| Topic: |
Science > Physics |
| User: |
"will" |
| Date: |
04 Nov 2006 04:30:02 AM |
| Object: |
freshmen physics problems |
Hi, I have two poblems from freshman Physics that I am puzzling over.
If anyone has insight into these I would be grateful to hear from you.
First, in a problem concerning circlular motion, a car moving in a
circular non-banked path. It was said in Sears and Zemansky that since
there is no motion in the radial direction (because the car in moving
uniformly in a circle) the frictional force that is required to keep
the car in a circular path is static and not kinetic friction. That
seems like an odd way to see it. I understand that the force is being
applied in the radial direction and since their is no change in radius
that we can say that the (frictional) force must be acting in the
radial direction where there is no movement but still it seems strange.
I guess the best way to resolve the issue would be if it were possible
to measure the frictional force and see if it is closer to that gotten
when a static coefficient is used as compared to that which is gotten
when a kinetic coefficient is used. Has this been done? Is there a
clearer way to see this?
The second question is with regard to apparent weight in a moving
elevator. A 50 kg woman stands on a bathroom scale while riding in the
elevator. The elevator is originally moving downward at 10 m/s. It
slows to a stop with constant acceleration in a distance of 25 m. What
is the reading on the scale during the acceleration. We can do this
problem by looking at the normal force exerted by the scale on the
woman since it is equal in magnitude to the force the woman exerts on
the scale by the third law. In that case the second law provides us
with the fact that the sum of the vertical forces exerted on the woman
are the normal force plus -mg and that this sum is equal to ma. We can
then solve for n since the other values are known.
But it is not necessarily intuitive to focus on the forces on the
woman. It seems more intuitive to focus on the scale and the force(s)
acting on it. In that case we have the forces acting on the scale are
the woman's weight , whose magnitude is again -mg. But at this point I
am confused about what to do from here.
Thanks for any help.
.
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| User: "Martin Hogbin" |
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| Title: Re: freshmen physics problems |
04 Nov 2006 03:43:09 PM |
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"will" <uujou@yahoo.com> wrote in message news:1162636201.881113.76780@h48g2000cwc.googlegroups.com...
Hi, I have two poblems from freshman Physics that I am puzzling over.
If anyone has insight into these I would be grateful to hear from you.
First, in a problem concerning circlular motion, a car moving in a
circular non-banked path. It was said in Sears and Zemansky that since
there is no motion in the radial direction (because the car in moving
uniformly in a circle) the frictional force that is required to keep
the car in a circular path is static and not kinetic friction. That
seems like an odd way to see it. I understand that the force is being
applied in the radial direction and since their is no change in radius
that we can say that the (frictional) force must be acting in the
radial direction where there is no movement but still it seems strange.
I think the point being made is that the tyres ar rolling on the road rather
than slipping thus the rubber does not move with respect to the road
and the stsic rather than dynamic coefficient of friction is applicable.
Neither is that well defined
The second question is with regard to apparent weight in a moving
elevator. A 50 kg woman stands on a bathroom scale while riding in the
elevator. The elevator is originally moving downward at 10 m/s. It
slows to a stop with constant acceleration in a distance of 25 m. What
is the reading on the scale during the acceleration. We can do this
problem by looking at the normal force exerted by the scale on the
woman since it is equal in magnitude to the force the woman exerts on
the scale by the third law. In that case the second law provides us
with the fact that the sum of the vertical forces exerted on the woman
are the normal force plus -mg and that this sum is equal to ma. We can
then solve for n since the other values are known.
But it is not necessarily intuitive to focus on the forces on the
woman. It seems more intuitive to focus on the scale and the force(s)
acting on it. In that case we have the forces acting on the scale are
the woman's weight , whose magnitude is again -mg. But at this point I
am confused about what to do from here.
The woman's weight does not act directly on the scales it is the reaction
force to the force that the scales exert on the woman that acts on the
scales. You have just calculated this.
