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Here's the replies to my question;
If a crane lifts a weight and the crane is standing on a giant force plate then the GRF measured by the force plate will increase as the weight accelerates to a certain terminal velocity. Once this velocity is reached and the weight displaces vertically at a constant velocity, what will be the total weight / force measured by the force plate. Lets say the weight is 200kg and the crane weight = 2000kg = total 2200kg and the acceleration of the weight is 2m/s/s and terminal velocity is 4m/s.
There appears to be no acceleration of the weight after the initial acceleration but since gravity is acting on it there must be a force acting opposite to gravity to keep it moving vertically therefore there must be an acceleration that increases the total force and yet there is a constant velocity which implies no force increase. Which is it?
Dave,
Not sure if this I read your question correctly, but I believe that the answer is that the net force exerted by the crane during acceleration is 2m/s/s. Since you have downward force of gravity, the total upward forces exerted by the crane is 11.8 m/s/s (2+gravity), yielding a net force of 2 m/s/s. At some velocity, the crane will automatically cut off any acceleration greater than the acceleration necessary to maintain constant velocity. So, at this point, the NET force is zero; but the TOTAL upward force is 9.8m/s/s. So the crane will switch the upward force from 11.8m/s/s to 9.8m/s/s when the velocity reaches 4 m/s.
Does this help?
-Geoff
Hi Dave,
There are two forces acting on the crane/weight system - the gravitational force (mg) and the inertial force (ma). The GRF will equal the gravitational force + the inertial force. In your example, the mg will be constant, and the ma will vary depending upon the acceleration of the crane + weight COM. At constant velocity, the ma will be zero. The mg is what keeps the weight moving. However, no acceleration means no additional forces are acting on the system.
I hope this makes sense,
John
"Terminal velocity" is typically used to refer to the speed of a falling
object at which the wind resistance force equals the gravitational
force. When these forces are in balance there is no acceleration and the
velocity is constant.
In your example (and neglecting wind resistance), when the crane lifts
the weight the GRF would exceed the combined weight of the crane and
object lifted while it is accelerating upward. When the object lifted
reaches a constant velocity the GRF would again return to the combined
weight of the crane and object.
Does that clarify things at all?
Brian Schulz, Ph.D.
Biomechanics Researcher
HSR&D/RR&D Center of Excellence, Maximizing Rehabilitation Outcomes
James A. Haley Veterans' Medical Center
8900 Grand Oak Circle, Room 149
Tampa, FL 33637-1022
Phone: (813) 558-3944
Fax: (813) 558-7691
Dave, I don't think your problem works because you are accelerating up, against gravity, and terminal velocity basically looses its meaning. The drag force will be acting downward (opposing motion) and gravity will be doing the same. External work (the crane) is solely responsible for the upward acceleration and the only limit to the applied force is the power of your crane.
Terminal velocity generally refers to fluid forces (drag, buoyancy, etc.) balancing out the weight of an object, resulting in no acceleration. This could apply if you were lowering the weight in your example, i.e. if you lowered it fast enough, the cable would eventually go slack and the weight would be in free-fall.
I'm not a force-plate guy at all, so I'm not sure how much this helps, but hopefully it will clarify things a little.
Curt
Hello Dave,
The answer is the same regardless of the velocity of the weight. Think of the situation where the weight is not moving and velocity is 0. The force plate will measure the weight of the crane plus the weight. or 2200 kg.
Best regards,
Paul Bussman
paul.bussman@kistler.com
___________________________________
Kistler Instrument Corp.
75 John Glenn Drive, Amherst, NY 14228-2171, USA
Tel +1 716 691 5100 (direct 716 213 5781)
Fax +1 716 691 5226
http://www.kistler.com
We can assume that the crane is small in comparison to the extent of
the Earth's magnetic field, so that the gravitational force on the
mass being lifted is independent of the height. When there is no
acceleration of the mass (even if the mass is moving at a constant
velocity) the forceplate will experience a force equal to the combined
weight of the mass and the crane. The force will exceed the weight of
the crane and the mass initially and then reduce to their combined
weight as the acceleration falls to zero. A good way to think about
this is to replace the crane with a lift, the mass with your head and
torso and the force plate with your legs. When the lift starts, you
feel like your body is heavier, but once it's up to speed, you don't
feel any different.
