Where is my mistake in this classical Mechanical model?
gu...@hotmail.com
and y_vertical directions seems inacurate?
Example:
If the slope was on fricitionless ice, the slope and block would NOT
move horizontaly, thus technically wouldn't the x_horizontal force = 0
since ***INSTEAD*** if one uses a finger to simulate the same
x_horizontal force then the block and slope would indeed both move on
the ice ?? .. ??
Sue...
"Identifying Action and Reaction Force Pairs"
http://www.glenbrook.k12.il.us/GBSSCI/PHYS/Class/newtlaws/u2l4b.html
Sue...
Sorcerer
"Sue..." <suzyse...@yahoo.com.au> wrote in message news:1168007538.6...@51g2000cwl.googlegroups.com...
[...]
Hey Dennis! I've never seen an aether, have you?
http://www.quackwatch.org/01QuackeryRelatedTopics/pseudo.html
Russell
> The Classic block on a slope demonstrates forces in the x_horizontal
> and y_vertical directions seems inacurate?
You should have posted this to sci.physics.
And, you should give more details. There are many "classic"
block on slope problems; it's not clear which one you are
talking about, or what bodies your forces are supposed to be
acting on, etc.
But, I will try to guess.
Assuming that you mean a block sitting on a slope
such that friction prevents motion of the block, then
of course the resultant force on the block in the
horizontal direction (and vertical direction too) is zero.
How could it be otherwise, if neither the block nor the
slope moves?
But, it is still useful to analyze this zero resultant as
the sum of nonzero vectors: (1) the vertical pull of gravity,
(2) the frictional force at angle a from horizontal, and
(3) the normal force of slope pushing on the block at
a right angle to the frictional force.
Force 1 has no horizontal component, but forces 2 and
3 both *do*, and these components are equal and opposite.
I assume that you are objecting to some textbook's
treatment of this. Did you forget that you must consider
*both* the frictional force and the normal force? You
won't get balancing horizontal components if you consider
only friction. Maybe that was your blunder. Or perhaps,
you have a bad textbook, but I would suggest you reread
it carefully before coming to that conclusion.
To complete the analysis, forces 2 and 3 also both have
vertical components, and the sum of these vertical
components is equal and opposite to force 1.
>
> Example:
>
>
> If the slope was on fricitionless ice, the slope and block would NOT
> move horizontaly, thus technically wouldn't the x_horizontal force = 0
Yes, as I said, the resultant horizontal component of
force is zero.
> since ***INSTEAD*** if one uses a finger to simulate the same
> x_horizontal force then the block and slope would indeed both move on
> the ice ?? .. ??
If you try to "simulate" only part of the force, not the
sum of all the forces, then of course you change the
problem.
Mike
gu...@hotmail.com wrote:
> The Classic block on a slope demonstrates forces in the x_horizontal
> and y_vertical directions seems inacurate?
What forces act in the x_horizontal direction? There is only a gravity
force acting on the block in the vertical direction (y for this
problem).
> Example:
>
>
> If the slope was on fricitionless ice, the slope and block would NOT
> move horizontaly, thus technically wouldn't the x_horizontal force = 0
> since ***INSTEAD*** if one uses a finger to simulate the same
> x_horizontal force then the block and slope would indeed both move on
> the ice ?? .. ??
No, you confuse the different coordinate system used. Often, a
coordinate system with x-direction in the direction of the slope is
used and the gravity force is decomposed in the two directions, one
along the x' direction and another in the -y' direction. Only the
component in the x' direction is responsible for the change in kinetic
energy as th eother one is orthogonal to the path and does no work.
Mike
gu...@hotmail.com
ok so how come Fx = M_block x g x cos (slope_angle) doesn't move the
slope on the icy surface where as if we substitute this Fx force
exerted by the block on the slope with a finger or spring, only then
the slope would move on the ice?
A spring is also static and only becomes a kinetic force once the slope
begins to move on the icy surface...?
gu...@hotmail.com
why isn't it zero if we simply substitute Fx exerted by the block on
the slope with a finger or a spring?
> > since ***INSTEAD*** if one uses a finger to simulate the same
> > x_horizontal force then the block and slope would indeed both move on
> > the ice ?? .. ??
>
> If you try to "simulate" only part of the force, not the
> sum of all the forces, then of course you change the
> problem.
The sum of all the forces exerted on the slope are below here:
Part A:
The block does NOT slide down the slope (friction)....instead it is
***BOTH*** the block and slope which the block is on which are both on
a frictionless ice surface
The block I believe exerts a force on the slope of:
Fx= M_block x g x cos (slope_anlge)
Fy = M_block x g x sin (slope_angle)
********************************************************************************
Part B:
So if one replaces Fx and Fy exerted by the block instead by two
fingers (or springs) of the same two vertical + horizontal forces the
block and slope would NOW slide on the ice
So what is the reason why the forces in Part A do not move the slope
and block on the ice .....where as Part B does move the block on the
ice??
