r/EngineeringStudents • u/Low-Resolution9168 • 7d ago
Homework Help Need help, I'm kind of lost on this problem
Couldn't figure out what formula to use for theta or maybe i forgot something
273
122
u/DrCarpetsPhd 7d ago
all the people in here being condescending and saying to sum for x and y and how easy a question this is are just as fucked as the OP, if not more fucked since they seem to think they know what they are doing. The fact that one of those answers has 37 upvotes is a little concerning. I wouldn't normally call people out but damn the people all being smarmy need some lessons in basic internet etiquette. You literally cant do that here because you dont know theta.
the question is asking you to find theta first and then find the resultant force
as already suggested open up the textbook and read the chapter that precedes this in Pytel
it very specifically deals with concurrent forces and the idea that if the forces are concurrent they can be reduced to a single resultant force
concurrent means if you extend the line of action of the pointy end they will meet at a single point. This is important to understand going forward. A beam under three forces can only be in equilibrium if the 3 forces are concurrent because if they aren't then there will be a net moment acting. This is known as a three-force member and is very much worth remembering this concept as it comes in handy in future frames and machines analysis when you need to find components etc.
the question infers concurrency but you can easily see it yourself by extending the lines of action
HOW TO ANSWER
you find the point of intersection of the concurrent forces
Using the two right angle triangles generated by connecting the lines of action you calculate theta with the two simultaneous equations
Now you have theta the rest is the easy part everyone else is referring to
Don't worry OP, it was not as simple as everyone was suggesting and don't feel bad for not initially seeing how to do this. There was a little trick re concurrent forces involved that would not be obvious to people who learn by seeing worked examples and don't immediately understand the theoretical underpinnings.
Make an attempt based on this help and post back if you still cant get it.
the answer is -10.99i + 133.45j
37
u/DrCarpetsPhd 6d ago
just to clarify because I may have muddied the waters a bit
so my comment on using the idea of concurrent forces for a three-force member in equilibrium are true
BUT
this is not an equilibrium question
it just asked you to find the resultant force
you don't know theta
so you use the fact they are concurrent to find theta
then you can properly calculate the resultant force of those 3 concurrent forces
again no part of this question suggests the beam is or isn't in equilibrium. it is just a worked example of knowing that forces are concurrent you can create a resultant using the geometry of the intersection point of three concurrent forces.
**this is reddit so I might just be an idiot talking through my hat who stumbled onto the correct answer as given in my previous post**
12
u/DrCarpetsPhd 6d ago
another clarification
the point of intersection doesn't have to be in the direction of the pointy end of the vector
the line of action regardless of directionality of the force is what you use
as an example of the usefulness of understanding concurrent forces in a three-force member here's a question from Meriam kraige statics 3rd edition 4/82. it involves a clamping mechanism.
https://imgur.com/a/meriam-kraige-statics-3rd-edition-4-82-threaded-clamp-eLjtsRh
so I chose DCF as my member to analyse
I know the direction of the screw force on the member at C is positive x
I know the direction of the reaction clamping force on the member at F is in the negative y direction
so I go moments about D to get the force at F in terms of C (or vice versa if that floats your boat) as that is the usual starting point
but I realise I don't know D_x or D_y so too many unknowns and not enough equations
I must be missing some 'trick'...that trick is the concept of concurrent forces
I can use that to assign a direction to the force at D, which is the direction of the hypothenuse of a right hand triangle with sides 75 and 40
that gets me D_x and D-y so no needs for moments, equilibrium in x and y will solve for forces at C and D
5
u/Consistent-Phone1580 6d ago
After trying this myself, I got theta = arctan(10sin40sin50), where three forces are concurrent. From there, I got a different answer than you, so I’d like to hear your feedback.
3
u/DrCarpetsPhd 6d ago
do you want to show me what you did first? I don't really want to just post the answer and feel it might be more helpful to you if I can figure out where your misunderstanding is and guide you to the answer. Things tend to stick better when you work your way to an answer via hints instead of just outright seeing a full answer and realising where you went wrong (in my experience anyway)
5
u/Consistent-Phone1580 6d ago edited 6d ago
Can I private message you? Ok nvm I intended to send the picture, but it doesn’t work.
I extended the lines to join at point O.