In the case of the crane and mass, there could be an additional
velocity-dependent force due to the drag from the air, but you'd have
to be lifting a very strange shaped mass, or lifting very fast to
notice it
)
I hope that helps.
--
Kevin Channon
Dave,
Here's a quick once over
F = MA + Mg + ma + mg
Where
F = force measured on force plate
M = mass of crane
m = mass of car
A = acceleration of crane == 0.0
a = acceleration of car == f(t)
g = acceleration due to gravity == constant
f(t) = 0 until the car starts moving, going positive for a bit, then dropping to zero as the car reaches its steady-state velocity.
Max F occurs at the peak of the "a" curve.
HTH,
Eric
Eric Fahlgren
Vice President, Product Development
---------------------------------------------
LifeModeler, Inc.
Bringing Simulation to Life
www.LifeMOD.com
+1 949 365 4163
I imagine you will get a lot of responses, but in case you don't...
You are correct that there is a force acting on the weight, and if the
weight has constant velocity opposite the direction of gravitational
acceleration, that force will be identically equal to the weight of
the object. When the weight is balanced by an external force, the
object will have constant velocity and thus no acceleration. Finally,
when the "weight" is moving at constant velocity the GRF will just be
2200kg*9.81m/s/s.
It might be helpful to think about a puck on ice. Applying an initial
force will cause the puck to move, but if not acted on by any other
external forces the puck will continue to slide with constant velocity
forever.
Best,
Jeff Bingham
Hi Dave,
I'm sure you'll get plenty of replies, but the correct answer is that when
the 200 kg weight is rising steadily at 4 m/s, the vertical GRF measured by
the force plate will be 21582 N, or weight of crane (19620 N) plus weight of
weight (1962 N). The forces acting on the weight are the weight due to
gravity (m*g) going down and the tension (T) in the cable going up.
After the initial acceleration, the acceleration is zero, so sum of forces =
m*a gives T - m*g = m*0. Thus the T simply equals 1962 N. This is all that
is required to keep the weight moving up. A greater tension would cause it
to accelerate and move up faster.
For completeness, when initially accelerating the weight, the equation would
be T - m*g = m*(2 m/s/s). This would mean that the tension in the cable
during acceleration is T = 200*2 + 200 * 9.81 = 2362 N, and the vertical GRF
would measure 21982 N.
--
Dennis Anderson
Ph.D. Candidate
Department of Engineering Science and Mechanics
Virginia Polytechnic Institute and State University
Kevin P. Granata Biomechanics Lab
Dave,
You are right: GRF for crane plus weight is 2200 kg* 9.8 N/kg when the
weight is rising at a constant speed , and when it is stationary. The
upward force on the weight, 200*9.8 N, is equal and opposite to the
downward force of gravity, so the weight moves at a constant speed,
which may be zero or nonzero.
One might object that this cannot be right, because when the weight is
rising, the center of mass of the system is rising, but when the weight
is stationary, the c.o.m. is fixed. Surely this different state of
affairs will cause a difference in the GRF.
Actually, no. It is true that work must be done to raise the system's
center of mass (rate of work=Mgv, where M and v are mass and velocity of
the weight, or mass of system and velocity of system's c.o.m.; you'll
get same answer either way), and it is true that no work is done when
the weight is not moving. But the GRFs are the same. Doing work need
not affect the GRF. For example, compressing a sideways spring while
standing on a force plate takes work, but I think we would agree that it
won't affect the GRF.
Bill
Dave, my humble opinion:
I would absolutely agree with you that the math does not jive with the intuitive concept. I have pondered this myself and come to this conclusion...
We already know for certain that the equation F=MA, what we as lay people use 99.999 percent of the time to describe physics around us, is not absolutely accurate (Einstein and all those types showed us that). We use it anyhow for the sake of simplicity, and because for most purposes it is accurate enough. If we are already using an equation that is not completely accurate for the sake of simplicity, why not continue simplicity and consider gravity a component of acceleration, even while gravity does not fit our definition of acceleration (change in velocity over time), because it easily explains the weight of the object.
Maybe you could say a lack of complete intuitive sense (and complete accuracy) is the price paid for the convenience.
To go a little overboard, but to make a point, I would add that until there is an equation for this so called "unified theory" (I am in no way claiming to understand physics on this sort of level, just using buzz words that you read about in science articles) there is no mathematical equation that completely agrees with intuition, IF intuition is considered to be a complete understanding of the universe.
