More advanced FBD problems-need some tips

[caddyguy109]

More advanced FBD problems--need some tips!

Okay, just got assigned 3 more FBD-based problems with which I not only have to do the normal calculations, but also use various computer programs of choice for additional calculations, plots, etc.
As such, these are a bit more advanced than the FBD problems I've done this far, and I would GREATLY appreciate some guidance/tips on successfully solving each one.

1)http://www.ihostphotos.com/show.php?id=182814"
2)http://www.ihostphotos.com/show.php?id=182816"
3)http://www.ihostphotos.com/show.php?id=182817"

I get the basic concepts, but each of these problems is a little more complicated (or, moreso, just are seeming that way right now) than I'm used to.
THANKS! Any and all suggestions welcome, as I start trying to work on these buggers...

 

[mathmike]

your page is to big to see you need to shrink it

 

[caddyguy109]

I know...it's fixed now^

 

[caddyguy109]

The sizes of the pages are fixed now. Anybody have tips/clues as to how to start and find each one? Don't want answers, just a running start and direction.

 

[caddyguy109]

Still haven't gotten very far. Anyone have any tips on resolving the geometries of each? That seems to be what's throwing me, to start with.

1) Can't figure out what/how gets combined to find the distance "a", to start with?
2) Angles...how to express theta as a function of M, W, and I?
3) Similar to two, but now how to express the Tension as a function of theta?

This probably isn't too difficult, but I'm just having trouble thinking how to start/layout the steps, especially for the ones with angles.

 

[lightgrav]

Geometries? I'd use the rectangular coordinates that houses are built on.

1) 0,0 at B; tan(theta) = 2R/a = (Z_p)/(X_p), L - sqrt(4R^2 + a^2)
it might be useful to take Moments around the z-line of action of P
at the x-line of action of the Pipe's F_N (beneath left side of a)
You know B_x already , so that will get you B_z
But there aren't any WRONG places the sum Moments around ...

2) at least it's a 90 degree angle! and frictionless ...
try taking Moments around the place where the two F_N intersect
did you sum F_x and F_z first?

3) this one's pretty easy. Moments around A, gives you T.
back-substitute into Sum F_x and F_z, to get reactions at pin.

 

[caddyguy109]

Okay, THANKS--that should get me going!

 

[caddyguy109]

Okay, finally got a chance to get back to these problems.

For #1:
a) Determining distance "a" as a function of known length "l" and the radius of tube "r":
You said: "0,0 at B; tan(theta) = 2R/a = (Z_p)/(X_p), L - sqrt(4R^2 + a^2)"

So, should I simplify down to the point that I get a "a=" with L & R on the on the right?

And with this:
"it might be useful to take Moments around the z-line of action of P
at the x-line of action of the Pipe's F_N (beneath left side of a)
You know B_x already , so that will get you B_z"

So, Bx = 0? Then how do I get Bz? Similarly, what is the z-line of action of P at the x-line of action of the Pipe's Normal Force--and how do I get moments from that?

Sorry to be so confused...it doesn't seem too hard but having so many unknowns is making me lost.

 

[caddyguy109]

Anybody? Am I just trying to simplify down that expression for the first part of #1, or what?

 

[caddyguy109]

I'm still confused with #1.

From this:

1) 0,0 at B; tan(theta) = 2R/a = (Z_p)/(X_p), L - sqrt(4R^2 + a^2)
it might be useful to take Moments around the z-line of action of P
at the x-line of action of the Pipe's F_N (beneath left side of a)
You know B_x already , so that will get you B_z
But there aren't any WRONG places the sum Moments around ...

Okay. So, to determine the distance a as a function of the known length L and radius r, in order to hold the rod in equilibrium with force P, am I taking this:

2R/a = L - sqrt(4R^2 + a^2)

and simplifying down so a is alone on one side basically? What about the (Pz) and (Px) mentioned above or the tan(theta)? I don't really need to include those for part a), do I?

 

[caddyguy109]

Should I FIRST be finding the total moments and total forces?

There's just something about this "easy" problem I'm not getting...