Martin Hogbin
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| User: "John C. Polasek" |
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| Title: Re: freshmen physics problems |
05 Nov 2006 10:11:00 PM |
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On 4 Nov 2006 02:30:02 -0800, "will" <uujou@yahoo.com> wrote:
Hi, I have two poblems from freshman Physics that I am puzzling over.
If anyone has insight into these I would be grateful to hear from you.
First, in a problem concerning circlular motion, a car moving in a
circular non-banked path. It was said in Sears and Zemansky that since
there is no motion in the radial direction (because the car in moving
uniformly in a circle) the frictional force that is required to keep
the car in a circular path is static and not kinetic friction. That
seems like an odd way to see it. I understand that the force is being
applied in the radial direction and since their is no change in radius
that we can say that the (frictional) force must be acting in the
radial direction where there is no movement but still it seems strange.
I guess the best way to resolve the issue would be if it were possible
to measure the frictional force and see if it is closer to that gotten
when a static coefficient is used as compared to that which is gotten
when a kinetic coefficient is used. Has this been done? Is there a
clearer way to see this?
At first glance, frictional force equals the centrifugal force for any
force up to the breakaway force which seems the same as being held in
by a cable. But in fact for a rubber tire, the part on the road is
flexed inward, and as it rotates out of contact, resumes its natural
dispositon. So the car would creep out to a larger radius if the
wheel were maintained tangent to the arc. In actual fact the wheel
has to be crabbed in a little to overcome this. Therefore, to maintain
a constant arc, the wheel will be doing a bit of cutting like the
tool on a lathe. Meanwhile, the tire will get war from this cycling.
The second question is with regard to apparent weight in a moving
elevator. A 50 kg woman stands on a bathroom scale while riding in the
elevator. The elevator is originally moving downward at 10 m/s. It
slows to a stop with constant acceleration in a distance of 25 m. What
is the reading on the scale during the acceleration. We can do this
problem by looking at the normal force exerted by the scale on the
woman since it is equal in magnitude to the force the woman exerts on
the scale by the third law. In that case the second law provides us
with the fact that the sum of the vertical forces exerted on the woman
are the normal force plus -mg and that this sum is equal to ma. We can
then solve for n since the other values are known.
But it is not necessarily intuitive to focus on the forces on the
woman. It seems more intuitive to focus on the scale and the force(s)
acting on it. In that case we have the forces acting on the scale are
the woman's weight , whose magnitude is again -mg. But at this point I
am confused about what to do from here.
Thanks for any help.
The car slows from 10m/s in 25m. The acceleration can be computed from
V = sqrt(2AX)
where V =10 and X = 25
100 = 2A*25
A = 2 m/ss or .2 gee
The woman would register 20% heavier.
John Polasek
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| User: "" |
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| Title: Re: freshmen physics problems |
06 Nov 2006 03:33:44 AM |
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John C. Polasek wrote:
At first glance, frictional force equals the centrifugal force for any
force up to the breakaway force which seems the same as being held in
by a cable. But in fact for a rubber tire, the part on the road is
flexed inward, and as it rotates out of contact, resumes its natural
dispositon.
....so the lateral force on the tire is stored within the tire as
elastic deformation, *assuming* the turn is not banked. The force is
released when the vehicle emerges from the turn.
So the car would creep out to a larger radius if the
wheel were maintained tangent to the arc. In actual fact the wheel
has to be crabbed in a little to overcome this.
You are now discussing the vagaries of wheel articulation and
suspension systems - a bit beyond the scope of the OP I believe. I
once examined a custom built automobile in which the drive wheels were
'amidships' on the left and right sides, and single wheels in the front
and back were used to steer. No 'crabbing' of the tires on the turn and
a *great* turn radius.
Therefore, to maintain
a constant arc, the wheel will be doing a bit of cutting like the
tool on a lathe.
