0

Now I don't know what to do from here. I don't know what numbers to plug in and to solve for what unknown.

Why haven't you included F_B in your equilibrium equations?

edit 2: and now I can see the image, I see you have that force..

2009-08-26 20:57:27

TheunforgivenII: Member; +12|6701

Flaming_Maniac wrote:
...

Three equations, three variables. Horizontal force, vertical force, theta. First two equations are horizontal and vertical force equilibrium, third is setting the discovered horizontal and vertical forces equal in terms of theta. Solve via substitution.

It is extremely annoying to do my best at helping, and then get no more a specific question about which part you didn't understand than "what?" to my entire response.

Am I suppose to set the horizontal and vertical force equilibrium equation to solve for T and then plug T into which ever equilibrium equation to solve for θ?

2009-08-26 21:00:20

Stubbee: Religions Hate Facts, Questions and Doubts; +223|7163|Reality

TheunforgivenII wrote:
Flaming_Maniac wrote:
...

Three equations, three variables. Horizontal force, vertical force, theta. First two equations are horizontal and vertical force equilibrium, third is setting the discovered horizontal and vertical forces equal in terms of theta. Solve via substitution.

It is extremely annoying to do my best at helping, and then get no more a specific question about which part you didn't understand than "what?" to my entire response.
Am I suppose to set the horizontal and vertical force equilibrium equation to solve for T and then plug T into which ever equilibrium equation to solve for θ?

as I said above and the answer is yes

2 equations and 2 unknowns (T and θ)

The US economy is a giant Ponzi scheme. And 'to big to fail' is code speak for 'niahnahniahniahnah 99 percenters'

2009-08-26 21:05:50

TheunforgivenII: Member; +12|6701

Stubbee wrote:
TheunforgivenII wrote:
Flaming_Maniac wrote:
...

Three equations, three variables. Horizontal force, vertical force, theta. First two equations are horizontal and vertical force equilibrium, third is setting the discovered horizontal and vertical forces equal in terms of theta. Solve via substitution.

It is extremely annoying to do my best at helping, and then get no more a specific question about which part you didn't understand than "what?" to my entire response.
Am I suppose to set the horizontal and vertical force equilibrium equation to solve for T and then plug T into which ever equilibrium equation to solve for θ?
as I said above and the answer is yes

2 equations and 2 unknowns (T and θ)

when I set the two equations together I get:

Tcos(θ)+Tsin(θ) = F + F₂θ₁ - F₂cosθ₁ + F

Now how do I solve for T? Both T's are on the same side with sin and cos

Last edited by TheunforgivenII (2009-08-26 21:06:35)

2009-08-30 10:43:19

TheunforgivenII: Member; +12|6701

Did I do this problem right?

Determine the tension developed in each wire used to support the 50-kg chandelier.

https://img54.imageshack.us/img54/576/problem320.jpg

Link D
ΣFy = BDsin45+CDsin60=50 (1)
ΣFx = BDcos 45+CDcos60=0 (2)

Link B
ΣFx = ABcos 30=BDcos45+BC (3)
ΣFy = ABsin30= BDcos45 (4)

solving for BD and CD:
BDsin45 + CDsin60 = 50(9.8)
=> BDsin45 = CDsin60 - 490
=> BD = (CDsin60 - 490) / sin45

BDcos45 + CDcos60 = 0
=> CDcos60 = -BDcos45
=> CD = (-BDcos45) / cos60

solving for AB and BC:
ABsin30 = BDcos45
=> AB = (BDcos45) / sin30

ABcos30 = BDcos45 + BC
=> (BDcos45/sin30)(cos30) = BDcos45 + BC
=> ((BD(cos45)(cos30)) / sin30) = BDcos45 + BC
=> BDcos45 - ((BD(cos45)(cos30)) / sin30) = BC
=> BC = ((BD(cos45)(sin30)) / sin30) - ((BD(cos45)(cos30)) / sin30)

Is that what the problem is asking for are those equations I just solved (highlighted)?

2009-08-30 10:45:11

FatherTed: xD; +3,936|6919|so randum

force of four members hehehhehehe

Small hourglass island
Always raining and foggy
Use an umbrella

2009-08-30 10:45:57

Uzique: dasein.; +2,865|6890

Following on from my original sex-themed post (see above):

Don't you find it disturbing that your diagram basically represents an electrical/mechanical female uterus?

You geeky scientists... Freud would have a field-day.

_{libertarian benefit collector - anti-academic super-intellectual}. _{http://mixlr.com/the-little-phrase/}

2009-08-30 12:50:59

WldctARCHe: Member; +9|5777|Kansas

Are you using a statics book written by an R.C. Hibbeler?

