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Doitwell

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- TL;DR Summary
- A variation of a well known decomposition of forces acting on a free body resting on a ramp, however it's now resting on either symmetric or asymmetric double ramp. What is the force value acting on each of the ramp "halves"?

Hi,

a newcomer here. I need to know the value(s) of the force(s) acting on each inclined face of a double ramp. Here's a visualisation of a double inclined ramp (just an example, could be asymmetric):

Imagine a body rests on the top faces, let's assume it's a circular bar (a cylinder) which exhibits contact with each face at the line near the middle of the V-shape height.

Let's first assume a symmetric ramp.

Now, I've drawn a section view with half of the forces (please excuse the lack of arrowheads - the dotted lines are just construction geometry, solid lines are geometry/forces):

The vector decomposition in itself seems pretty straightforward but I just don't know which forces contribute to the force exerted on each half of the ramp (or the reaction forces if you will). Or, at least, I'm not sure if I'm right.

I thought that we have a situation where the force in question should be the sum of F

and F

F

F

F

But then there's this other component of the F

Now that I come to think about it when typing - if the left side of the ramp was a wall perpendicular to the right side of the inclined plane then the reaction force acting on the left side would have been just F

and if the ramp was symmetric both those forces would be equal - 45deg inclination angle.

If we vary the inclination angle on one side the center of gravity moves. Unless the angle between the inclination planes is 90 deg I think I'm getting some fallacy...cause if I change an angle on one side only (left) the forces keep the value while the angle under which F

Could someone point me in the right direction please?

a newcomer here. I need to know the value(s) of the force(s) acting on each inclined face of a double ramp. Here's a visualisation of a double inclined ramp (just an example, could be asymmetric):

Imagine a body rests on the top faces, let's assume it's a circular bar (a cylinder) which exhibits contact with each face at the line near the middle of the V-shape height.

Let's first assume a symmetric ramp.

Now, I've drawn a section view with half of the forces (please excuse the lack of arrowheads - the dotted lines are just construction geometry, solid lines are geometry/forces):

The vector decomposition in itself seems pretty straightforward but I just don't know which forces contribute to the force exerted on each half of the ramp (or the reaction forces if you will). Or, at least, I'm not sure if I'm right.

I thought that we have a situation where the force in question should be the sum of F

_{n}and F

_{d}for each half.F

_{n}is the perpendicular component of the force of gravity acting on the body (0.5*mass*gravity*cos(30))F

_{s}is the parallel component of the force (0.5*mass*gravity*sin(30))F

_{d}is a decomposed component of F_{s}acting along the line of contact.But then there's this other component of the F

_{s}left behind.Now that I come to think about it when typing - if the left side of the ramp was a wall perpendicular to the right side of the inclined plane then the reaction force acting on the left side would have been just F

_{s}like so:and if the ramp was symmetric both those forces would be equal - 45deg inclination angle.

If we vary the inclination angle on one side the center of gravity moves. Unless the angle between the inclination planes is 90 deg I think I'm getting some fallacy...cause if I change an angle on one side only (left) the forces keep the value while the angle under which F

_{s}acts changes.Could someone point me in the right direction please?