How Do Ladders in Equilibrium Behave When Hinged and Resting on Rough Surfaces?

[hms.tech]

Homework Statement

The image in the attachment shows two uniform ladders (red and blue) in equilibrium resting on grass(green). The distance between A and B is "L" units.

The two ladders are freely hinged at C.

Because the ladders are uniform their Centers of mass lie at their mid points.

As the ladders are on rough grass we can conclude that friction forces act at A and B. (not shown in the diagram due to lack of space).

the friction at A, F_{A}, is directed towards B

the friction at B, F_{B}, is directed towards A.

The normal reactions at A and B are shown in the diagram.

Homework Equations

these are the rules i learned in my lesson on Equilibrium of forces.

Moment about any point is zero

Vector sum of the forces is zero

The Attempt at a Solution

I don't really think that the first rule is correct because clearly a force will act at the point C (although i don't know the direction or magnitude). According to my intuition a force will definitely act at C where the two ladders are hinged.
And if that is true (which it should be) then there is a net moment about A and about B.

Another thing which confuses me is :
"the line of action of all forces must pass through a common point for the system to be in equilibrium"

I don't understand this point either.
Why is this necessarily true and is it true in this particular case ?

 

[lewando]

I don't really think that the first rule is correct because clearly a force will act at the point C (although i don't know the direction or magnitude). According to my intuition a force will definitely act at C where the two ladders are hinged.
And if that is true (which it should be) then there is a net moment about A and about B.

If you isolate the hinge and draw an FBD, the sum of the forces must equal zero since the hinge is not accelerating.

Another thing which confuses me is :
"the line of action of all forces must pass through a common point for the system to be in equilibrium"

I don't understand this point either.
Why is this necessarily true and is it true in this particular case ?

Maybe this quote should be in some context. A bridge comes to mind as a counterexample. For 2-force systems, it is true-- if it were not, the system would begin to rotate.

 

[hms.tech]

"the line of action of all forces must pass through a common point for the system to be in equilibrium"

what would u say to this example :

The line of action of the combined weight passing through the center of mass of the system (ie the Center of mass of the two rods) coincides at a single unique point with both :
1. the line of action of the Total reaction at A (ie the vector sum of friction and normal at point A)
2. the line of action of the Total reaction at B (ie the vector sum of friction and normal at point B)

Does this hold true ?

 

[lewando]

Another thing which confuses me is :
"the line of action of all forces must pass through a common point for the system to be in equilibrium"

I don't understand this point either.
Why is this necessarily true and is it true in this particular case ?

To clarify, that rule is used to identify if a system is in moment equilibrium, however a system may be in moment equilibrium when the lines of force do notintersect (simple bridge example).

To explore the new example you describe (common center of mass), try doing a graphical analysis. A way to get the angles of the reaction force of the ground is to split the problem into 2 halves by installing a frictionless brick wall between the ladders and treating each side independently.