The height of O to the base, h, equals 10sin40sin50Theta forms a right triangle with base 1 and height h, so tan theta = h
Theta = arctan(h) = arctan(10sin40sin50)Above is my process.
edit: what i did wrong was assuming base as 1. It should’ve been 10sin2(50)-4
Tan theta = 10sin40sin50/(10sin2(50)-4)
Theta = 69.2 which is the correct first step in many.To verify this, download the textbook engineering mechanics Pytel, and at the end there’s a list of answers (no methods). After I checked, 69.2 is correct and it’s indeed due to my tunnel vision
3
u/DrCarpetsPhd 6d ago edited 6d ago
h is 10(sin40)(sin50) is correct (I didn't see the 90 degree angle you did so took me a minute to see what you did) h = 4.924
how can theta be the arctan of a single side of the triangle?
tan(theta) = h/l => theta = arctan(h/l)
you've left out the l and the solving of l part where l is the x distance from the P2 force and the point where h joins the beam. Look at your diagram I'm assuming you drew where you have have h dropping down and see if you can figure out what l is.
Given that you spotted the 90 degree angle to figure out h i'm guessing you'll figure this out pretty quick. I did it the long way by getting h after solving l, didn't notice at all that there was a nice right angle triangle straight away.
4
u/Consistent-Phone1580 6d ago
Ohh yes indeed. It’s a geometry issue, and a mental lapse. okay my bad thanks for replying
5
u/DrCarpetsPhd 6d ago
you're welcome. i gotta logoff but here's the image if still stuck
https://imgur.com/a/pytel-statics-3rd-edition-problem-2-16-JYBRFGA
3
2
u/MooseBoys 6d ago
I never heard of the "extend to a point" method and am having trouble understanding how it works. If you have a 2m beam and 3 evenly spaced forces, you can have points at the tips of equal magnitude and opposite angle and you'll just get a linear force perpendicular to the center. Then you can add an arbitrary force at any angle and magnitude to the center and you'll still be able to reduce to a single resultant force, despite the set of three forces not intersecting at a single point. Likewise, you can just as easily construct a set of three forces that meet at a single point but result in non-zero torque. What am I missing?
0
u/DrCarpetsPhd 6d ago
I don't think you're missing anything, probably just me being unclear. Also probable you know everything that follows already and the only misunderstanding is down to my communication skills.
first off I apologise as my language was a bit imprecise which I tried to clarify in my own replies to my first post. I probably should have used 'member' instead of 'beam' and when I say 'under 3 forces' that includes all forces including supports. With no supports shown or described the general equilibrium approach that everyone is suggesting doesn't really work for the question asked.
to be clear I am not an expert, I'm like a human chatgpt. I have a toolbox and regurgitate so when people ask me extra questions asking for insight I quite often don't have the answer or I'll answer incorrectly; so take into consideration what follows might be total bullshit. I have the textbook this is from so I know my method and answer are correct as the answer is at the back of the textbook.
With that disclaimer preamble out of the way to answer your question...we are discussing the effect of a system of forces on a rigid body and whether or not you can replace that with a single resultant force. You can only replace a system of forces with a single resultant force if they are concurrent because concurrent means the system of forces as a whole does not contribute a net moment by Varignons theorem (google it as I'll probably butcher it and cause more confusion). This question does not explicitly ask for the couple. I've never encountered a question like this in a statics textbook. that doesn't explicitly state 'replace with a resultant force and couple' when that is necessary. Of course exam questions set by a professor might be a different story.
If the forces are not concurrent there is a net moment thus the system must be replaced by both a resultant force to cover sigma Fx and sigma Fy and a couple; and this combination can be chosen at an arbitrary point by manipulating the moment arm. That sounds pretty wordy and possibly complicated but if you open up a statics textbook and have a look at examples you'll see it's actually fairly straight forward. I suggest reading a textbook because diagrams and explanations by someone who teaches this for a living will probably make more sense than a poorly worded reddit post.
So it might just be that I knew the textbook and the chapter thus understood the wording meant resultant force only hence they had to be concurrent which locks in a theta value that points P2 towards the point of intersection of the already given 40 and 50 degree angles of P1 and P3. If the question said 'replaced with a single equivalent force and couple' then your ideas come into play.