Regards,
Greg
Hello Mr. Smith,
After the weight has reached its terminal velocity, the crane is only
pulling on the crate with a force that is equal to its own weight. At
terminal velocity the force plate should read 2200kg * 9.81 m/s^2 =
21,582 N - which is the same as when the weight is hanging at rest
from the crane. When the crate is accelerating, the force exerted on
the force plate will be the same as the static case with the
additional force due to acceleration: F = (2000 kg + 200 kg)*9.81
m/s^2 + 2 m/s^2 * 200 kg = 21,982 N. This is assuming no air
resistance, an infinitely stiff cable, no load sway, and a whole lot
of other details that in real life may change these numbers.
Hope this helps,
Matt
Dave,
I have my students do simple experiments that are similar to your
question. I have my students stand on a bathroom scale while riding up
and down in an elevator. The scale reading is the same as one's body
weight only under special circumstances which are when standing
perfectly still (zero acceleration, zero velocity) or when moving up or
down at a constant speed (zero acceleration, constant non-zero
velocity). Otherwise, if the elevator is accelerating, the scale reading
will be different than one's body weight. If we call the upward
direction positive, the scale reading will be larger than body weight
when the elevator accelerates upward to begin its ascent. Then after a
period of constant velocity the scale reading will be smaller than body
weight as the elevator slows down near the top. Coming down it is just
the opposite. The scale first decreases as the elevator begins is decent
and then increases above body weight at the end as the elevator slows
down as it reaches the bottom. It is a really easy experiment to do and
the students learn about Newton's second law and how ground reaction
force is not the same thing as body weight.
In the case of your crane, the explanation is the same. In order to
accelerate the object upward the ground reaction force must be larger
than the total weight of the system. But this is only occurs for a brief
time initially to get the object moving upward. Once the acceleration
stops and velocity is constant, the GRF is once again equal to the total
weight of the system. Then when slowing down near the top, the GRF is
briefly less than the total weight of the system. The overall impulse is
exactly zero since there is no overall change in momentum of the system
from bottom to top, assuming velocity is zero at the bottom and at the
top.
Hope this helps.
--Rick
Richard N. Hinrichs, Ph.D.
Dept. of Kinesiology
Arizona State University
P.O. Box 870404
Tempe, AZ 85287-0404
(1)480-965-1624 (office)
(1)480-965-8108 (fax)
hinrichs@asu.edu (email)
http://www.public.asu.edu/~hinrichs (personal web page)
http://kinesiology.clas.asu.edu (Dept. web page)
If the crane is holding the weight and the weight is not moving, the cable supporting the weight is supplying the force to prevent the weight from falling. If the weight is moving upward at a constant velocity the cable is still supplying the same force to hold the weight. It takes no additional force to raise the weight at constant velocity. It only takes force to get it to that velocity. Therefore, the force plate shows higher force during the period of acceleration, but the force must drop back again to 2200 x 9.81 N once the velocity is constant.
Theodore
Dave -
After the initial acceleration the measured GRF will be equal to the
GRF when the weight is stationary. If you draw a free body diagram,
you'll see that the force applied by gravity (mg) must be opposite in
direction and equal in magnitude to the force applied by the cable
(-mg). Thus the total force on the weight is zero, the acceleration
is zero, and the body continues at its present velocity. The
stationary case and the constant velocity case are identical
dynamically, one is just constructed in an inertial reference frame
that is moving relative to the ground.
Hope this clarifies,
Stuart
Pax!
Just go back to Newton's
m*a = F.
For the lifted body (m) you have when a = 0:
0 = m*a = F = F_grav - T = m*g - T
where T = tension of the wire connecting mass and crane = m*g = 200*9.86
N in your example.
Force acting on the crane is Mg + T = 2200 * 9.86 N in your example and
this is equal in magnitude to the GRF.
One can test these things by standing on a force plate with a real time
force display and stand up from squatting position or lifting a weight
with the arm -- one sees significant changes in GRF only when doing
*jerky* motions.
What is sometimes confounding is that an Earth laboratory is not an
inertial system and we have to *correct* for this by using the
gravitational force. The other method is to imagine (like Einstein) that
the laboratory is in a accelerating rocket.
Best regards
Frank Borg
When the crane acheives its final constant velocity, the GRF from the plate should be the same as if the person were back on the ground.
Generally, if the entire system is not accelerating from our (inertial) point of reference, there is no NET force acting externally on any component of the system.
