Free body diagram of a truncated cone?

[kronoskar]

Hi,
I have a question about basic statics.

I have heard from someone that the forces acting on a truncated cone in a hole of corresponding geometry is different from an ordinary block sliding down a wedge, since the normal force on one side of the cone will be affected by the normal force on the other side. I've had some difficulties concerning this when trying to figure out the frictional forces occurring when the cone is turned/screwed/pressed downards into the hole. I've attached a jpg showing a truncated cone wedged into a hole of a corresponding geometry (drawing in 2D but problem probably to be solved in 3D), the cone is symmetrical.

That is, will the normal force on each side of the cone be larger than each of the horisontal force (for example mg)?

For block sliding down a wedge, the relation between the normal force and the horisontal force is something like this:

http://i359.photobucket.com/albums/oo31/tanzl/freebodydiagram1.jpg

Would gladly appreciate the help!

Brgd

Oskar

 

[Simon Bridge]

Welcome to PF;
Not sure I understand you - the normal forces around the cone have to sum to the weight like, um, normal.

 

[Ritz_physics]

If the cone is symmetrical about it's vertical axis, the normal force on each side will be the same. By normal force, i mean the reaction force of the surfaces of the hole on the cone which are perpendicular to the surface.

 

[kronoskar]

Thanks for your replies Simon and Ritz physics!
I will try to be more precise, which of the two diagrams are correct, the one to the right or the one to the left? The diagrams show the reaction force of the surface of the hole on ONE of the sides, but of course (in 2D) there would be a similar diagram for the other side).

Oskar

 

[Simon Bridge]

Draw both components of the weight, not just the normal, and you'll see ;)

The weight equals the normal force plus the tangential force.

What you are doing is putting x-y axis on the slope, so that the x-axis points along the slope and the y-axis is normal to the slope; then resolving the weight force into components.

 

[kronoskar]

Simon: So you mean the picture to the right is correct?
Cuz, if so the truncated cone would be the same as for a wedge (as one could believe), I felt this what a kind of "dummy question" since the weight always equl the normal force plus the tangential force (and not that the normal force equals the weight plus the tangential force). But an person with many years in the field claimed that the free body diagram to left would be true for a truncated cone in a hole. I never got a proper explanation to why so I thought that maybe I had miss out some of the fact (like in 3D, that the oppposite sides are "pressing" the truncated cone inwards and affects the normal force to be bigger). Would the result look different if the truncated cone is turned/twisted inside the hole? I.e. would the resulting friction force result in another normal force?

 

[Simon Bridge]

Technically the weight exerts a pressure over the entire area.

But you can picture it as a wedge between two supports.

Twisting without pushing down just slides the cone in it's hole - this motion would be opposed by the friction which will be given by the normal force times a coefficient and act tangentially to the surface.