Calculate Fb Reaction at Point A for 30° Triangle

[fysik]

hello
The below structure (the triangle) is connected to fixed point A that is a hinge (can rotate freely).

Fc force is perpendicular to (ac) and Fb is horizontal. If the (bac) angle is 30 degrees, calculate the reaction at A.

My attempt:

Since Fc is perpendicular to (ac) it doesn't apply any force across the line (ac) so it doesn't create any reaction at a.
Fb creates a reaction at a, since if we analyse it in two axis (the ab axis and the perpendicular to ab) a part of it (across the ab axis) creates a reaction (the same but with different direction) and the perpendicular to (ab) part does not create any reaction at a at all.
But how to find the part of Fb on the (ab) axis?
And is my approach correct?

Thanks

 

[haruspex]

Are you sure this is the entire question? I see no way to determine the angle ab makes to Fb, and the system of forces shown cannot be stable, yet no masses are given. (It would accelerate infinitely fast.)

 

[fysik]

sorry the angle of Fb in regard to bc is 30 degrees

And no masses are given.

Well, the question implies that the structure is in equilibrium, thus the Fc should be in opposite direction. But does that matter? We are talking about instant reaction in a, and I think this is the same regardless the direction of Fc. Am I wrong?

 

[SteamKing]

hello
The below structure (the triangle) is connected to fixed point A that is a hinge (can rotate freely).
View attachment 83203
Fc force is perpendicular to (ac) and Fb is horizontal. If the (bac) angle is 30 degrees, calculate the reaction at A.

My attempt:

Since Fc is perpendicular to (ac) it doesn't apply any force across the line (ac) so it doesn't create any reaction at a.

Are you sure about this?

It's like saying if I step on a bathroom scale, I should weigh nothing because my weight is acting perpendicular to the scale.

Fb creates a reaction at a, since if we analyse it in two axis (the ab axis and the perpendicular to ab) a part of it (across the ab axis) creates a reaction (the same but with different direction) and the perpendicular to (ab) part does not create any reaction at a at all.

Forces don't magically vanish. A force can be opposed by another force of equal magnitude applied in the opposite direction; then the net force is zero.

But how to find the part of Fb on the (ab) axis?

You don't need to do this. See the discussion about Fb and Fc above.

Remember, forces are vector quantities. The net force is the vector sum of the component forces.

 

[fysik]

no, I am saying if you place the bathroom scale vertically and step on it, you will weight nothing

 

[fysik]

what force does the Fc create on A?
it only rotates the rod, it doesn't presses it!

are you going to drive me crazy? what force does Fc exerts on A?
this is madness!

the A is the point of rotation for the force Fc so there is no movement on it! there is no force applied at that point! are you serious that Fc exerts force on A?

 

[fysik]

ok, so you mean that the Fc will create the same Fc in A, in the opposite direction and parallel axis passing from A?

 

[SammyS]

what force does the Fc create on A?
it only rotates the rod, it doesn't presses it!

are you going to drive me crazy? what force does Fc exerts on A?
this is madness!

the A is the point of rotation for the force Fc so there is no movement on it! there is no force applied at that point! are you serious that Fc exerts force on A?

Are you talking to yourself?

If you are responding to some particular post, or some portion of a post, it's helpful to use the "reply" feature, or the "quote" feature.

 

[fysik]

Are you talking to yourself?

If you are responding to some particular post, or some portion of a post, it's helpful to use the "reply" feature, or the "quote" feature.

so each of the forces are transferred into A intact? ie. with the same magnitude and the same direction?
can you produce a sketch that shows how each of Fb, Fc are transferred on A?

 

[SteamKing]

so each of the forces are transferred into A intact? ie. with the same magnitude and the same direction?
can you produce a sketch that shows how each of Fb, Fc are transferred on A?

Rather than the bathroom scale, perhaps a seesaw is a better illustration:

​

The blue seesaw beam is clearly able to rotate about the fulcrum (the blue triangle), and if the masses resting on the seesaw beam are evenly balanced, there will be no rotation clockwise or counterclockwise. The fulcrum point is also supporting the weight of the red barrels and the green boxes (not to mention the weight of the beam itself), otherwise, the whole apparatus would be moving in some direction.

That's what we mean when we say something is in static equilibrium: there is zero net force and zero net moment acting on the system.

