Help! M2 Moments problem.

Hello!

I am stuck on a problem, I've taken it as far as I can but no progress

A uniform rod of length 2𝑎 runs from A to B and has mass 𝑘𝑚 where 𝑘=3. The rod is fixed at the point A and is freely hinged to another uniform rod of length 2𝑎 and mass 𝑚 at the point B. Both rods are in equilibrium with the first rod horizontal and the second inclined at angle 𝜃=30∘ to the vertical, with its unhinged end in contact with the rough floor. The coefficient of friction between the rod and the floor is 𝜇.

https://isaacphysics.org/api/images/content/questions/physics/mechanics/statics/level4/figures/Statics_hinged_rod.svg

So I labelled all the forces like so in this picture:

https://isaacphysics.org/api/images/content/questions/physics/mechanics/statics/level4/figures/Statics_hinged_rod2.svg

And by resolving and equating the forces I have

4mg = M + R
N = F

Taking pivot around A I have
(11/2)mg + Fsqrt(3) = 3R

Moments about B
2M + Fsqrt(3) = (5/2)mg + R

3M + Nsqrt(3) = (13/2)mg

So I need to find minimum mew, so that F <= mewR

Now when I try to solve the equations, all I end up with is (11/2)mg + Fsqrt(3) = 3R and really don't know how to progress.

Any Ideas?

Reply 1

h2shin

Oh above, after moments about B, I took moments about C as well.

They all just give me the same thing, I i suppose it's reassuring all the equations are consistent, just no way to find the mew!

Reply 2

h2shin

No one with any ideas?

Reply 3

ghostwalker

Original post by h2shin

No one with any ideas?

When mu is a minimum, you'll have

F=\mu R

Your problem is that you are treating the whole structure ABC as one thing and that's not generating enough information. BUT each of the two parts AB, BC is itself in static equilibrium.

E.g. You can take moments about B for just AB, which will give you M, and thus you can work out R from your initial equation. And so on.

On a quick calculation I got mu to be:

Spoiler

(edited 9 years ago)

Reply 4

h2shin

Original post by ghostwalker

When mu is a minimum, you'll have

F=\mu R

Spoiler

AHH!! Right. I thought that was to do with it but the random x's and y's were confusing the hell out of me as well. thanks!

So it's exactly like a connected newton's third law system problem in M1 just in two dimension! with forces at B just cancelling out when you look at the whole system and taken out of the picture when I take pivot around B to equate the torque.

So simple but you helped me get there :smile:

(edited 9 years ago)

Reply 5

ghostwalker

Original post by h2shin

AHH!! Right. I thought that was to do with it but the random x's and y's were confusing the hell out of me as well. thanks!

np.

PS: I used the equations you posted without checking them to work out my answer.

Reply 6

h2shin

Original post by ghostwalker

np.

PS: I used the equations you posted without checking them to work out my answer.

I got the same answer as you :smile:

The internal forces all cancelled out / were not relevant to the torque that's being calculated.

Thanks so much, this question was annoying me the whole day.

Just because the connected objects problem was taken from 1 dimension to 2 it just completely threw me off. argh.