So consider the person: if they are not accelerating, and we already know the force of gravity is constant, then the force applied from the plate they are standing on is equal in magnitude and direction to the force due to their weight (200*9.81 N).
-Andrew Kraszewski
Research Engineer
Hospital for Special Surgery
Dave,
After the weight reaches terminal velocity, if it remains moving in a constant speed, the force applied on the weight from crane = the force applied from gravity (m*g). Newton's first law states "In the absence of force, a bosy either is at rest or moves in a straight line with constant speed."
I'm not sure if this answers your question,
Hsinyi Liu
Just like ridding in an elevator. You feel heavier as the elevator accelerates upwards. No change once constant velocity is reached, then lighter as the elevator decelerates to a stop at the higher floor.
During acceleration up, measurement of forceplatform in N is 200kg * G + 200*2m/s/s + 2000kg *G
No additional force to the 200kg*G + 2000kg*G is required once constant velocity is reached.
During decleration as mass nears the top of the crane the weight measured will be less. 200kg * G - 200*?m/s/s + 2000kg *G
Regards,
Matthew Brodie PhD
I think you may be confusing the weight force due to the acceleration
(g) and an acceleration (a) of a body when writing Newton's law F = ma
and/or not separating the crane and weight masses correctly.
For your example, if the weight is not moving OR at a constant velocity
v=4 m/s, then for both cases acc of the weight is zero. Thus, the force
plate would read:
F = (2000 + 200)kg * 9.81 m/s^2 = 21850 N, which is the sum of the crane
& weight masses multiplied by the acceleration of gravity.
If the crane accelerated the weight (and I think in your case, only the
weight is accelerated) with a=2 m/s^2, then the force plate would read F
= 21850 +/- (200 kg * 2 m/^2), depending on direction
(positive/negative, up/down, same as g / opposite of g) of the weight's
acceleration.
Hope that resolves it, so that concept & math are in agreement.
Warmest regards.
Jen
Hi Dave,
I often find free-body diagrams helpful in sorting through questions like the one you raised. But first, some helpful "ground rules:"
I will avoid calling the load lifted by the crane a "weight," because this is often used to refer to the force of gravity on such a load. Note also that, in this context, weight should not be measured in kg, as this is a unit of mass. The weight would better be expressed in Newtons, as this is a unit of force (mass x acceleration, or kg-m/s^2). This may seem like semantics, especially for colleagues in exercise physiology, but it adheres to appropriate concepts in physics.
FORCES
A free-body diagram of the load will have a force vector pointing downward (mg) that represents the force of gravity (i.e. the weight of the load). It will also have a force vector pointing upward equal in magnitude to (mg), that represents the crane's cable connection. Because these two forces are equal and opposite, there is no net force on the load, and it continues at zero acceleration (constant velocity) in the direction it was moving (upward) when the forces just canceled each other.
(Note that during the initial upward acceleration, the force at the cable connection would have been greater than (mg).)
A free-body diagram of the crane at the same time (i.e., the constant velocity period) will include a force vector pointing downward at the cable connection, equal in magnitude to (mg). It will also include a force vector pointing downward equal to the weight of the unloaded crane (Mg). Since neither the load nor the crane are accelerating, the reaction force from the ground on the crane/load system must be upward, and equal to (mg + Mg). Hence, the force plate will measure a force of (mg + Mg) while the load is moving with constant velocity.
MOMENTS
The load will rotate such that the upward cable force vector and the downward weight vector are co-linear. This will result in no net moment on the load, and therefore no additional rotation of the load.
It is unlikely that the crane/load system will be able to rotate. So, the ground reaction force (mg + Mg) will have a center-of-pressure beneath the center-of-mass for the combined crane/load system.
CENTER OF MASS (COM)
The location of the COM of the crane/load system will not be constant in this example, because the mass of the load is moving upward at constant velocity. I believe this means the change in location of the COM will also be at constant velocity, i.e., at zero acceleration, so this will not change the measured ground reaction force of (mg + Mg).
CAVEAT
All of this assumes that there are negligible accelerations of the mechanism that causes the load to move upward with a constant velocity. If there were large internal masses associated with this mechanism, and they were accelerating, this could cause the COM of the crane/load system to accelerate, with accompanying changes in the forces I noted above.
That's how I see it...
