# Determine forces acting on double inclined plane

• B
• Doitwell
Doitwell
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 Fn
and Fd for each half.
Fn is the perpendicular component of the force of gravity acting on the body (0.5*mass*gravity*cos(30))
Fs is the parallel component of the force (0.5*mass*gravity*sin(30))
Fd is a decomposed component of Fs acting along the line of contact.

But then there's this other component of the Fs 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 Fs 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 Fs acts changes.

Could someone point me in the right direction please?

The lack of arrowheads cannot be excused. They indicate the direction of the force the arrows represent which is necessary for writing the sum (or resultant) force correctly. Without them, a free body diagram (FBD), of the kind you are trying to show, is difficult to interpret and write equations for.

The tails of the arrows are also important because they need to be drawn at the point of the extended body where the force acts. This is important for determining torques (or moments).

Just to make sure what you are asking. On the right is shown a correctly drawn FBD for a cylinder of weight ##W## resting on an asymmetric angle. Assumed given are angle ##\theta## and the angle that the weight forms with the vertical direction.

Is your goal to find the contact forces ##N_1## and ##N_2##?

Is this a homework problem?

Hi,
Thanks for taking the time to answer my question. Yes, indeed those are the values I'm searching for.
I've omitted the arrowheads because the software does not support them OOTB and the problem definition is rather simple. I'll make up for that next time I'm uploading something.

Oh, sorry I was writing early in the morning and I missed the other question.

No, it's not a homework problem.

Doitwell said:
Could someone point me in the right direction please?
You have 3 forces acting on the body that need to add up to zero (vectorially). The same applies to the moments generated by those 3 forces (e.g. around the bodies center of mass).

To get a unique solution, you would need to assume the contact is friction-less, which constrains the directions of contact forces to the surface normals, so you have 2 unknowns (contact force magnitudes) and 2 equations (force and moment equilibrium).

Last edited:
Allright, I know the equilibrium rule. The issue is I don't know which vectors take part in the solution so I can't write them down and solve...
In all cases where inclination angle is different from 45o (or 0) the gravity force decomposes to normal force component and parallel component which does not act along the line of contact on the other side of the prism. Should the parallel component be further decomposed and what happens with the other remaining portion of the force?

Doitwell said:
The issue is I don't know which vectors take part in the solution so I can't write them down and solve...
Just look at the diagram provided by @kuruman :

Assuming no friction and circular shape, all moments around C are zero, so you are left with the force vector equilibrium:

W + N1 + N2 = 0

When you split this equation into horizontal and vertical components (using known angles and trigonometric functions), you get 2 equations, and two unknowns (magnitudes of N1 and N2).

Juanda and kuruman
Look at the drawing on the right. It is self-explanatory. The unit vectors ##~\mathbf{\hat t}_1##, ##\mathbf{\hat n}_1## are in directions parallel and perpendicular to the incline on the left and likewise we have ##~\mathbf{\hat t}_2##, ##\mathbf{\hat n}_2## for the incline on the right.

Angle ##\theta## is the half-angle of the double-inclined plane and angle ##\varphi## is the angle by which the bisector of the double-inclined plane is tipped away from the vertical.

The goal is to express ##\mathbf{\hat n}_1## and ##\mathbf{\hat n}_2## in terms of the Cartesian unit vectors ##~\mathbf{\hat x}## and ##~\mathbf{\hat y}##. To do this, we first find ##~\mathbf{\hat t}_1## and ##~\mathbf{\hat t}_2## and then find the perpendicular vectors in the right direction. Formally, a 2D vectors is written as $$\mathbf{A}=A\cos\!\alpha~\mathbf{\hat x}+A\sin\!\alpha\mathbf~{\hat y}$$where angle ##\alpha## increases counterclockwise from the positive x-axis.
For ##\mathbf{\hat t}_1##
$$\alpha=2\theta+90^{\circ}-(\theta+\varphi)=90^{\circ}+(\theta-\varphi)\implies \begin{cases} \cos\!\alpha=-\sin(\theta-\varphi) \\ \sin\!\alpha=\cos(\theta-\varphi) \end{cases}$$Thus, $$\mathbf{\hat t}_1=-\sin(\theta-\varphi)~\mathbf{\hat x}+\cos(\theta-\varphi)~\mathbf{\hat y}.$$To find ##\mathbf{\hat n}_1##, you swap the components of ##\mathbf{\hat t}_1## and change the sign of one of them making sure that the normal points in the desired direction. $$\mathbf{\hat n}_1=\cos(\theta-\varphi)~\mathbf{\hat x}+\sin(\theta-\varphi)~\mathbf{\hat y}.$$Note that ##\mathbf{\hat n}_1\cdot \mathbf{\hat t}_1=0## which means that the two are perpendicular and that both components of ##\mathbf{\hat n}_1## are positive as expected.

Finally, you can write $$\mathbf{N}_1=N_1\mathbf{\hat n}_1=N_1\left[\cos(\theta-\varphi)~\mathbf{\hat x}+\sin(\theta-\varphi)~\mathbf{\hat y}\right].$$I leave it up to you to find ##\mathbf{N}_2## and finish as suggested by @A.T. in post #8.

A.T. said:
Just look at the diagram provided by @kuruman :
View attachment 347442

Assuming no friction and circular shape, all moments around C are zero, so you are left with the force vector equilibrium:

W + N1 + N2 = 0

It might also help to visualize the equation above as vector addition (vector triangle.):

N1 + N2 = -W

This makes it clear that the component magnitudes must satisfy the following:

N1X = N2X
N1Y + N2Y = W

So given the elevation angles a1 and a2for N1 and N2 respectively, we get:

N1 cos(a1) = N2 cos(a2)
N1 sin(a1) + N2 sin(a2) = W

which needs to be solved for the force magnitudes N1 and N2.

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Doitwell said:
...
In all cases where inclination angle is different from 45o (or 0) the gravity force decomposes to normal force component and parallel component which does not act along the line of contact on the other side of the prism. Should the parallel component be further decomposed and what happens with the other remaining portion of the force?

A.T. said:
It seems to me, that in this problem you won't get a unique solution if friction is allowed:
https://en.wikipedia.org/wiki/Statically_indeterminate
I believe that you are correct.

I may be wrong by seeing static friction in this case as irrelevant, only because the geometry is such that it makes any relative sliding movement impossible.
Therefore, the value and direction of the reaction vectors seem to be independent from the magnitude of the coefficient of friction.

Lnewqban said:
I believe that you are correct.

I may be wrong by seeing static friction in this case as irrelevant, only because the geometry is such that it makes any relative sliding movement impossible.
Even though no motion is possible in the stable equilibrium state, stresses may be present. Those stresses may have arisen when the system was assembled.

Suppose, for instance that the two flat ramps were slightly separated. The ball is nestled between. Then the ramps are slid slowly together. The ball would then settle under some compressive stress from the static friction.

Contrariwise, if the ramps started tightly together and were then slowly and slightly separated, there would be some tensile stress (or at least a reduction in the compressive stress from the normal forces).

Nasty non-ideal effects like chattering could plague an attempt to accurately model such things.

Juanda and Lnewqban

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