Statics is this problem easy or hard?

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The discussion revolves around a statics problem involving a uniform beam supported by a wall and a wire, where participants debate the difficulty of solving for equilibrium conditions. The main challenge identified is the lack of known angles in the diagram, which complicates the calculation of forces and moments. Participants suggest drawing force diagrams and using trigonometric relationships to find unknown lengths and angles. There is confusion about how to apply equilibrium equations effectively without complete information, leading to frustration among those attempting to solve the problem. Ultimately, the conversation highlights the importance of visualizing forces and applying systematic approaches to tackle statics problems.
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Statics...is this problem easy or hard?

Yes, I know, easy and hard are relative statements, but I mean easy or hard compared to the other stuff I posted here-- if anyone remembers.

Homework Statement

A uniform beam, AB, whose length is 2 meters and mass is 10 kg is supported in its tip, A, by a smooth vertical wall and at its other tip B - he's tied to the wall by a wire BC, as depicted in the drawing. At what distance, AC, you tie the distance to the wall so the beam is at equilibrium? As well, calculate the tension on the wire (T), and the reaction force (R) of the wall at point A[PLAIN]http://img341.imageshack.us/img341/2428/240np.jpg

The Attempt at a Solution



I just want to ask you guys for ideas. I can't find the angles of the damn thing. There's only one triangle in this diagram whose angles and lengths I can find-- that's AB as the hypotenuse of a right triangle whose other angles are 45 degrees. Without angles, what am I to do? I tried parallelograms... nothing. Moments won't get me anywhere as I don't have enough info with respect to the angles...you might say that without angles I got nothing... so no parallagram...no Pythagoras...no tricks? Help... a lot of folks in my class are struggling with that problem...
 
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Hi Dory! :smile:
Dory said:
I just want to ask you guys for ideas. I can't find the angles of the damn thing. There's only one triangle in this diagram whose angles and lengths I can find-- that's AB as the hypotenuse of a right triangle whose other angles are 45 degrees. Without angles, what am I to do? I tried parallelograms... nothing. Moments won't get me anywhere as I don't have enough info with respect to the angles...you might say that without angles I got nothing... so no parallagram...no Pythagoras...no tricks? Help... a lot of folks in my class are struggling with that problem...

So far as the geometry is concerned, I don't see what the difficulty is :confused: … you add AC to the bottom bit, and you have the side of another right-angled triangle.

Draw the forces on the diagram

(why haven't you done so already?) …

you know that the moments on the beam must add to zero (about any point), so look at the forces on the diagram, and decide which point do you think you should take moments about to make it easiest?
 


Holy s***! I didn't see that. I add AC to the bottom bit! That's friggin' brilliant :D That's how I discover AC right? I'll do at home... at work now...once I got the geometry I'll solve it :)
 


In order to add AC to the bottom bit, I'd have to assume that the horizontal imaginary line connecting B to the wall is equal to AC...how do I know that for sure?
 
Dory said:
In order to add AC to the bottom bit, I'd have to assume that the horizontal imaginary line connecting B to the wall is equal to AC...how do I know that for sure?

Sorry, Dory, I have no idea what you're talking about :confused: …

if you draw that horizontal line (BD say),

then you have two right-angled triangles, BCD and BAD.
 


Right, but what good are these triangles? I mean, I can find out everything about the smaller right triangle, but the bigger right triangle-- I only have its BD and no angle! So, 1 length and no angle...still stuck. Or am I missing something.
 
Dory, I get the impression that you don't like drawing forces on diagrams. :redface:

You need to get used to this!

Draw the forces on the beam (only) …

how many are there?

what do they look like?

do they cross? at how many points? and where? :wink:
 


There are 3! Na (wall's normal), Bc (rope pulling) and mass x gravity (split in middle)...there, tiny-tim..you know I'll always make diagrams for you if you only ask! :) [sorry, can't use the scanner this late so quality sucks]

http://img716.imageshack.us/img716/9664/there11.jpg

Uploaded with ImageShack.us

I just wasn't sure how it helps me see the picture...should I build a triangle with these forces you suggest and I have the angles?

