Re: Apocalypse NOW!

<assorted insanity snipped>

> Now I am going to explain again what I am trying to say.
>
> In above figure AB and CB are V-shaped spring of same length and
> stiffness and both springs are in relaxed state initially. Angle ABC
> is "solid angle" . Let angle ABC be 60 degree in this figure(for the
> sake of explanation only, in actual Action Device this angle will be
> very small).

I'm assuming that by "solid angle", this angle is somehow made rigid.
Slightly odd, but it can be done. You could make each of the members AB and
CB out of two concentric tubes, able to slide in and out, with an internal
spring. Is that kind of what you have in mind?

>
> At t = 0, we apply same magnitude of force on point A and C and we
> pull point A of spring AB towards point E in the direction of line BE
> which makes "60 degree" angle with X axis and we pull point C of
> spring CB towards point F in the direction of line BF which also makes
> "60 degree" angle with X axis. And we are pulling point A and C in
> such a way that these points must stretch or extend to point E and F
> resp. on x axis.
>
> Please note that we are pulling points A and C in the direction which
> makes 60 degree angle with X axis. We are NOT pulling point A and C in
> "downward" direction.
>

But you are, as everybody has tried to point out ad nauseam. Using your
cartesian system, and some notation I introduced in an earlier post, the
force exterted by the springs AB and CB on the point B can be decomposed
into an {x,y} vector as:

{ -F sin(alpha), -F cos(alpha) } and
{  F sin(alpha), -F cos(alpha) }

respectively. The two springs exert no net force along the x axis on B,
but -2Fcos(alpha) along the y axis (the minus sign indicates that the force
is towards the origin, or "downwards" in your chosen orientation.

> At t = t, spring AB is stretched and point A of spring AB reaches to
> point E on X axis. Also at t = t, spring CB is stretched and point C
> of spring reaches to point F on X axis.
>

yes, we got that.

> We will find that point B has not shifted its position along Y axis.
> It remains where it was at t = 0. Because if point B shifts its
> position along Y axis, to say, point B', angle AB'C (or EB'F, because
> point A coincides with point E and point C coincides with point F)
> will be different from angle ABC i.e. greater than 60 degree. But as
> stated above, "angle ABC is solid angle" which does not change due to
> forces acting at point B.

this implies that in order to maintain your "solid angle" (which seems to be
validated by your explanation to mean "rigid") there is a torque exterted at
B on each of the members to maintain that.
Nonetheless, just because point B is constrained not to move in your frame
of reference, doesn't mean the force goes away. That was what I was talking
about before. The fact that B doesn't move relative to E and F just means
that you have redefined the boundary of your closed system. You are relying
on an external force at E and F to prevent *the whole thing moving*.

When you throw in the other components of your device needed to complete the
closed system with no external forces applied, it all exactly cancels out.

>
> Yes, point B will shift its position in space but certainly not along
> Y axis or in XY plane. Third vector will be produced at point B
> direction of which will be perpendiculer to XY plane.
>
> Am I right or wrong?

wrong. Where is a force perpendicular to the XY plane in your explanation?

Krill

>
> -Abhi.