Show Line of action of suspended rods relative to points on the rods

[Richie Smash]

Homework Statement

The figure shown consists of two rods at right angles to each other and each is uniform of 2m length.
D and E are the midpoints of AC and AD
Draw the line of action of the weight of the arrangement relative to A, B,C, D and E when it is
(a) Suspended from A and in equilibrium
(b) Suspended from D and in equilibrium

(ii) Clearly indicate the position of the centre of gravity of the arrangement relative to A,B,C,D and E

Homework Equations

The Attempt at a Solution

On my diagram I have drawn one big line from A and I believe that would be the line of action?

The arrows in red indicate the lines of actions from each point, however they want the line of action of the entire weight of the arrangement, so clearly the centre of gravity will indicate where this weight is

From there I can draw the line of action I believe, I know that the centre of gravity is in the middle of the uniform object but I only have the centre of Gravity for the two individual planks, I'm not sure how to get the CG for the entire thing.

Also I'm not sure how to drawn the two different lines of actions It is rather confusing.

But actually, in my opinion I would say that the centre of gravity of the entire thing is at point A?

I edited the photo actually, and just have drawn the line of action when it is suspended at A and the centre of gravity, indicated in blue

 

[haruspex]

On my diagram I have drawn one big line from A and I believe that would be the line of action?

Case a), case b), or both?

I would say that the centre of gravity of the entire thing is at point A?

Certainly not.
How do you find the common mass centre of two equal masses if you know their mass centres?

 

[Richie Smash]

The line was for case (a)

You find the common mass centre by adding the mass *length of each plank divided b the sum of their masses?

 

[haruspex]

You find the common mass centre by adding the mass *length of each plank divided b the sum of their masses?

That works in one dimension, but this is two dimensional.
For equal masses, where the individual masses are known, it is quite simple.

 

[Richie Smash]

the centre of gravity is

Xcg= (Mass1*Length from reference point 1+Mass2*Length from reference point 2)/ Mass1+Mass2

How do I picture the lines of action someone help :(

 

[haruspex]

Xcg= (Mass1*Length from reference point 1+Mass2*Length from reference point 2)/ Mass1+Mass2

That is true in one dimension, but it is not at all clear what you mean by that in two dimensions.

Here's a clue: if you want the mass centre of an assemblage, all you need to know is the mass and mass centre of each body in the assemblage. I.e. you can treat each body as a point mass.
Can you apply that to the two rods?

 

[Richie Smash]

OK, well if each body is a mass point, then (an entire rod * the mass centre, +the next entire rod * the mass centre,) /the two entire rods added mass

 

[haruspex]

(an entire rod * the mass centre, +the next entire rod * the mass centre,) /the two entire rods added mass

That is still completely unclear. How is anyone supposed to multiply a rod by its mass centre? It is meaningless.
If you meant "mass of entire rod * position of its mass centre", how are you turning the position into a number?

In co-ordinates, say A=(0,0), B=(0,2), C=(2,0).
Where is the mass centre of AB?
Where is the mass centre of AC?
Where is the mass centre of the two combined?

 

[Richie Smash]

AB is (1,0)
AC is (0,1)

The mass centre of the two is (1/2,1/2)

 

[haruspex]

AB is (0,1)
AC is (1,0)

The mass centre of the two is (1/2,1/2)

Right!
So can you now draw how it would hang from A and from D?

 

[Richie Smash]

This is from A

 

[haruspex]

This is from A

Yes.

 

[Richie Smash]

this is from D

 

[haruspex]

this is from D

Correct.