I would've said it simply kinetic pressure versus static gravity....
but the springs (or fingers) only become kinetic once the slope and its
block start to move on the ice?
Mike
That is a different kind of question, i.e. whether the slope will move
or not. I only replied on the coordinates issue before. The motion of
the slope/block system has to do with conservation of momentum. When
the block slides down the slope the center of mass of the system
changes and the parts move so that momentum is conserved according to:
F12 = -F21 which is Newton's third law, and F12 is the force exerted vy
the block on the slope and F21 is the force exerted by the slope on the
block and these forces are equalk in the absence of an external force.
Or:
d(MsVs)/dt = -d(MbVb)/dt ----> MsVs+MbVb = constant, where Ms is the
mass od the slope, Vs the velocity of the slope, Mb the mass of the
block and Vb the velocity of the block.
Note that the above equation is vectorial, Vs and Vb are velocities. It
is also a non-linear problem since F12 - -F21 is dependent on nthe
position of the block on the slope.
In addition, to solve the problem you must consider other conditions
such as conservation of mechanical energy to solve the problem.
>
> A spring is also static and only becomes a kinetic force once the slope
> begins to move on the icy surface...?
I do not uinderstand what you want to say but an externally applied
force is a different problem and momentum is not conserved. In such
case it is better to use Lagrangian Mechanics and derive the
Lagrange-Euler equation sof motion with a forcing term.
Mike
Russell
But Fx is zero! Not much to substitute.
>
> > > since ***INSTEAD*** if one uses a finger to simulate the same
> > > x_horizontal force then the block and slope would indeed both move on
> > > the ice ?? .. ??
> >
> > If you try to "simulate" only part of the force, not the
> > sum of all the forces, then of course you change the
> > problem.
>
> The sum of all the forces exerted on the slope are below here:
>
> Part A:
>
>
> The block does NOT slide down the slope (friction)....instead it is
> ***BOTH*** the block and slope which the block is on which are both on
> a frictionless ice surface
>
>
> The block I believe exerts a force on the slope of:
>
>
> Fx= M_block x g x cos (slope_anlge)
> Fy = M_block x g x sin (slope_angle)
Why do you believe that? I'm having a hard time
seeing what diagram you might have drawn to
erroneously arrive at those figures. The correct
results (assuming x horizontal and y vertical) are:
Fx = 0
Fy = Mg, the full weight of the block.
Consider: if you were carrying the slope-and-block
system in your hands, would you expect you could
lighten your load just by rotating the slope to change
the angle? Obviously not, as long as the block stayed
put of course. So your angle-dependent Fy *can't* be
right.
Similarly, your intuition about Fx is actually correct --
it can't be nonzero, or the slope-and-block system
would try to move horizontally out of your hands (or
sideways on the ice). Fx=0 has to be right.
>
>
> ********************************************************************************
>
>
>
> Part B:
>
>
> So if one replaces Fx and Fy exerted by the block instead by two
> fingers (or springs) of the same two vertical + horizontal forces the
> block and slope would NOW slide on the ice
Sure, if you used *your* Fx and Fy. But they're wrong.
>
>
> So what is the reason why the forces in Part A do not move the slope
> and block on the ice .....where as Part B does move the block on the
> ice??
If Fx and Fy were really as you say, then the slope-and-
block system *would* move in Part A. That should have
been a big hint that your calculation was wrong.
Tom Roberts
> The Classic block on a slope demonstrates forces in the x_horizontal
> and y_vertical directions seems inacurate?
Well your description is most definitely insufficient! And your
conclusion is most definitely inaccurate....
> If the slope was on fricitionless ice, the slope and block would NOT
> move horizontaly,
Not true.
I assume that the slope is an independent wedge resting on frictionless
ice that is perfectly horizontal, and that the block rests on its
inclined upper surface which is also made of frictionless ice. Gravity
is present. Initially both block and wedge are held unmoving on the ice,
then released simultaneously.
Given those assumptions, when released the block will descend down the
inclined slope of the wedge, pushing it to the side; the wedge and block
will move horizontally apart such that the total center of mass remains
unmoving. It is not possible to do this with the block moving down a
vertical line -- it must move both vertically and horizontally.
> thus technically wouldn't the x_horizontal force = 0
No. After the block reaches the horizontal ice, assuming nothing
untoward happens, the block and wedge will continue moving apart,
forever (obviously an idealization).
> since ***INSTEAD*** if one uses a finger to simulate the same
> x_horizontal force then the block and slope would indeed both move on
> the ice ?? .. ??
If you change the problem the results change. But indeed, both block and
slope move on the ice for your original configuration.
Tom Roberts
Russell
> gu...@hotmail.com wrote:
> > The Classic block on a slope demonstrates forces in the x_horizontal
> > and y_vertical directions seems inacurate?