Hmm, let me see... okay... I tried building all sorts of triangles... I know one angle is 45 degrees but it's not a right triangle...don't have the other two...
 
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Dory said:
There are 3! Na (wall's normal), Bc (rope pulling) and mass x gravity (split in middle)...there, tiny-tim..you know I'll always make diagrams for you if you only ask! :)

I know! :biggrin:

ok, now you have 3 lines of force …

they can meet in 3 points, or in one point …

which is it? :wink:
 
  • #10


Dory said:
[PLAIN]http://img716.imageshack.us/img716/9664/there11.jpg[/QUOTE]

Completely OT, but I really think you should post this picture here.
 
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  • #11


(Thanks Borek I did :) )

Tiny-tim, I've already built that triangle in the picture. Do see it? It's a right triangle with the hypotenuse being BC. Problem is, I only got the fact it's a right triangle. The rest of the angles aren't given to me. Pleassssssse can you throw me a bigger bone?
 
  • #12
Dory, why did you not answer my last post? :confused: …

ok, now you have 3 lines of force …

they can meet in 3 points, or in one point …

which is it? :wink:​
 
  • #13


That one point is between C and B... I THINK. Technically you can move vectors around so I'm not sure as to significance of where they meet. I do see the significance of builing a triangle with them. Those 3 points are C, B, and the straight line I've drawn in red that connects B to the wall.
 
  • #14
Dory said:
Technically you can move vectors around …

No!

You can move free vectors (like velocity) around (that's how we make vector triangles of velocities),

but you can't move bound vectors (like force) around …

a bound vector (I'm not sure that's the correct name, btw :redface:) has a line which is part of the vector …

a force applied to the edge of something does not have the same effect as a force applied to the centre.
… so I'm not sure as to significance of where they meet.

Suppose a body in equilibrium has exactly 3 forces acting on it, and suppose that the three lines of force meet in 3 different points …

choose one of the points, and take moments (torques) of forces about that point …

what do you find? :wink:​
 
  • #15


First off I used my better trigonometry to find AC = 1.41

Now I used the Law of Sines to find out that the angle between CB and AB is 24.11.

Now, being very excited and feeling close to the solution, I do this:

Sum of all moments on A = 0 ; Tsin(24.11) x 2 - mgcos(45) x 1 = 0

Then I see that the result of T is way off to what T should be ... then I get a myocardial infraction. :(

Is my equation correct?
 
  • #16
Dory said:
First off I used my better trigonometry to find AC = 1.41

Now I used the Law of Sines to find out that the angle between CB and AB is 24.11.

Now, being very excited and feeling close to the solution, I do this:

Sum of all moments on A = 0 ; Tsin(24.11) x 2 - mgcos(45) x 1 = 0

Then I see that the result of T is way off to what T should be ... then I get a myocardial infraction. :(

Is my equation correct?

hmm …

i] how did you find AC = 1.41 (= √2) ?

ii] how did you get 24.11° ?

iii] how about answering my last question?
 
  • #17
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  • #18
Now you've swapped over A and C !

Are you trying to confuse me? :biggrin:
Dory said:
I used parallelograms to find out AC = 1.41

But I don't understand how you knew it was a parallelogram. :confused:
To answer your last question I thought I did by doing sum of all moments on A! That's one of the points where the vectors meet. I don't mean to disapppoint you, I'm trying! See, I'm scanning and writing and ****...it's a tough problem!

But only one force (on the beam) goes through A (either A :rolleyes:) …

Try the point where the normal force and the weight of the beam meet. :smile:
 
  • #19


Oops, I swapped A and C by accident! Heh.

How did I know it was a parallelogram? I just drew parallel triangles :) No? You saying I someone cheated? Humphff! ;)

Normal force and weight of the beam meet at mg, thus:

Sum of all moments on mg = 0 ; -Na(cos45) x 1 + T(sin24.11) x 1 = 0
Sum of all forces of X = 0 ; Na - T(sin 45+24.11) = 0

Hmm...but according to this T = 0 ...
 