 

[lightgrav]

Sum of Forces in the x direction = 0 (this *IS* statics, no?)
there's a P in the -x direction, anf there's no friction in the pipe.
So the Bx must cancel Px.

 

[caddyguy109]

Okay.

Yeah, it's Statics alright...too bad I tend to forget the simple things.

What about finding the factor a as a function of L and R? Am I headed the right direction with that or no?

 

[lightgrav]

I think the best way to involve parameters like "a" and "R"
is just to write the torque equations. They'll be messy enough,
you don't need to resort to involved geometry (Pythagoras, trig)
to include the important parameters.

If some problem asks you to write an important quantity
as a function of an UNimportant variable,
tell them to walk off a cantilever.
(an important variable will show up all by itself
as you write the statics Force component and torque eq'ns)

by the way, once you know Bz, then Fz=0 gets you az.
but you have to know how far down the force P is applied.
That's the only reason to do this geometry work.

 

[caddyguy109]

Okay...so:

Summing the forces:
FB=0 -> (-Px + Bx)i + (Py + By - Ay)j ?

Did I forget the sin and cos?

 

[lightgrav]

The Force P is horizontal, so Py = 0.
Don't forget the weight of the rod.

 

[caddyguy109]

After doing the sum of forces, got:
Ay= -49jN
By= 49jN

Then finding the moments about B, as an example, would this be right?
MA=0 -> [(L sin(theta))j x -Pi] + [Byj x (sqrt(4R^2 + a^2))i]

Right?

 

[lightgrav]

I'm pretty sure 10 kg weighs 98N (-j) here, so By + Ay = 98N .
otherwise ... looks like these are moments around A ...
where is the 98N thru c.o.m.?

 

[caddyguy109]

I'm pretty sure 10 kg weighs 98N (-j) here,
otherwise ... looks like work about to be accomplished!

Yeah, I had:
F=0 -> (By-Ay-W)j=0
(By-Ay)j = Wj
(By-Ay)j = 98jN
Then:
By = 49jN
Ay = -49jN

Okay, does look like I'm FINALLY getting somewhere on this problem. But, to once again go back to part a), how would I actually write out the equation where a is a function of L and R? Like in the very beginning, what I was talking about, or what?

The big point of this assignment (3 problems) is to resolve certain components into functions of other components, and then plot that data using a couple given ranges and stuff.

 

[caddyguy109]

Okay, got this total moment:

(P*L*sin(theta) - By*cos(theta)*sqrt(4R^2 + a^2) + (W*L/2)*cos(theta))k N-m = 0

I have By as calculated previously, but do I really get out of this total moment? Is it how I get my a = function of... value?

 

[lightgrav]

be very careful when you start writing Ay = - 49N, meaning upward!
how do you expect the pipe's compression to support half the rod?

Torque equations will have these variables related in them.

Don't forget weight of the Rod, in torque equations,
else there won't be any "scale constant" set (only relative F's)

 

[caddyguy109]

Hmm...I thought the Ay= -49jN just meant the downward Normal Force of the tube on the rod at A. How could the (-) mean upward?

Now that I have the total moment (I fixed it to include the W factor ^), is this what I solve for A?

 

[lightgrav]

you started out saying
Sum Fy: By - Ay - W = 0 , implying that Ay is positive downward.
Then somehow you set Ay and By to have the same magnitude,
ended with Ay = - 49N ... negative meaning opposite of presumed.
This is just enough to cancel By . If Ay ISN'T upward, what holds W?

 

[caddyguy109]

Okay, figured out the whole (+/-) Ay thing.

 

[caddyguy109]

Now just simplify down the sum of Moments about B to get "a" as a function of the others?

By the way, this was my total sum of moments:

(P*L*sin(theta) - By*cos(theta)*sqrt(4R^2 + a^2) + (W*L/2)*cos(theta))k N-m = 0

resulting in (I think) this:

a= [(P*L*sin(theta) + (W*L/2)*cos(theta))/By*cos(theta)] - 2R

Hmm...well that WOULD work, since I know W and By and am given P and L in order to plot a(R), but what the heck do I do with the angles theta?