FB
Frank L Buczek Jr, PhD
Branch Chief, HELD/ECTB
Coordinator, MSD Cross Sector Program
National Institute for Occupational
Safety and Health (NIOSH)
1095 Willowdale Road MS 2027
Morgantown, WV 26505
304-285-5966 voice, 304-285-6265 fax
fbuczek@cdc.gov
Good one Frank
Dear Dave Smith,
As for your main question I can tell you that the force you ought to be
measuring at a constant lifting speed is indeed 2200kg. At constant
speed the lifting force (force on the cable) is equal to the gravity force.
What you might not have considered yet is that the lifting acting by the
crane can cause some low frequency (probably From the standpoint of the weight, you are quite right that there is an
external force applied by the crane even when the weight is moving at
constant velocity; this external force is equal and opposite to the weight
and keeps the weight in its state of constant velocity. In other words,
the sum of the forces on the weight equal its mass times
acceleration (equal to zero in this state of constant velocity) and thus
the (negative) gravity force of the weight is balanced by the equal and
opposite (positive) force applied by the crane.
>From the standpoint of the crane, it has the weight pressing down on it,
just as it would if the weight were stationary. Contrast this to the
situation in which the weight is free-falling along its track (or
whatever connection it has to the crane). In this free-fall scenario, the
crane exerts no force on the weight, and the GRF reflects only the crane.
Anyhow, back to the constant-velocity scenario: the crane has the weight
pressing down on it (i.e. the crane is supporting the weight), just as it
would in a static situation. Thus in the constant-velocity state, the GRF
equals the sum of the crane+weight, just as in a static state.
An alternative way of approaching this problem is to consider the center
of mass of the combined crane+weight system, since the GRF equals the sum
of the combined weight and the product of the combined mass times the
acceleration of the COM of the combined system. One could plot the COM
throughout the entire scenario using the following equation:
COM position = [(Crane position)*(2000kg) + (Weight
position)*(200kg)]/2200kg
where Crane position = constant
Weight Position = initial position + (1/2)(2 m/s/s)*t^2 from t=0 to t=2
sec; and
Weight position = (Height at 2 sec) + (4 m/s)*(t-2 sec) after t=2 sec
Taking the first and second derivative of the COM position would show that
the COM of the crane+weight system is accelerating only when the weight is
accelerating, and the COM is moving at a constant velocity (i.e. zero
accel) when the weight is at terminal velocity. Thus, with no
acceleration of the COM, the GRF must equal the combined weight.
I hope this helps (and I hope that it's correct, too!). Thanks again for
the post.
-Dave ( another good one thanks Dave G)
Dear Dave,
according to Newton's first law, you need only external forces to accelerate something. The 2200kg (N) your force plate form registered is based on the gravitation of the earth. There is no additional need of forces, because there is no acceleration.
If you look at your experiment with horizontal forces it should be more clear.
On the other hand, you may also compare people with different weight. They all should have different "acceleration forces" if they stand on a force plate form. They do not have it. What your force plate register is the different force of gravity that exists between bodies with different mass.
Greetings from
Ferdi
___________________________________
Dr. Ferdinand Tusker
TU-München
Fakultät für Sportwissenschaft
Tel.: 089 289 24575
Interesting question - I think it's easier to imagine that the ground that
the crane sits on is flat, and it's accelerating upwards at 9.81 m/s per
second. Easier to intuitively picture than the Earth's mass warping space
and time so that it *seems* like the flat ground is accelerating upward.
When the 200kg mass is accelerating at 2 m/s/s, then for those two seconds,
the force measured by the force plate will be 2200kg*9.81m/s/s +
200kg*2m/s/s, because both accelerations are upward. Once 4m/s velocity is
reached, then the weight will go down to 2200kg*9.81m/s/s because velocity
requires no force to maintain. (The crane will only need force,
200kg*9.81m/s/s, to counter the continued 9.81m/s/s acceleration of the flat
Earth).
Would you forward other answers to me? I look forward to hearing if I'm
totally wrong.
Andy Lammers
Superb thanks Andy
Dave, Just do free body diagrams of the box (the
"weight" being lifted) and of the crane. The vertical
ground reaction force will always be total weight of
crane and the box plus the mass of the box times its
acceleration (positive acceleration defined as
upward).
