What is the force exerted on cylinder A in a V-shaped trough?

[Bradracer18]

Homework Statement

Ok, my problem has 3 smooth homogeneous cylinders(A,B,C), which are stacked in a V shaped trough.(the cylinders...look more like circles) are also all touching each other and each is touching a surface of the trough. The trough angles are Left = 30 deg and right = 45 deg. Each cylinder has a diameter of 500mm and a mass of 100kg. Find the force exerted on cylinder A by the inclined surfaces. From Left to Right...they are labeled B,A,C.

Homework Equations

None yet...well, maybe W=mg

The Attempt at a Solution

Ok, so I drew a free body diagram, separating each cylinder. I've got each force drawn out...they include the forces normal to the surface(Nb, NaL(left), NaR(right), Nc) I've also drawn in the force pointing strait down on each cylinder, of 100kg. And, finally each cylinder has a force pointing in the direction of the cylinder which touches it(they are equal and opposite) called(Nab, Nac,etc).

So, starting on cylinder B.

Nbsin30 + Nabcos30 = 0
Nbcos30 + Nabsin30 -100(9.8) = 0

I'm not sure how to solve these equations...or, what to do next(I'm sure its the same for Nc...but I don't know what to do for Na).
Thanks,
BradView attachment cylinders.bmp

 

[Bradracer18]

Is this problem not understandable...or what do I need to do, to get help?

 

[Bradracer18]

Still no replies...I'm sending it back to the top. If you are reading this, and can't understand the problem...please tell me so I can try to explain it better. I need help.

 

[Kurdt]

As you say its all about resolving forces. Cylinder A exerts a greater force than just its weight on the two planes because of the two cylinders above it.

For cylinder B the force parallel to the plane and acting on A is: [tex] mg \sin (\theta) = F_{BA}[/tex]

I'm sure you can fill in the rest.

 

[Bradracer18]

I guess I'm still confused on this one. I've got the answers...so that isn't so much as a concern for me...its learning how to do this.

Answers are FLeft = 1304N and Fright = 829NIts not this is it...because I can't get this to work either.

Fba = mgsin(30)
Fac = mgsin(45)
FaL = mgcos(30)
FaR = mgcos(45)

Then just add up the left(ba and aL) and the right(ac and aR)...forces?

 

[Kurdt]

The geometry of the problem comes into play a lot. If you draw a more accurate diagram you may see why it is a lot more involved.

 

[Bradracer18]

I've got a accurate diagram in front of me, on my paper. I've got it all separated, and shows every force.

I guess I'm still stuck thinking that I do this:

For cylinder B(and how it effects A)

Fbsin(30) + Fabcos(30) = 0 ---which would be the Ftotal in the x direction for B.

Fbcos(30) + Fabsin(30) -100(9.8) = 0 ---Ftotal in y direction.

What is wrong with this thinking...and if its right...I guess I don't know how to solve it...for those force totals.

Also, where does the diameter come into play here...later?

 

[Kurdt]

Let me just ask you does the direction if the force from cylinder B on cylinder A form a right angle with the surface of the right hand plane?

 

[Bradracer18]

umm I'm not sure exactly what you are asking(if you are asking seriously...or trying to get me to think)...if its the last one...then I'm going to say no...I'd say they form a 95 deg angle between them.

The reason I am going the route above, is because my instructor somewhat started the problem...and I can't pick up from his point.

 

[Kurdt]

If the direction of the force does not form a right angle with the right hand plane then you can't simply add them can you? And the angle is not 95 degrees.

 

[Bradracer18]

ok...I think I've got it figured out. I must have had a brain fart...seems easy now. I found an example in the book(different chapter)...and it helped some too. Anyways thanks for all the help Kurdt.

Can you help with my other post(about the Force on a collar)...that is, if you have time...and can understand my pic.

Thank you!