 

[fysik]

I am embarrassed now
but draw for me please the forces acting on A

 

[SteamKing]

I am embarrassed now
but draw for me please the forces acting on A

That's the thing. The forces acting on A are already drawn. All you have to do is transfer each force, Fb and Fc to point A. The magnitude and direction of each force will be the same, regardless of the location of each force on the diagram.

 

[fysik]

yes, I am very embarrassed now
why I couldn't think of it?
to be honest, I imagined that this could be the case
but the fact that the point A is not moving, made me think that there is no force acting on it
so if the A wasn't fixed, each of the AB or AC rods (if they weren't connected at A) would rotate with axis of rotation their middle point?
and in case of AB, there would be also some horizontal movement along with the rotation? but in AC there would be only rotation and no movement?

 

[fysik]

anyone?

 

[SteamKing]

yes, I am very embarrassed now
why I couldn't think of it?
to be honest, I imagined that this could be the case
but the fact that the point A is not moving, made me think that there is no force acting on it
so if the A wasn't fixed, each of the AB or AC rods (if they weren't connected at A) would rotate with axis of rotation their middle point?
and in case of AB, there would be also some horizontal movement along with the rotation? but in AC there would be only rotation and no movement?

Just because something isn't moving doesn't necessarily imply that no force is acting on it. There may be a restraint against any motion taking place in response to any forces which may be applied.

If you go up to a stone wall and push on it with both hands, does the wall move? If the wall doesn't move, does this mean that your hands (and by extension, you) are not applying any force to the wall?

In the mechanism shown in the OP, any linear motion was restrained by the pin at point A, but the mechanism was free to rotate about this point. If we analyze the mechanism as a free body, because the mechanism isn't free to translate (but can rotate), then the equations of static equilibrium require that no net force be acting on the mechanism. If no net force is acting on the mechanism, but there are forces applied to it, then this implies that there is a force developed at pin A which is equal in magnitude and opposite in direction to the sum of all the other forces acting on the mechanism. This force at pin A is called the reaction to the other applied forces.

 

[fysik]

if we have a stick AC on its own (not fixed or connected at all) and a force like Fc is acting, what will happen to that stick?
I imagine it will rotate by its center and not translate?

also, the same with stick AB and Fb, what will happen?
I imagine it will rotate and translate at the same time?

 

[haruspex]

if we have a stick AC on its own (not fixed or connected at all) and a force like Fc is acting, what will happen to that stick?
I imagine it will rotate by its center and not translate?

No. Assuming it has some mass, it will translate according to the standard equation, F=ma. Because the force is not through the mass centre it will also rotate.
If it has no mass then it is physically impossible to apply a nonzero force to it.

 

[fysik]

No. Assuming it has some mass, it will translate according to the standard equation, F=ma. Because the force is not through the mass centre it will also rotate.
If it has no mass then it is physically impossible to apply a nonzero force to it.

are you sure it will translate? I think it should rotate only! because the force is perpendicular to the axis that connects the mass center and the point of action of the force Fc!

 

[SammyS]

are you sure it will translate? I think it should rotate only! because the force is perpendicular to the axis that connects the mass center and the point of action of the force Fc!

The object will rotate about point a, but if the center of mass is located anyplace besides at a, then it also translates. What path does the center of mass follow ?

 

[fysik]

why will it rotate around a? I told you that it is not connected at a or any other point at all!
plus the center of mass is obviously the center of it
why would it rotate around a?
also why would it translate?

 

[SammyS]

why will it rotate around a? I told you that it is not connected at a or any other point at all!
plus the center of mass is obviously the center of it
why would it rotate around a?
also why would it translate?

Are we working on the original problem which was as follows?

hello
The below structure (the triangle) is connected to fixed point A that is a hinge (can rotate freely).

Fc force is perpendicular to (ac) and Fb is horizontal. If the (bac) angle is 30 degrees, calculate the reaction at A.

It clearly says.

" fixed point A that is a hinge (can rotate freely) "

I take it to mean that A and a are used interchangeably.

 

[fysik]

I think we solved that problem and proceeded with this one:
https://www.physicsforums.com/threads/vectors-of-forces.812763/#post-5103198

 

[haruspex]

why will it rotate around a? I told you that it is not connected at a or any other point at all!
plus the center of mass is obviously the center of it
why would it rotate around a?
also why would it translate?

If you have a uniform rod AB, mass m, length 2L (either drifting in space or lying on a smooth table, say), and apply a force F at B at right angles to the rod, you will have the linear force equation F=ma and the torque equation $FL=mL^2\alpha/3$. There will be translation and rotation.