  • #20
Dory said:
Normal force and weight of the beam meet at mg, thus:

Sum of all moments on mg = 0 ; -Na(cos45) x 1 + T(sin24.11) x 1 = 0

where's mg ? :confused:

anyway, if the normal force goes through mg, how can its moment about that point be anything other than 0 ? :redface:
 
  • #21


Dory said:
How did I know it was a parallelogram? I just drew parallel triangles :) No?

Upper triangle hypotenuse is NOT parallel to the beam.
 
  • #22


tiny-tim said:
where's mg ? :confused:

anyway, if the normal force goes through mg, how can its moment about that point be anything other than 0 ? :redface:

Well then, think of it as sin(45) -- it's the same score anyway. It's the vector perpendicular to the beam so it has an arm of 1.

And mg is at the center of the beam...

Upper triangle hypotenuse is NOT parallel to the beam.

Yes...*embarrassed*...I see that now.

*long breath* This question is too much of a roughie. I keep going back to the drawing board and get nothing. I think I'll try to work with other problems with you guys when they'll come up...I'm really getting nowhere with this. You have no idea how excited I was when I felt I was about to solve it...but then...nothing!
 
  • #23
Dory said:
And mg is at the center of the beam...

Then the normal force does not go through mg.

The normal force and the weight meet at the point vertically above mg and level with C …

take moments (of all the forces on the beam) about that point.
 
  • #24


I see what you want me to do... I got the lengths by using the triangle at the bottom, but the problem is if I don't have the angle of T, I still can't solve it!

Please pretty please show me how you solve it? I promise I won't make it a habit it's just this horrific exercise...just this one time! (After looking at the solution I'll understand your logic)
 
  • #25
Dory said:
I see what you want me to do... I got the lengths by using the triangle at the bottom …

Sorry, I have no idea what you mean. :confused:

Please show us your diagram, including the lines of the normal force and the weight force, mark the points where they meet, and take moments about that point.
 
  • #26


Dory: I can be wrong, but it seems to me like you are trying to find geometry first, physics later. It won't work. You know system is at equilibrium, write equations describing the equilibrium using unknown yet angles/lengths, and solve for them.
 
  • #27


Fair enough, but this is all I can get going following you tiny-tim :P
 

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  • #28
Nice diagram! :smile:

ok, if ∑M = 0, then what is the contribution of T to that? :wink:
 
  • #29


Heh, thanks, thought you might enjoy it ;)

Well, to answer your question, there's a tiny distance I think between your imaginary point and T's moment-causing vector to this point, but I don't know the angle, and I don't know T! How is it solvable again? *scratches head*
 
  • #30
but you know that ∑M = 0 !

sooo … ? :smile:
 
  • #31


So...T equals... 0? *scratches head*...
 
  • #32
No, the moment of T about that point equals 0,

sooo … ? :smile:​
 
  • #33


Can't be that T=0! Answer says differently! Unless I mistranslate something in the question... are you saying that T=0?
 
  • #34
The moment of T is zero.
 
  • #35
Dory: Let the vertical distance between point A and C be an unknown. You have three unknowns, and you can write three equilibrium equations. The summation of moment equation can be about any point you prefer. After you create your three equilibrium equations, solve the three equations for the three unknowns.
 
  • #36


I can't get to the point of 3 equations and 3 unknowns because I don't have AC and I don't have the angle...here, look, this is where I keep getting stuck at:

Sum of all moments on A = 0 = T(sin alpha) x 1.41 - mg x 1.41/2 = 0
Sum of all moments on B = 0 = -NA x 1.41 + mg 1.41/2 = 0
Sum of all moments on mg = 0 ; -Na(cos45) x 1 + T(sin alpha) x 1 = 0

I don't know how to include AC is one of the unknowns...regardless, since the angle is also an unknown I don't see how can this be solved...
 