Only the articulated wheels that do the steering. Examine the tracks
left by a standard four-wheeler in dirt or mud sometime. You can
easily tell which marks were made by the front wheels and which by the
back.
Meanwhile, the tire will get war from this cycling.
"wear"?
Tom Davidson
Richmond, VA
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| User: "John C. Polasek" |
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| Title: Re: freshmen physics problems |
06 Nov 2006 08:45:45 AM |
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On 6 Nov 2006 01:33:44 -0800, "tadchem@comcast.net"
<tadchem@comcast.net> wrote:
John C. Polasek wrote:
At first glance, frictional force equals the centrifugal force for any
force up to the breakaway force which seems the same as being held in
by a cable. But in fact for a rubber tire, the part on the road is
flexed inward, and as it rotates out of contact, resumes its natural
dispositon.
...so the lateral force on the tire is stored within the tire as
elastic deformation, *assuming* the turn is not banked. The force is
released when the vehicle emerges from the turn.
No, it is a continuous process.
The contact with the road is an oval patch and the cycling occurs as
each patch is continually replaced by the next fresh unstressed piece
of rubber ready to take up the load and deflect. It restores as it
leaves the work zone.
So the car would creep out to a larger radius if the
wheel were maintained tangent to the arc. In actual fact the wheel
has to be crabbed in a little to overcome this.
The wheel has to be crabbed by the angle A = dR/R so instead of the
wheel plane normal to the radius, you have to cut it in by A. You can
see this from a diagram.
Imagine we are set for radius R and are turning and there is zero
centrifugal force.
Now turn on the centrifugal force, and the tire rubber will deflect
outward by dR so it's now on a radius R + dR. Now draw the line of the
wheel plane from R + dR, and angle it inward so it's now tangent to
your original arc of R.
If you take the trouble to draw this you will see a big right triangle
split into two smaller ones and the angle is A = dR/R.
This would be true even with iron tires because everything, without
exception, has a compliance given by 1/Youngs.
You are now discussing the vagaries of wheel articulation and
suspension systems - a bit beyond the scope of the OP I believe. I
once examined a custom built automobile in which the drive wheels were
'amidships' on the left and right sides, and single wheels in the front
and back were used to steer. No 'crabbing' of the tires on the turn and
a *great* turn radius.
Therefore, to maintain
a constant arc, the wheel will be doing a bit of cutting like the
tool on a lathe.
Only the articulated wheels that do the steering. Examine the tracks
left by a standard four-wheeler in dirt or mud sometime. You can
easily tell which marks were made by the front wheels and which by the
back.
Meanwhile, the tire will get war from this cycling.
"wear"?
warm
Tom Davidson
Richmond, VA
John Polasek
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| User: "PD" |
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| Title: Re: freshmen physics problems |
05 Nov 2006 10:45:48 AM |
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On Nov 4, 4:30 am, "will" <u...@yahoo.com> wrote:
Hi, I have two poblems from freshman Physics that I am puzzling over.
If anyone has insight into these I would be grateful to hear from you.
First, in a problem concerning circlular motion, a car moving in a
circular non-banked path. It was said in Sears and Zemansky that since
there is no motion in the radial direction (because the car in moving
uniformly in a circle) the frictional force that is required to keep
the car in a circular path is static and not kinetic friction. That
seems like an odd way to see it. I understand that the force is being
applied in the radial direction and since their is no change in radius
that we can say that the (frictional) force must be acting in the
radial direction where there is no movement but still it seems strange.
I guess the best way to resolve the issue would be if it were possible
to measure the frictional force and see if it is closer to that gotten
when a static coefficient is used as compared to that which is gotten
when a kinetic coefficient is used. Has this been done? Is there a
clearer way to see this?
Yes, there is a simple way to see this.
Kinetic friction is when two surfaces slide over each other -- a
slipping action. When a tire rolls, it doesn't slip on the road. The
bottom surface of the tire meets the road but does not slip on it.