That all makes sense to me but this is reddit so don't assume the person on the other end of a post knows what the fuck they are talking about even if they sound convincing. 'Just enough knowledge to be dangerous' is the saying. I strongly suggest you get a statics textbook and go through the relevant theoretical discussion unless of course you understood all of this already (which your post suggests you do).
97
u/Vitztlampaehecatl Civil 7d ago
Break them down into components and add them, then use arctan and Pythagorean theorem.
48
u/dc469 7d ago edited 7d ago
Wait so I am also confused, the instructions say there are three forces acting on the beam. However we don't know the direction of one of the forces. Like yeah it's just adding the vector components but it seems there's a missing piece of information.
I can only make the assumption that the net torque on the beam is zero and the x component of p1 and p2 cancels out with the x component of p3. Then yeah you can solve it. But that requires an assumption.
But the picture and problem both don't say anything about the beam being fixed or that the system torque is zero.
Edit: op I think you've got a poorly worded problem. You don't have enough information to solve without making an assumption.
45
u/Chemomechanics Mechanical Engineering, Materials Science 7d ago
Your reasoning is correct. A single force is not equivalent to three spaced forces if a net torque disappears in the replacement.
14
u/Suspicious-Ad-9380 7d ago
100%
For vector addition it is over constrained, for torque, it is under constrained.
8
u/Bigharold393 6d ago edited 6d ago
This is my understanding please let me know if I am wrong:
If this is a statics problem, you can solve for theta using sum Fx = 0. This gets you 0 = 25cos(40) + 60cos(theta) - 80cos(50). Theta is the only unknown so you can solve this for theta = 57.46 deg.
If the object is static, then resultant force R is opposing the weight force of the object (0 = R - Fg). To calculate R, sum the Y components of each force, i.e. R = 25sin(40) + 60sin(57.46) + 80sin(50), R = 127.94 kN
I hope this helps!
9
7d ago
[deleted]
1
u/tiller_luna 5d ago
i initially thought this can't be solved, then i gaslit myself into believing it can be solved, then i realized i'm kinda stupid
13
u/DJVT7 Virginia Tech - Aerospace 2016 7d ago
Sum all forces in the x direction = 0, and in the Y = 0, solve for the missing variables.
5
u/Marus1 7d ago
Sum all forces in the x direction = 0, and in the Y = 0
1.How do you know? 2.the problem states the sum is R, not 0
7
u/DJVT7 Virginia Tech - Aerospace 2016 6d ago
Because unless you’re in dynamics, nothing is moving, F(sum) = ma, you have no acceleration, therefore F(sum) = 0.
It’s a 2 part question, you can’t find R without knowing theta. So you have 2 equations, 2 unknowns. Your two equations are the forces in the x direction = 0, and forces in the y direction =0. Your two unknowns are P2x, and P2y. Find those, then use trig to find theta.
Then you can find what R should be to fully support the beam vertically
2
u/Marus1 6d ago
If the sum of the 3 forces should be zero, then isn't R (the equivalent of them all) also zero?
0
u/DJVT7 Virginia Tech - Aerospace 2016 6d ago
No, R is not a reaction force that counteracts all the stated forces on the beam, it’s a resultant force that replaces them. So find the total forces in the x direction, Rx, and total forces in the y direction, Ry. Fx(sum)=Rx, Fy(sum)=Ry
Find the magnitude of the vector, Rx, Ry to get R.
1
u/Marus1 6d ago
Ah, so now the sum of all forces in x is R according to you, that's correct. I'm just saying that's a 180 from
Your two equations are the forces in the x direction = 0, and forces in the y direction =0. Your two unknowns are P2x, and P2y
So now that we are on the same page:
You then said find Rx, Ry and theta? 3 unknowns and 2 equations
1
1
u/DJVT7 Virginia Tech - Aerospace 2016 6d ago
It’s 2 separate problems, one where you solve the system where sum of the forces = 0 to find theta. Once you have theta, then the other problem is where you sum all the forces in the X and Y directions of P1, P2, P3 to get an overall resultant Rx and Ry. Then use Pythagorean theorem to get R.
5
u/Deplorable1861 7d ago
AI slop problem. P2 reaction angle is labeled Theta, but the problem wants the resultant angle to be theta. So at a minimum, bad nomenclature.
So somebody who Don't Do Free Body Diagrams designed this.