 

[caddyguy109]

Anyone?

As a refresher, here's the first problem again:
http://www.ihostphotos.com/show.php?id=182814"

My sum of forces, resulted in finding:
By = 49jN
Ay = -49jN

Then for sum of moments about point A, I got this (which is where the starts):

(P*L*sin(theta) - By*cos(theta)*sqrt(4R^2 + a^2) + (W*L/2)*cos(theta))k N-m = 0

then, because I need the distance "a" as a function of L and R, I simplified that down to this:

a= [(P*L*sin(theta) + (W*L/2)*cos(theta))/By*cos(theta)] - 2R

a= [(P*L*tan(theta) + (W*L/2))/By] - 2R

Which is where my confusion really builds. I found By and W, and am given a range for R values and then values for P and L, but NO angle info. So what the heck do I do with the thetas?

 

[caddyguy109]

------------

 

[caddyguy109]

Can anybody help me out on the moment troubles? Seems right, except for the fact that I have no theta specification in the problem--thus, since I need to make plots with some of the data, I'm missing something.

Should I even have the angles?

 

[caddyguy109]

Anyone?

 

[Gokul43201]

Some comments :

1. The reaction at B acts normal to the length of the rod.
2. The reaction at A acts vertically downwards.

Calling the angle between the rod and the vertical, X (so that tanX = a/2r) gives :

A + W = BsinX - - - (vertical forces)

P = BcosX - - - (horizontal forces)

Bd = (Wl/2)sinX + (Pl)cosX - - - (moments about A, with d^2 = a^2 + 4r^2)

 

[caddyguy109]

Some comments :

1. The reaction at B acts normal to the length of the rod.
2. The reaction at A acts vertically downwards.

Calling the angle between the rod and the vertical, X (so that tanX = a/2r) gives :

A + W = BsinX - - - (vertical forces)

P = BcosX - - - (horizontal forces)

Bd = (Wl/2)sinX + (Pl)cosX - - - (moments about A, with d^2 = a^2 + 4r^2)

Okay. So, what do I then use for the X value?

What does the Bd stand for?

 

[Gokul43201]

1. You only need values of sinX and cosX, which are respectively $a/\sqrt{a^2 + 4r^2}$ and $2r/\sqrt{a^2 + 4r^2}$.

2. A and B are the normal reaction forces by the pipe on the rod at those corresponding points (sorry about the terrible choice of notation). Perhaps I should have called them $N_A, ~N_B$.

3. d is the length of the rod inside the pipe. $d = \sqrt{a^2 + 4r^2}$.

 

[caddyguy109]

1. You only need values of sinX and cosX, which are respectively $a/\sqrt{a^2 + 4r^2}$ and $2r/\sqrt{a^2 + 4r^2}$.

2. A and B are the normal reaction forces by the pipe on the rod at those corresponding points (sorry about the terrible choice of notation). Perhaps I should have called them $N_A, ~N_B$.

3. d is the length of the rod inside the pipe. $d = \sqrt{a^2 + 4r^2}$.

FINALLY, something that makes sense!

Now, to arrange it all...

Now, I'm completely lost on the second one and a little on the third:
2) http://www.ihostphotos.com/show.php?id=182816"

This was said here:
"at least it's a 90 degree angle! and frictionless ...
try taking Moments around the place where the two F_N intersect
did you sum F_x and F_z first?"

but I just can't, for whatever reason, still figure out what the heck to do. ?

3) http://www.ihostphotos.com/show.php?id=182817"

Again, I was told this:
"this one's pretty easy. Moments around A, gives you T.
back-substitute into Sum F_x and F_z, to get reactions at pin."

and did this so far, until I tried to figure out what the tension TBE was and realized something was wrong:
http://www.ihostphotos.com/show.php?id=192075"

Anything in this one REALLY stick out to you?

 

[caddyguy109]

Anybody?

 

[caddyguy109]

Any help on the second two problems, based on what I said two posts up??