When a tire *skids* (as when you stomp on the brakes and lock the
wheels), that's kinetic friction. And since kinetic friction is
typically lower than what you can accomplish with static friction, this
is good motivation for anti-lock brakes.
The second question is with regard to apparent weight in a moving
elevator. A 50 kg woman stands on a bathroom scale while riding in the
elevator. The elevator is originally moving downward at 10 m/s. It
slows to a stop with constant acceleration in a distance of 25 m. What
is the reading on the scale during the acceleration. We can do this
problem by looking at the normal force exerted by the scale on the
woman since it is equal in magnitude to the force the woman exerts on
the scale by the third law. In that case the second law provides us
with the fact that the sum of the vertical forces exerted on the woman
are the normal force plus -mg and that this sum is equal to ma. We can
then solve for n since the other values are known.
But it is not necessarily intuitive to focus on the forces on the
woman. It seems more intuitive to focus on the scale and the force(s)
acting on it. In that case we have the forces acting on the scale are
the woman's weight , whose magnitude is again -mg. But at this point I
am confused about what to do from here.
The question is what the scale *reads*. What the scale reads is the
force it has to apply to the object that's sitting on it. If, for
example, the scale operates according to Hooke's law, then there is
some deflection or compression that is proportional to the force the
scale exerts, and it is this deflection or compression that the scale
reports (though the numbers on the scale are calibrated to force units,
not deflection units).
When you gently stand on a scale, it tells you what force it has to
provide to keep you from penetrating it. In this case, the force it
provides happens to equal your weight, since by doing so, it puts you
in equilibrium.
When you jump on a scale from a small stool, however, the scale has to
provide more force because it also has to slow you down. This is why
the needle jumps so sharply. You don't weigh anything more, but in this
scale the force the scale provides is not equal to your weight.
Thanks for any help.
.
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| User: "Herman Trivilino" |
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| Title: Re: freshmen physics problems |
12 Nov 2006 02:21:16 PM |
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On Nov 4, 4:30 am, "will" <u...@yahoo.com> wrote:
Hi, I have two poblems from freshman Physics that I am puzzling over.
If anyone has insight into these I would be grateful to hear from you.
First, in a problem concerning circlular motion, a car moving in a
circular non-banked path. It was said in Sears and Zemansky that since
there is no motion in the radial direction (because the car in moving
uniformly in a circle) the frictional force that is required to keep
the car in a circular path is static and not kinetic friction. That
seems like an odd way to see it. I understand that the force is being
applied in the radial direction and since their is no change in radius
that we can say that the (frictional) force must be acting in the
radial direction where there is no movement but still it seems strange.
I guess the best way to resolve the issue would be if it were possible
to measure the frictional force and see if it is closer to that gotten
when a static coefficient is used as compared to that which is gotten
when a kinetic coefficient is used. Has this been done? Is there a
clearer way to see this?
The tires have to slide if the car moves in a radial direction. As long as
the car maintains its circular motion, the tires won't slide. You can see
this by examining the tracks left by the car. You get an imprint of the
tire's tread pattern when the car doesn't slide.
The second question is with regard to apparent weight in a moving
elevator. A 50 kg woman stands on a bathroom scale while riding in the
elevator. The elevator is originally moving downward at 10 m/s. It
slows to a stop with constant acceleration in a distance of 25 m. What
is the reading on the scale during the acceleration. We can do this
problem by looking at the normal force exerted by the scale on the
woman since it is equal in magnitude to the force the woman exerts on
the scale by the third law. In that case the second law provides us
with the fact that the sum of the vertical forces exerted on the woman
are the normal force plus -mg and that this sum is equal to ma. We can
then solve for n since the other values are known.
You have a clear and concise knowledge of the solution. If you are indeed a
freshman student, they you are to be complimented on your mastery of the
subject matter. Very few freshman students are able to do this.