  • #37
Hi Dory! :smile:

(btw, it's moments about a point (or axis), not on a point :wink:)
Dory said:
I can't get to the point of 3 equations and 3 unknowns because I don't have AC and I don't have the angle...here, look, this is where I keep getting stuck at:

Sum of all moments on A = 0 = T(sin alpha) x 1.41 - mg x 1.41/2 = 0
Sum of all moments on B = 0 = -NA x 1.41 + mg 1.41/2 = 0
Sum of all moments on mg = 0 ; -Na(cos45) x 1 + T(sin alpha) x 1 = 0

I don't know how to include AC is one of the unknowns...regardless, since the angle is also an unknown I don't see how can this be solved...


There's usually no reason to take moments (of forces on the same body) about two different points (the second equation usually doesn't give any extra information that can't more easily be obtained from a linear equation) …

take moments about one point, and use linear components of force to get the other equations …

in this case, the moments about A (i don't think yours are correct, btw) together with the two equations for x and y components should do it :wink:

however, it would be simpler in this case to notice that all three forces must go through the same point (otherwise the moment about anyone meeting point would be non-zero) …

so you can re-draw the diagram so that the wire goes through the meeting point of the nomal force and the weight, and finding the angle then become a pure geometry question! :smile:

try it both ways​
 
  • #38


try it both ways

Last time I've taken this advice it didn't work out well for either me or my boyfriend, but I'll try it again just for you ;)...

[hehe...sorry...had to go there]
Okay, back to the problem-- I've tried taking the easier route by drawing a triangle and solving it in a geometrical fashion. Yes, I know where those points meet, shown in my last uploaded img. I have a right triangle. I know mg, and I know the angle opposite to T (90 degrees). I don't know anything else.

http://img39.imageshack.us/img39/9876/geotry.jpg

Uploaded with ImageShack.us

With respect to sum of all forces - you're right, I should've used these equations instead, but I still don't see how they let me find the angle...I hope I'm not being lazy here... I'm really trying my best to solve it but there's a limit to how many times a girl can write statics equations and fail! Regardless, if you say it can be solved in a geometrical fashion I rather check that out instead of working myself to death again with equations. I still don't see how I can find the angle.
 
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  • #39
Hi Dory! :smile:
Dory said:
Last time I've taken this advice it didn't work out well for either me or my boyfriend, but I'll try it again just for you ;)...

[hehe...sorry...had to go there]

ohhh … I assumed you were a John :rolleyes: …

being a goldfish, I naturally immediately thought of http://en.wikipedia.org/wiki/John_Dory" :biggrin:
Okay, back to the problem-- I've tried taking the easier route by drawing a triangle and solving it in a geometrical fashion. Yes, I know where those points meet, shown in my last uploaded img. I have a right triangle. I know mg, and I know the angle opposite to T (90 degrees). I don't know anything else.

Do you mean https://www.physicsforums.com/attachment.php?attachmentid=31884&d=1296732455", in post #27, with the cute picture of you?

But your T there doesn't go through the black dot … the forces should all meet there, which is immediately gives you tanα = 2. :smile:
With respect to sum of all forces - you're right, I should've used these equations instead, but I still don't see how they let me find the angle...I hope I'm not being lazy here... I'm really trying my best to solve it but there's a limit to how many times a girl can write statics equations and fail! Regardless, if you say it can be solved in a geometrical fashion I rather check that out instead of working myself to death again with equations. I still don't see how I can find the angle.

Let's review the options:

You can take moments about that dot, which gives you α geometrically, and then you can find T from the y components.

You can take moments about some other point, such as A, which gives you an equation for T and α, and then again use the y components to get another equation for T and α (or use a vector triangle ).
 
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  • #40


tiny-tim said:
Hi Dory! :smile:


ohhh … I assumed you were a John :rolleyes: …

being a goldfish, I naturally immediately thought of http://en.wikipedia.org/wiki/John_Dory" :biggrin:

And you didn't think of Dory without the John? :)

dory.jpg


Do you mean https://www.physicsforums.com/attachment.php?attachmentid=31884&d=1296732455", in post #27, with the cute picture of you?