Peter
If a crane lifts a weight and the crane is standing on a giant force plate then the GRF measured by the force plate will increase as the weight accelerates to a certain terminal velocity. Once this velocity is reached and the weight displaces vertically at a constant velocity, what will be the total weight / force measured by the force plate. Lets say the weight is 200kg and the crane weight = 2000kg = total 2200kg and the acceleration of the weight is 2m/s/s and terminal velocity is 4m/s.
There appears to be no acceleration of the weight after the initial acceleration but since gravity is acting on it there must be a force acting opposite to gravity to keep it moving vertically therefore there must be an acceleration that increases the total force and yet there is a constant velocity which implies no force increase. Which is it?
Dave,
Not sure if this I read your question correctly, but I believe that the answer is that the net force exerted by the crane during acceleration is 2m/s/s. Since you have downward force of gravity, the total upward forces exerted by the crane is 11.8 m/s/s (2+gravity), yielding a net force of 2 m/s/s. At some velocity, the crane will automatically cut off any acceleration greater than the acceleration necessary to maintain constant velocity. So, at this point, the NET force is zero; but the TOTAL upward force is 9.8m/s/s. So the crane will switch the upward force from 11.8m/s/s to 9.8m/s/s when the velocity reaches 4 m/s.
Does this help?
-Geoff
Hi Dave,
There are two forces acting on the crane/weight system - the gravitational force (mg) and the inertial force (ma). The GRF will equal the gravitational force + the inertial force. In your example, the mg will be constant, and the ma will vary depending upon the acceleration of the crane + weight COM. At constant velocity, the ma will be zero. The mg is what keeps the weight moving. However, no acceleration means no additional forces are acting on the system.
I hope this makes sense,
John
"Terminal velocity" is typically used to refer to the speed of a falling
object at which the wind resistance force equals the gravitational
force. When these forces are in balance there is no acceleration and the
velocity is constant.
In your example (and neglecting wind resistance), when the crane lifts
the weight the GRF would exceed the combined weight of the crane and
object lifted while it is accelerating upward. When the object lifted
reaches a constant velocity the GRF would again return to the combined
weight of the crane and object.
Does that clarify things at all?
Brian Schulz, Ph.D.
Biomechanics Researcher
HSR&D/RR&D Center of Excellence, Maximizing Rehabilitation Outcomes
James A. Haley Veterans' Medical Center
8900 Grand Oak Circle, Room 149
Tampa, FL 33637-1022
Phone: (813) 558-3944
Fax: (813) 558-7691
Dave, I don't think your problem works because you are accelerating up, against gravity, and terminal velocity basically looses its meaning. The drag force will be acting downward (opposing motion) and gravity will be doing the same. External work (the crane) is solely responsible for the upward acceleration and the only limit to the applied force is the power of your crane.
Terminal velocity generally refers to fluid forces (drag, buoyancy, etc.) balancing out the weight of an object, resulting in no acceleration. This could apply if you were lowering the weight in your example, i.e. if you lowered it fast enough, the cable would eventually go slack and the weight would be in free-fall.
I'm not a force-plate guy at all, so I'm not sure how much this helps, but hopefully it will clarify things a little.
Curt
Hello Dave,
The answer is the same regardless of the velocity of the weight. Think of the situation where the weight is not moving and velocity is 0. The force plate will measure the weight of the crane plus the weight. or 2200 kg.
Best regards,
Paul Bussman
paul.bussman@kistler.com
___________________________________
Kistler Instrument Corp.
75 John Glenn Drive, Amherst, NY 14228-2171, USA
Tel +1 716 691 5100 (direct 716 213 5781)
Fax +1 716 691 5226
http://www.kistler.com
We can assume that the crane is small in comparison to the extent of
the Earth's magnetic field, so that the gravitational force on the
mass being lifted is independent of the height. When there is no
acceleration of the mass (even if the mass is moving at a constant
velocity) the forceplate will experience a force equal to the combined
weight of the mass and the crane. The force will exceed the weight of
the crane and the mass initially and then reduce to their combined
weight as the acceleration falls to zero. A good way to think about
this is to replace the crane with a lift, the mass with your head and
torso and the force plate with your legs. When the lift starts, you
feel like your body is heavier, but once it's up to speed, you don't
feel any different.
In the case of the crane and mass, there could be an additional
velocity-dependent force due to the drag from the air, but you'd have
to be lifting a very strange shaped mass, or lifting very fast to
notice it
)I hope that helps.