>
> Well your description is most definitely insufficient! And your
> conclusion is most definitely inaccurate....
Amen to both of those. Especially the first.
>
>
> > If the slope was on fricitionless ice, the slope and block would NOT
> > move horizontaly,
>
> Not true.
But it's true *if* there is sufficient friction between the
block and the slope.
>
> I assume that the slope is an independent wedge resting on frictionless
> ice that is perfectly horizontal, and that the block rests on its
> inclined upper surface which is also made of frictionless ice.
The OP stated (elsewhere) that there is friction between
the block and the slope. The most thorough way to treat
the problem is to generalize it, but -- as has happened
before in this ng -- it seems you have treated one extreme
case and I've treated the other.
(Of course if the block moves at all, the slope will also
move on the ice.)
Gravity
> is present. Initially both block and wedge are held unmoving on the ice,
> then released simultaneously.
>
> Given those assumptions, when released the block will descend down the
> inclined slope of the wedge, pushing it to the side; the wedge and block
> will move horizontally apart such that the total center of mass remains
> unmoving. It is not possible to do this with the block moving down a
> vertical line -- it must move both vertically and horizontally.
You're absolutely right about this, and I'll add, this
is closer to what I'd call the "classic" block on slope
problem than the one I treated. Perhaps even, this
is the problem the OP *should* have asked about.
But, my interpretation of his several posts is that he
didn't.
>
>
> > thus technically wouldn't the x_horizontal force = 0
>
> No. After the block reaches the horizontal ice, assuming nothing
> untoward happens, the block and wedge will continue moving apart,
> forever (obviously an idealization).
Yes, though of course by then, the force really *is* zero.
gu...@hotmail.com
I'm pretty sure Fx is not zero, there is both a vertical and horizonal
force exerted ON THE SLOPE by the block.
Example:
a 45 degree slope will feel MORE HORIZONTAL force by the block then a
88 degree slope, as well and 88 degree slope will feel very LITTLE
vertical force from the block.
gu...@hotmail.com
Good reply, so there is an Fx and Fy force exerted by the block, but
lets say the slope wasn't made of ice and it prevented the block from
sliding down, either way Fx and Fy would still be the same value as
before, it's still not
clear what is the exact logic that if I replaces those two forces
(Fx,Fy) by a two spring or finger Fx & Fy forces only then the block
would move on the ice (since the force exerted by the block is not a
resultant force, only the slope's force is)?
Russell
Yes, but I explained to you in some detail in my *first*
reply to you that this horizontal force is zero in the case
where there is enough friction to prevent movement of
the block.
Remove that friction, and yes, both the block and the
slope will move, as Tom Roberts explained.
> Example:
>
> a 45 degree slope will feel MORE HORIZONTAL force by the block then a
> 88 degree slope, as well and 88 degree slope will feel very LITTLE
> vertical force from the block.
Given sufficient friction (and that would be hard to
accomplish at 88 degrees) the block still wouldn't
move. The vertical component of the friction would
be large and the horizontal component of it would
be small, and these components would be positive
in a direction *opposed* to the components of
force that you describe above. Net force both
vertically and horizontally would be zero, as it must
be (since we are saying the block doesn't move).
gu...@hotmail.com
Ok but if the block was static, its Fx force exerted by the block on
the slope IS NOT a resultant force (only the reactive force of the
slope is a resultant force): meaning if I remove the block then as well
the Fx force of the slope is zero (doesn't remain the same) BECAUSE
it's a resultant force.
What I'm getting at, lets say the slope was BORN BLIND (no eyes):
It feels both an Fx and Fy force so much so that it might break
exactly according the value of Fx versus the strength of it's surface.
Now SIMPLY by replacing the block and providing the EXACT same value of
Fx and Fy forces (two springs or two fingers for Fx and Fy) then the
blind slope would move on the ice (the slope would say hey I'm blind
but the same VALUE of force I felt on me as before for some reason this
time is making me move on the ice but NOT BEFORE)...and as I said prior
the force of the block is NOT a RESULTANT force???
Russell
> >
>
[context? please learn some posting netiquette]
> Ok but if the block was static, its Fx force exerted by the block on
> the slope IS NOT a resultant force (only the reactive force of the
> slope is a resultant force): meaning if I remove the block then as well
> the Fx force of the slope is zero (doesn't remain the same) BECAUSE
> it's a resultant force.
I couldn't parse your sentence above. I have told you
repeatedly that the resultant force is zero, so the reactive
force is also zero and your point (if I am guessing it right)
is moot.
*What* is holding the block so that it is "static", as you
describe it? Friction. Is there some section in your
textbook that explains that friction is a force? Please
read it.
You are trying to analyze this problem as if friction
didn't have to be included in the sum of forces acting
on the block. This is why you always come up with
the wrong answer.