But your T there doesn't go through the black dot … the forces should all meet there, which is immediately gives you tanα = 2. :smile:

But then, there is no point where all the forces meet! Presuming my diagram is correct, those forces will go on forever and never meet at 3 places. There are 2 forces that meet at 2 points, though...

Let's review the options:

You can take moments about that dot, which gives you α geometrically, and then you can find T from the y components.

If I get there and just there that'll be good enough.

You can take moments about some other point, such as A, which gives you an equation for T and α, and then again use the y components to get another equation for T and α (or use a vector triangle ).

This I've tried, but I'll put it on the backburner for now.. I want to get the geometry way right first. Sounds easier... "sounds"...
 
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  • #41
Dory said:
And you didn't think of Dory without the John? :)

Oh, that Dory! :smile:

But … isn't she famous for being pretty … but a bit dumb? :redface:
But then, there is no point where all the forces meet! Presuming my diagram is correct, those forces will go on forever and never meet at 3 places. There are 2 forces that meet at 2 points, though...

"presuming"?! :rolleyes:

Dory, your diagram is wrong! :smile:
… I want to get the geometry way right first. Sounds easier... "sounds"...

It is easier …

but you seem to have such an aversion to drawing forces on diagrams that you can't bring yourself to drawing three together! :redface:
 
  • #42


tiny-tim said:
Oh, that Dory! :smile:

But … isn't she famous for being pretty … but a bit dumb? :redface:

hehe...

Hey um... *scratches head* I...feel... strangely stung...

"presuming"?! :rolleyes:

Dory, your diagram is wrong! :smile:

Right, the diagram (where I've drawn myself) is wrong, but the last pic I've uploaded is of a triangle presuming that all 3 points meet somewhere between C and B. I only know for a fact it's a right triangle since NA and MG are straight lines...what of T... how can I find it out if I only have 1 angle (90) and 1 force known.

It is easier …

but you seem to have such an aversion to drawing forces on diagrams that you can't bring yourself to drawing three together! :redface:

No no...no aversion! here!
 

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  • #43
Dory said:
No no...no aversion! here!

Then why is your diagram so wonky?​

Dory, I've decided that you need a lesson in drawing :wink: …

Draw the wall first.

Draw the beam at 45°, to go exactly 8 squares horizontally (as you've done).

Draw the vertical line through the centre of the beam (yes! it can be drawn vertically! :rolleyes:).

Draw the horizontal line from the top of the beam, to meet the vertical line.

Then say to yourself … "Since the beam is in equilibrium, the only three forces on it must all go through the same point, so I must draw the string through that point, if I can force myself to do it :redface: … yes I can! I can! I can! :-p" …

and then draw the string through that point, to the wall. :wink:

You should now have two triangles on the same base, one with twice the height of the other :smile:​
 
  • #44


Okay, I've redrawn it. Here, do you mean the triangles I filled in in the uploaded pic?

I still don't see how can I find out their angles!
 

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  • #45
No, I meant the two triangles you get if you use the horizontal line from the bottom of the beam to the wall as the base of the triangles.
 
  • #46


Are these the two triangles then?
 

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  • #47
No! :smile:

The two triangles both with the same baseline, one with hypotenuse along the beam, the other along the string. :smile:

(oh, and make the baseline go from the end of the string, not from the adjacent corner of the beam that's almost a square away! :rolleyes:)
 
  • #48


Are these?
 

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  • #49
Hunky-dory! :biggrin:

ok. now you know the base and the heights of both triangles …

sooo … ? :smile:​
 
  • #50


Okay, fine, so now I got the ratios in meters of the triangle I need. So, I use the ratios in Newtons now to find for T. When I do that T = 109 [N]...the answer book says T = 109 [N]... which means that... wait a second.. I'm getting the same answer as the answer book... YEAAAAAAAAAAAAAAAAA BABYY :DD :D :D

//

;)


The entire solution posted here...thank you so much tiny-tim! I even wrote you a special thanks on the paper :) :) :)
 

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