But it is not necessarily intuitive to focus on the forces on the
woman. It seems more intuitive to focus on the scale and the force(s)
acting on it. In that case we have the forces acting on the scale are
the woman's weight , whose magnitude is again -mg. But at this point I
am confused about what to do from here.
If you draw a free-body diagram of the scale, there are two forces. The
downward force exerted on the scale by the woman (the reading on the scale
gives the magnitude of this force, and it is NOT equal to mg). And the
upward force exerted on the scale by the elevator's floor. The net force
exerted on the scale is the vector sum of these two forces, and it equals
the mass times the acceleration of the scale. Since you don't know the mass
of the scale, you have no way of using this scheme to solve the original
problem.
I read some of the other responses you got to your original post, and some
of them seem to contain some obfuscations. I'll add my two cents worth.
When something interacts with the scale, we get a reading on the scale.
It's an indication of the strength of the force associated with that
interaction. But the interaction is symmetrical, in accordance with
Newton's Third Law. It is no more, or less, correct to say that the scale
measures the force exerted ON it, than to say it measures the force exerted
BY it.
The woman exerts a force on the scale, the scale exerts a force on the
woman. These two forces are equal in magnitude to each other and, by
design, to the reading on the scale.
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| User: "Timo A. Nieminen" |
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| Title: Re: freshmen physics problems |
04 Nov 2006 02:40:31 PM |
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On Sat, 4 Nov 2006, will wrote:
Hi, I have two poblems from freshman Physics that I am puzzling over.
If anyone has insight into these I would be grateful to hear from you.
First, in a problem concerning circlular motion, a car moving in a
circular non-banked path. It was said in Sears and Zemansky that since
there is no motion in the radial direction (because the car in moving
uniformly in a circle) the frictional force that is required to keep
the car in a circular path is static and not kinetic friction. That
seems like an odd way to see it. I understand that the force is being
applied in the radial direction and since their is no change in radius
that we can say that the (frictional) force must be acting in the
radial direction where there is no movement but still it seems strange.
I guess the best way to resolve the issue would be if it were possible
to measure the frictional force and see if it is closer to that gotten
when a static coefficient is used as compared to that which is gotten
when a kinetic coefficient is used. Has this been done? Is there a
clearer way to see this?
Firstly, the simple view of friction taught in freshman physics is really
an approximation that applies to the friction between two smooth hard
surfaces. The road is not smooth, and the tyre is neither smooth nor hard.
However, it's a useful approximation.
Q1: Is the part of the tyre that's touching the road moving?
A1: No, therefore static friction.
Further: Kinetic friction does work, static friction doesn't, since
there's no movement at the point of contact. Roll a cardboard tube on a
smooth floor. Slide a cardboard box. Which goes further?
Q2: Kinetic friction opposes motion. Kinetic friction would be in the
direction opposite to the motion. Since centripetal force is at
right-angles to the motion, can centripetal force be provided by kinetic
friction?
The second question is with regard to apparent weight in a moving
elevator. A 50 kg woman stands on a bathroom scale while riding in the
elevator. The elevator is originally moving downward at 10 m/s. It
slows to a stop with constant acceleration in a distance of 25 m. What
is the reading on the scale during the acceleration. We can do this
problem by looking at the normal force exerted by the scale on the
woman since it is equal in magnitude to the force the woman exerts on
the scale by the third law. In that case the second law provides us
with the fact that the sum of the vertical forces exerted on the woman
are the normal force plus -mg and that this sum is equal to ma. We can
then solve for n since the other values are known.
But it is not necessarily intuitive to focus on the forces on the
woman. It seems more intuitive to focus on the scale and the force(s)
acting on it. In that case we have the forces acting on the scale are
the woman's weight , whose magnitude is again -mg. But at this point I
am confused about what to do from here.