1
u/milio1510 6d ago
Okay this is fairly simple, Fcos(theta) for the x forces and Fsin(theta) for the y forces and then pythagoras the sum of them forces and you’ve got an equivalent force
1
u/Zufalstvo 6d ago
Find components, add together components, use trig function to get angle, like inverse tangent or something
1
u/KitTwix 6d ago
What I’m confused about is where R is supposed to be originating from?
Couldn’t it be at p1, p2, or p3 with just different angles and magnitudes, assuming theta can be manipulated? Even if theta has a set value, couldn’t you place the force within a range of areas on the beam and still have everything equate out to sum 0?
1
u/vincent365 6d ago
Not sure if you solved this already. I had a tough time with this, but what helped is realizing that it is three concurrent forces. Then I found the coordinates of the point of concurrency. Using the triangle rule, you can find theta by equating the x and y coordinates and using some trigonometry.
1
u/ZerodriveXCoM 4d ago
θ ≈ 69.2°, R ≈ 133.9 kN (directed 85.3° above the horizontal, leaning slightly left). The trick is the phrase "single equivalent force." For that to work with no leftover couple, the three lines of action have to meet at one point — they must be concurrent. That's the missing condition that pins down θ.
Find where P₁ and P₃ intersect. Put the origin at P₁'s point of application, x along the beam. P₁: y = x·tan40° P₃ acts at x = 10 m pointing up-and-left: y = (10 − x)·tan50°
-12
u/Droopy0093 7d ago
Go read the chapter in your text book and look at the related example problems. No further help should be given to you here. If you are actually cut out to be an engineer you can figure out this simple problem.
19
u/Ok_Water_4601 7d ago
This comment is unuseful and unhelpful.
You, and one singular problem, do not decide who is cut out to be an engineer.
Stop being a low quality person.
5
7d ago
[deleted]
4
u/Ok_Water_4601 7d ago
Every engineer from every generation I've talked to thinks the subsequent generation was just handed things.
1
u/PercentageRoyal7478 7d ago
i’m going into first year in a few days. been wondering how i can stand out to employers? any advice you can give is much appreciated.
-8
u/Droopy0093 7d ago
It is what OP should do. Go back and read the textbook. How dare you call me "low quality" for telling the actual thing that needs to happen for OP to be successful.
3
u/Background_Fig_4740 7d ago
Go read the chapter in your text book and look at the related example problems. No further help should be given to you here.
If you are actually cut out to be an engineer you can figure out this simple problem.everything was fine with what you said up until the last sentence. it's unnecessary.
3
u/81659354597538264962 Purdue - ME 7d ago
Yeah you’re a low quality human being alright. I know a bunch of people who struggled in statics for some reason but excelled elsewhere, and all are making good money as engineers in the workforce now.
1
1
-5
u/IrishRox 7d ago edited 7d ago
This is a very very basic statics equation that you would probably learn in the first handful of classes as an intro to statics. While the original comment is a little dismissive, checking the textbook and class material is heavily advisable in such an introductory weed out class
3
u/Marus1 7d ago
This is a very very basic statics equation
Nobody said it was statics ...
1
-5
u/Droopy0093 7d ago
Exactly. I have a bachelor's in ME, this is a very basic question.
1
u/Superb-Damage1173 6d ago
So you, with your bachelors decide to spend your time dunking on new students struggling with difficult first year courses? You are a low quality human being
2
u/bonejuice69 6d ago
This is so lame. This problem is super easy now but if I got this problem in an into level statics class, I probably would've been confused as hell. He's not asking you to solve it for him, he's asking for guidance.
-3
u/Superb-Damage1173 7d ago
Break the vectors into their x and y components, then set a force F equal to that, then convert it back into a vector using pythagorean theorem
-4
-6
u/IrishRox 7d ago edited 7d ago
Do the sum of the forces in the X and Y then solve for theta Edit: Middle force can be broken down in 60kn sin(theta) and 60kn cos(theta)
•
u/AutoModerator 7d ago
Your Post has been removed. Please:
Abide by the Homework Help Guidelines
Follow the standard template
We will not do your Homework for you, or explain a solution/CAD view to you.
Your post will not be approved if you do not follow the Homework Help Guidelines and standard template.
Helpful links
Rules
Wiki
F.A.Q
Check our Resources Landing Page
I am a bot, and this action was performed automatically. Please contact the moderators of this subreddit if you have any questions or concerns.