2009-08-30 13:31:35

TheunforgivenII: Member; +12|6701

WldctARCHe wrote:
Are you using a statics book written by an R.C. Hibbeler?

Yupp the 12th edition

2009-08-30 13:59:49

WldctARCHe: Member; +9|5777|Kansas

Okay... I was using the 11th edition when I went through my statics, so the problems are almost identical

Link D
∑Fy = BDsin(45) + CDsin(30) = 50*(9.8) (#1#)
∑Fx = -BDcos(45) + CDcos(30) = 0 (#2#)

Link B
∑Fx = BDcos(45) + BC = ABcos(30) (#3#)
∑Fy = ABsin(30) = BDcos(45) (#4#)

Using these equations I believe everything is solvable to a actual value.
Solving (#2#) you get BDcos(45) = CDcos(30) => BD = CD*(cos(30)/cos(45)) (#5#)

Plugging (#5#) into (#1#) you solve for CD
CD*(cos(30)/cos(45))*(sin(45)) + CDsin(30) = 50*(9.8) => CD = 358.7N

From here just plug in the value of CD into (#1#) or (#2#)
BD = 439.3 N

BD into (#4#) you get
AB = 621.3 N

BD & AB into (#3#) you get
BC = 227.4 N

P.S. There should be answers in the back of the book to check these against.

2009-08-30 14:24:06

TheunforgivenII: Member; +12|6701

WldctARCHe wrote:
Okay... I was using the 11th edition when I went through my statics, so the problems are almost identical

Link D
∑Fy = BDsin(45) + CDsin(30) = 50*(9.8) (#1#)
∑Fx = -BDcos(45) + CDcos(30) = 0 (#2#)

Link B
∑Fx = BDcos(45) + BC = ABcos(30) (#3#)
∑Fy = ABsin(30) = BDcos(45) (#4#)

Using these equations I believe everything is solvable to a actual value.
Solving (#2#) you get BDcos(45) = CDcos(30) => BD = CD*(cos(30)/cos(45)) (#5#)

Plugging (#5#) into (#1#) you solve for CD
CD*(cos(30)/cos(45))*(sin(45)) + CDsin(30) = 50*(9.8) => CD = 358.7N

From here just plug in the value of CD into (#1#) or (#2#)
BD = 439.3 N

BD into (#4#) you get
AB = 621.3 N

BD & AB into (#3#) you get
BC = 227.4 N

P.S. There should be answers in the back of the book to check these against.

Thank you. I'll be looking over this problem

2009-08-30 15:12:43

TheunforgivenII: Member; +12|6701

WldctARCHe wrote:
Okay... I was using the 11th edition when I went through my statics, so the problems are almost identical

Plugging (#5#) into (#1#) you solve for CD
CD*(cos(30)/cos(45))*(sin(45)) + CDsin(30) = 50*(9.8) => CD = 358.7N

how did you solve for CD to equal 358.7? I've been trying but I keep getting 717.4 N

2009-08-30 15:30:03

WldctARCHe: Member; +9|5777|Kansas

CD*(cos(30)/cos(45))*(sin(45)) + CDsin(30) = 50*(9.8) ==> Factor CD out of Left hand side
CD * {[ {cos(30)/cos(45)} * {sin (45)} ] + sin(30)}= 50*9.8 ==> cos(45) and sin(45) are same thus [{cos(30) * (cos(45)} / sin(45)] = cos(30)
CD * [ {cos(30)} +{ sin(30)}] = 490 ==> Divide both sides by (cos(30) + sin(30))
CD = 490 / (cos(30) + sin(30)) ==> Solve
CD = 358.70

Are the answers to this problem in the back of the book?

2009-08-30 15:34:18

TheunforgivenII: Member; +12|6701

WldctARCHe wrote:
CD*(cos(30)/cos(45))*(sin(45)) + CDsin(30) = 50*(9.8) ==> Factor CD out of Left hand side
CD * {[ {cos(30)/cos(45)} * {sin (45)} ] + sin(30)}= 50*9.8 ==> cos(45) and sin(45) are same thus [{cos(30) * (cos(45)} / sin(45)] = cos(30)
CD * [ {cos(30)} +{ sin(30)}] = 490 ==> Divide both sides by (cos(30) + sin(30))
CD = 490 / (cos(30) + sin(30)) ==> Solve
CD = 358.70

Are the answers to this problem in the back of the book?

Thanks again and the answers aren't in the back, but I trust you

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ATG wrote:

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TheunforgivenII wrote:

Bevo wrote:

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TheunforgivenII wrote:

Flaming_Maniac wrote:

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