You are trying to focus on the scale and the forces on it - in particular,
the force exerted by the woman on the scale. However, you don't know the
mass of the scale, or the normal force exerted by the elevator on the
scale. Two unknowns makes it difficult. What about the woman? You know her
mass, you know her acceleration, and you know her weight (w=mg). The only
unknown quantity is the force exerted on her by the scale, so you can find
this easily.
As you say, by Newton 3, the magnitude of this force is the same as the
force she exerts on the scale (which is what the scale measures) - to
find one is to find the other. The focus is on the scale; the woman is the
most convenient way to find the numbers. Since you don't know the mass of
the scale, what else can you do? Even if you did know the mass of the
scale, you'd still need to use the woman to find F_woman_scale first
before you could find F_elevator_scale. Given that the question only asks
for the first, why bother to find the second? It might be good practice.
Guess a suitable mass for the scale, and find F_elevator_scale. Don't
forget the weight of the scale!
Now, the woman is the first object of attention since the forces on her
are the simplest part of the problem. To start with the simplest part of
the problem is often useful - teaching this strategy of problem solving is
part of the aim of the question. You're not just learning physics, you're
also learning to solve physics problems (and other problems), which is not
the same thing.
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
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| User: "" |
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| Title: Re: freshmen physics problems |
04 Nov 2006 07:37:33 AM |
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will wrote:
Hi, I have two poblems from freshman Physics that I am puzzling over.
If anyone has insight into these I would be grateful to hear from you.
First, in a problem concerning circlular motion, a car moving in a
circular non-banked path. It was said in Sears and Zemansky that since
there is no motion in the radial direction (because the car in moving
uniformly in a circle) the frictional force that is required to keep
the car in a circular path is static and not kinetic friction. That
seems like an odd way to see it. I understand that the force is being
applied in the radial direction and since their is no change in radius
that we can say that the (frictional) force must be acting in the
radial direction where there is no movement but still it seems strange.
"Kinetic friction" between two solid objects (tire / road) is also
called "sliding friction." Typically is varies with the relative
velocities of the two objects, among other things. At the point of
contact between the tires and the surface ('where the rubber meets the
road') there is no sliding unless you are drift racing or some such.
The second question is with regard to apparent weight in a moving
elevator. A 50 kg woman stands on a bathroom scale while riding in the
elevator. The elevator is originally moving downward at 10 m/s. It
slows to a stop with constant acceleration in a distance of 25 m. What
is the reading on the scale during the acceleration.
As long as the elevator is moving at a constant velocity, there is no
additional acceleration. When the elevator 'slows to a stop' there is
a new acceleration introduced. The elevator is applying a force to the
woman who, in turn, applies an equal and opposite force to the elevator
(*through* the bathroom scale). This force is in addition to her own
weight. The scale, of course, cannot distinguish between the reaction
force and the gravitational force, so it registers the simple sum of
the forces.
HTH
Tom Davidson
Richmond, VA
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| User: "" |
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| Title: Re: freshmen physics problems |
05 Nov 2006 05:57:58 AM |
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In article <1162647453.762690.153860@i42g2000cwa.googlegroups.com>,
"tadchem@comcast.net" <tadchem@comcast.net> wrote:
will wrote:
<snip>
The second question is with regard to apparent weight in a moving
elevator. A 50 kg woman stands on a bathroom scale while riding in the
elevator. The elevator is originally moving downward at 10 m/s. It
slows to a stop with constant acceleration in a distance of 25 m. What
is the reading on the scale during the acceleration.
As long as the elevator is moving at a constant velocity, there is no
additional acceleration. When the elevator 'slows to a stop' there is
a new acceleration introduced. The elevator is applying a force to the
woman who, in turn, applies an equal and opposite force to the elevator
(*through* the bathroom scale). This force is in addition to her own
weight. The scale, of course, cannot distinguish between the reaction
force and the gravitational force, so it registers the simple sum of
the forces.
If the OP doubts you, he could always do the experiment. It might
make the neighbors next to the elevator go nuts but...
/BAH
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