# More advanced FBD problems-need some tips!

Anyone?

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

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?

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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?

Anyone?

Gokul43201
Staff Emeritus
Gold Member

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)

Gokul43201 said:

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?

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Gokul43201
Staff Emeritus
Gold Member
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. $d = \sqrt{a^2 + 4r^2}$.

Gokul43201 said:
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" [Broken]

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" [Broken]

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" [Broken]

Anything in this one REALLY stick out to you?

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Anybody???

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

Anybody???

Can anyone help me figure out #2:
2) http://www.ihostphotos.com/show.php?id=182816 [Broken]

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, get it going. Where the heck do I take the sum of moments about? And how do I then express the angle (theta) corresponding to equilibrium as a function of M, W, and L???

This is the LAST question for a while, I swear

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