--
Kevin Channon
Dave,
Here's a quick once over
F = MA + Mg + ma + mg
Where
F = force measured on force plate
M = mass of crane
m = mass of car
A = acceleration of crane == 0.0
a = acceleration of car == f(t)
g = acceleration due to gravity == constant
f(t) = 0 until the car starts moving, going positive for a bit, then dropping to zero as the car reaches its steady-state velocity.
Max F occurs at the peak of the "a" curve.
HTH,
Eric
Eric Fahlgren
Vice President, Product Development
---------------------------------------------
LifeModeler, Inc.
Bringing Simulation to Life
www.LifeMOD.com
+1 949 365 4163
I imagine you will get a lot of responses, but in case you don't...
You are correct that there is a force acting on the weight, and if the
weight has constant velocity opposite the direction of gravitational
acceleration, that force will be identically equal to the weight of
the object. When the weight is balanced by an external force, the
object will have constant velocity and thus no acceleration. Finally,
when the "weight" is moving at constant velocity the GRF will just be
2200kg*9.81m/s/s.
It might be helpful to think about a puck on ice. Applying an initial
force will cause the puck to move, but if not acted on by any other
external forces the puck will continue to slide with constant velocity
forever.
Best,
Jeff Bingham
Hi Dave,
I'm sure you'll get plenty of replies, but the correct answer is that when
the 200 kg weight is rising steadily at 4 m/s, the vertical GRF measured by
the force plate will be 21582 N, or weight of crane (19620 N) plus weight of
weight (1962 N). The forces acting on the weight are the weight due to
gravity (m*g) going down and the tension (T) in the cable going up.
After the initial acceleration, the acceleration is zero, so sum of forces =
m*a gives T - m*g = m*0. Thus the T simply equals 1962 N. This is all that
is required to keep the weight moving up. A greater tension would cause it
to accelerate and move up faster.
For completeness, when initially accelerating the weight, the equation would
be T - m*g = m*(2 m/s/s). This would mean that the tension in the cable
during acceleration is T = 200*2 + 200 * 9.81 = 2362 N, and the vertical GRF
would measure 21982 N.
--
Dennis Anderson
Ph.D. Candidate
Department of Engineering Science and Mechanics
Virginia Polytechnic Institute and State University
Kevin P. Granata Biomechanics Lab
Dave,
You are right: GRF for crane plus weight is 2200 kg* 9.8 N/kg when the
weight is rising at a constant speed , and when it is stationary. The
upward force on the weight, 200*9.8 N, is equal and opposite to the
downward force of gravity, so the weight moves at a constant speed,
which may be zero or nonzero.
One might object that this cannot be right, because when the weight is
rising, the center of mass of the system is rising, but when the weight
is stationary, the c.o.m. is fixed. Surely this different state of
affairs will cause a difference in the GRF.
Actually, no. It is true that work must be done to raise the system's
center of mass (rate of work=Mgv, where M and v are mass and velocity of
the weight, or mass of system and velocity of system's c.o.m.; you'll
get same answer either way), and it is true that no work is done when
the weight is not moving. But the GRFs are the same. Doing work need
not affect the GRF. For example, compressing a sideways spring while
standing on a force plate takes work, but I think we would agree that it
won't affect the GRF.
Bill
Dave, my humble opinion:
I would absolutely agree with you that the math does not jive with the intuitive concept. I have pondered this myself and come to this conclusion...
We already know for certain that the equation F=MA, what we as lay people use 99.999 percent of the time to describe physics around us, is not absolutely accurate (Einstein and all those types showed us that). We use it anyhow for the sake of simplicity, and because for most purposes it is accurate enough. If we are already using an equation that is not completely accurate for the sake of simplicity, why not continue simplicity and consider gravity a component of acceleration, even while gravity does not fit our definition of acceleration (change in velocity over time), because it easily explains the weight of the object.
Maybe you could say a lack of complete intuitive sense (and complete accuracy) is the price paid for the convenience.
To go a little overboard, but to make a point, I would add that until there is an equation for this so called "unified theory" (I am in no way claiming to understand physics on this sort of level, just using buzz words that you read about in science articles) there is no mathematical equation that completely agrees with intuition, IF intuition is considered to be a complete understanding of the universe.
