Normal Force between two concentric cylinders

[Joshua Smith]

Homework Statement

a cylinder sits inside of another fixed cylinder without friction and an internal force acts on the cylinder as shown in the diagram. Draw the normal force vector and write down the force and moment equations.

I drew in my normal vector to counteract the A force vector and wrote out the force and moment balance equations but when I did a check by plugging in numbers I got that the normal force causes a moment on the inner cylinder. What did I do wrong here?

Homework Equations

Force balance, moment balance

In the picture the black dash is the thin outer cylinder

The Attempt at a Solution

∑Fx = 0, ∑Fy = Ay - Ny = 0, ∑Fz = Az - Nz = 0,
∑Mx = (Az * you - Ay * za) + (Nz * r cosθ - Ny * r sinθ) = 0,
∑Mx = ( -Ay * za) + (Nz * r cosθ - Ny * r sinθ) = 0,

 

[haruspex]

A normal force is one that is perpendicular to the plane of contact between two bodies. In this case it would be radial to the cylinders.

 

[Joshua Smith]

A normal force is one that is perpendicular to the plane of contact between two bodies. In this case it would be radial to the cylinders.

so In that case it doesn't matter where the A force originate, just its angle? it would like this then?

 

[haruspex]

so In that case it doesn't matter where the A force originate, just its angle? it would like this then?
View attachment 112026

Yes, that's better.
There is a a mistake n the relevant equations you posted. They made a certain assumption which might not apply here.

 

[Joshua Smith]

Yes, that's better.
There is a a mistake n the three relevant equations you posted. They all made a certain assumption which might not apply here.

since the outer cylinder is fixed the inner cylinder can't move so the forces should sum to zero, but it could rotate about x

 

[haruspex]

since the outer cylinder is fixed the inner cylinder can't move so the forces should sum to zero, but it could rotate about x

Right.
So what equations do you now have?

 

[Joshua Smith]

Right.
So what equations do you now have?

There will be a moment about the x-axis caused by the y component of the A force

∑Fx = 0, ∑Fy = Ay - Ny = 0, ∑Fz = Az - Nz = 0,
∑Mx = -Ay * za = 0

 

[haruspex]

There will be a moment about the x-axis caused by the y component of the A force

Yes, so why did you write this:

∑Mx = -Ay * za = 0

?

For the linear forces, you need to get the x and y components of A in terms of A, θ, az and r.

 

[Joshua Smith]

Yes, so why did you write this:

?

For the linear forces, you need to get the x and y components of A in terms of A, θ, az and r.

oops. I copy pasted and modified which led to me forgetting to take out the = 0.

 

[Joshua Smith]

I have a question about this with friction. If you added friction between the two cylinders then there would be a tangential force acting at the point where the normal force touches to counteract the rolling of the cylinder but then the force balance wouldn't work out? How is this rectified?
∑Fn = A - N = 0, ∑Ft = -uN ~= 0,
∑Mx = -Ay * za + uN * r

 

[haruspex]

then the force balance wouldn't work out

You have assumed the normal force would continue to act along the same radius.