Statics is this problem easy or hard?

[Femme_physics]

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...

 

[tiny-tim]

Hi Dory!

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 … 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?

 

[Femme_physics]

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 :)

 

[Femme_physics]

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?

 

[tiny-tim]

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 …

if you draw that horizontal line (BD say),

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

 

[Femme_physics]

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.

 

[tiny-tim]

Dory, I get the impression that you don't like drawing forces on diagrams.

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?

 

[Femme_physics]

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...

 

[tiny-tim]

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!

ok, now you have 3 lines of force …

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

which is it?

 

[Borek]

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

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

 

[Femme_physics]

(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?

 

[tiny-tim]

Dory, why did you not answer my last post? …

ok, now you have 3 lines of force …

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

which is it? ​

 

[Femme_physics]

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.

 

[tiny-tim]

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 ) 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? ​

 

[Femme_physics]

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?

 

[tiny-tim]

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?

 

[Femme_physics]

I used parallelograms to find out AC = 1.41

See here

http://img94.imageshack.us/img94/4237/thisishowu.jpg

:D

I got 24.11 degrees by knowing the following information:

http://img40.imageshack.us/img40/8922/scannorz.jpg

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!

 

[tiny-tim]

Now you've swapped over A and C !

Are you trying to confuse me?

I used parallelograms to find out AC = 1.41

But I don't understand how you knew it was a parallelogram.

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 ) …

Try the point where the normal force and the weight of the beam meet.

 

[Femme_physics]

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 ...

 

[tiny-tim]

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 ?

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

 

[Borek]

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

Upper triangle hypotenuse is NOT parallel to the beam.

 

[Femme_physics]

where's mg ?

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

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!

 

[tiny-tim]

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.

 

[Femme_physics]

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)

 

[tiny-tim]

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.

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.

 

[Borek]

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.

 

[Femme_physics]

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

 

[tiny-tim]

Nice diagram!

ok, if ∑M = 0, then what is the contribution of T to that?

 

[Femme_physics]

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*

 

[tiny-tim]

but you know that ∑M = 0 !

sooo … ?

 

[Femme_physics]

So...T equals... 0? *scratches head*...

 

[tiny-tim]

No, the moment of T about that point equals 0,

sooo … ? ​

 

[Femme_physics]

Can't be that T=0! Answer says differently! Unless I mistranslate something in the question... are you saying that T=0?

 

[tiny-tim]

The moment of T is zero.

 

[nvn]

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.