How forces act on a shape and its centre of Mass (

Nov 2019
17
0
London
Hello all

I am now trying to understand how a uniformally distributed force and a non uniformally distributed force acts on a shape.

If I had an oddly shaped object for which I found the centre of mass and its surface area.

I then applied a uniformly distributed force of 100N on that object's surface, then would I be correct in saying that the uniformly distributed force of 100N is converted to an equivalent point force by multiplying the area by the force. The equivalent point force then acts through the objects centre of mass.

Below is a picture of what I am trying to understand:-


eplpplp.JPG


Would this illustration be correct?

Thank you.
 
Oct 2017
578
297
Glasgow
No, the impact of an evenly distributed force over a surface area will not be equal to a concentrated force at the centre of mass. To help see this, consider for example that, in panel 1, the effect of having a distributed set of forces could impose a torque, causing the object to rotate. This cannot happen in panel 2 because the perpendicular distance between the force and the COM is zero.
 
Jun 2016
1,198
565
England
Your illustration would only be correct if the centre of mass is coincident with the Centre of Force.
The Centre of Force is determined by the sum of the forces acting over all regions of the shape,
since you have said that the force is uniform, this dictates that the Centre of Force will coincide with the Geometrical Centroid.

In your picture, the Centre of Mass is clearly offset from the Geometrical Centroid, and thus from the Centre of Force.
thus, since the Centre of Force is offset from the Centre of Mass, there will be a rotation imparted to the object.
 
Nov 2019
17
0
London
Your illustration would only be correct if the centre of mass is coincident with the Centre of Force.
The Centre of Force is determined by the sum of the forces acting over all regions of the shape,
since you have said that the force is uniform, this dictates that the Centre of Force will coincide with the Geometrical Centroid.

In your picture, the Centre of Mass is clearly offset from the Geometrical Centroid, and thus from the Centre of Force.
thus, since the Centre of Force is offset from the Centre of Mass, there will be a rotation imparted to the object.
Hi Woody

Thank you for your response.

I think I'm more confused then I was to start with, I have made things more confusing myself so my apologies.

If I can step back a moment; I am trying to determine the method to convert a uniformly distributed force being applied over a shapes surface area into an equivalent point force that acts through the centre of mass and the centroid (where centre of mass and centroid are the same point).

I have amended my diagram to look like:-

UDL.JPG

Is there a method of converting a uniformly distributed force of 100N being applied to a surface area 0f 120m^2 into an equivalent point force that acts through the centre of mass and the centroid?

This is what I am trying to determine.
 
Nov 2019
17
0
London
No, the impact of an evenly distributed force over a surface area will not be equal to a concentrated force at the centre of mass. To help see this, consider for example that, in panel 1, the effect of having a distributed set of forces could impose a torque, causing the object to rotate. This cannot happen in panel 2 because the perpendicular distance between the force and the COM is zero.

Hi Benit13

Thank you for your response, your comment about a torque being introduced due to the perpendicular distance from the force to COM is correct and does make sense.

So, if I were to amend the diagram to look like:-



UDL.JPG


Is there a method of converting a uniformly distributed force of 100N being applied to a surface area 0f 120m^2 into an equivalent point force that acts through the centre of mass and the centroid?

This is what I am trying to determine. - is this even possible?

Thanks
 
Jun 2016
1,198
565
England
Split the shape down into smaller, simpler shapes.
Simpler shapes for which the centroid is easy to find.
The centroid of the complex shape will simply be the average of the centroids of the simpler shapes.

How is your maths?
Are you familiar with vector arithmetic?
For a 2 dimensional shape, you need two values to describe each position on the shape,
these two values form position "vectors" describing the position of any point on the shape (relative to some arbitrarily chosen "origin")
The average of the position vectors of the centroid of each of the simple shapes will then give the position vector of the centroid of the complex shape (relative to your arbitrarily chosen "origin").

To look at your uniformly distributed Force it is perhaps useful to step back and think how this force would be generated.
Force = Pressure multiplied by the Area over which it acts.
So the uniformly distributed force implies a uniform pressure applied at all points on the surface of the shape.
1 Newton per square meter of pressure is also called one Pascal.
So your uniform pressure would be \(\displaystyle 100N/120m^2 = 0.833333\)... Pascals.

To answer you central question, the answer is yes, the effect of a single 100N force acting at the centroid of the shape would be equivalent to that of a uniformly distributed pressure (of 0.83333... Pascals) acting over \(\displaystyle 120m^2.\)
 
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Nov 2019
17
0
London
Split the shape down into smaller, simpler shapes.
Simpler shapes for which the centroid is easy to find.
The centroid of the complex shape will simply be the average of the centroids of the simpler shapes.

How is your maths?
Are you familiar with vector arithmetic?
For a 2 dimensional shape, you need two values to describe each position on the shape,
these two values form position "vectors" describing the position of any point on the shape (relative to some arbitrarily chosen "origin")
The average of the position vectors of the centroid of each of the simple shapes will then give the position vector of the centroid of the complex shape (relative to your arbitrarily chosen "origin").

To look at your uniformly distributed Force it is perhaps useful to step back and think how this force would be generated.
Force = Pressure multiplied by the Area over which it acts.
So the uniformly distributed force implies a uniform pressure applied at all points on the surface of the shape.
1 Newton per square meter of pressure is also called one Pascal.
So your uniform pressure would be \(\displaystyle 100N/120m^2 = 0.833333\)... Pascals.

To answer you central question, the answer is yes, the effect of a single 100N force acting at the centroid of the shape would be equivalent to that of a uniformly distributed pressure (of 0.83333... Pascals) acting over \(\displaystyle 120m^2.\)

Thank you Woody.

May I ask one final follow on question which is:-

If I had a shape whose centre of mass and centroid were located at different points on the shape and a uniformly distributed load was applied across the entire surface area of the shape then if the uniformly distributed load was converted to an equivalent point load then would that point load act through the centre of mass or at the centroid?
 
Jun 2016
1,198
565
England
The load is distributed uniformly over the shape
so it is the geometry of the shape that dictates the centre of action of the force,
so the centre of action of the force will be the centroid of the shape.

If the centre of mass of the shape is offset from the centre of force (the centroid of the shape),
then the resultant motion of the shape will include a rotation due to the imbalance between these two points.