Regards,
Greg
Hello Mr. Smith,
After the weight has reached its terminal velocity, the crane is only
pulling on the crate with a force that is equal to its own weight. At
terminal velocity the force plate should read 2200kg * 9.81 m/s^2 =
21,582 N - which is the same as when the weight is hanging at rest
from the crane. When the crate is accelerating, the force exerted on
the force plate will be the same as the static case with the
additional force due to acceleration: F = (2000 kg + 200 kg)*9.81
m/s^2 + 2 m/s^2 * 200 kg = 21,982 N. This is assuming no air
resistance, an infinitely stiff cable, no load sway, and a whole lot
of other details that in real life may change these numbers.
Hope this helps,
Matt
Dave,
I have my students do simple experiments that are similar to your
question. I have my students stand on a bathroom scale while riding up
and down in an elevator. The scale reading is the same as one's body
weight only under special circumstances which are when standing
perfectly still (zero acceleration, zero velocity) or when moving up or
down at a constant speed (zero acceleration, constant non-zero
velocity). Otherwise, if the elevator is accelerating, the scale reading
will be different than one's body weight. If we call the upward
direction positive, the scale reading will be larger than body weight
when the elevator accelerates upward to begin its ascent. Then after a
period of constant velocity the scale reading will be smaller than body
weight as the elevator slows down near the top. Coming down it is just
the opposite. The scale first decreases as the elevator begins is decent
and then increases above body weight at the end as the elevator slows
down as it reaches the bottom. It is a really easy experiment to do and
the students learn about Newton's second law and how ground reaction
force is not the same thing as body weight.
In the case of your crane, the explanation is the same. In order to
accelerate the object upward the ground reaction force must be larger
than the total weight of the system. But this is only occurs for a brief
time initially to get the object moving upward. Once the acceleration
stops and velocity is constant, the GRF is once again equal to the total
weight of the system. Then when slowing down near the top, the GRF is
briefly less than the total weight of the system. The overall impulse is
exactly zero since there is no overall change in momentum of the system
from bottom to top, assuming velocity is zero at the bottom and at the
top.
Hope this helps.
--Rick
Richard N. Hinrichs, Ph.D.
Dept. of Kinesiology
Arizona State University
P.O. Box 870404
Tempe, AZ 85287-0404
(1)480-965-1624 (office)
(1)480-965-8108 (fax)
hinrichs@asu.edu (email)
http://www.public.asu.edu/~hinrichs (personal web page)
http://kinesiology.clas.asu.edu (Dept. web page)
If the crane is holding the weight and the weight is not moving, the cable supporting the weight is supplying the force to prevent the weight from falling. If the weight is moving upward at a constant velocity the cable is still supplying the same force to hold the weight. It takes no additional force to raise the weight at constant velocity. It only takes force to get it to that velocity. Therefore, the force plate shows higher force during the period of acceleration, but the force must drop back again to 2200 x 9.81 N once the velocity is constant.
Theodore
Dave -
After the initial acceleration the measured GRF will be equal to the
GRF when the weight is stationary. If you draw a free body diagram,
you'll see that the force applied by gravity (mg) must be opposite in
direction and equal in magnitude to the force applied by the cable
(-mg). Thus the total force on the weight is zero, the acceleration
is zero, and the body continues at its present velocity. The
stationary case and the constant velocity case are identical
dynamically, one is just constructed in an inertial reference frame
that is moving relative to the ground.
Hope this clarifies,
Stuart
Pax!
Just go back to Newton's
m*a = F.
For the lifted body (m) you have when a = 0:
0 = m*a = F = F_grav - T = m*g - T
where T = tension of the wire connecting mass and crane = m*g = 200*9.86
N in your example.
Force acting on the crane is Mg + T = 2200 * 9.86 N in your example and
this is equal in magnitude to the GRF.
One can test these things by standing on a force plate with a real time
force display and stand up from squatting position or lifting a weight
with the arm -- one sees significant changes in GRF only when doing
*jerky* motions.
What is sometimes confounding is that an Earth laboratory is not an
inertial system and we have to *correct* for this by using the
gravitational force. The other method is to imagine (like Einstein) that
the laboratory is in a accelerating rocket.
Best regards
Frank Borg
When the crane acheives its final constant velocity, the GRF from the plate should be the same as if the person were back on the ground.
Generally, if the entire system is not accelerating from our (inertial) point of reference, there is no NET force acting externally on any component of the system.
So consider the person: if they are not accelerating, and we already know the force of gravity is constant, then the force applied from the plate they are standing on is equal in magnitude and direction to the force due to their weight (200*9.81 N).
-Andrew Kraszewski
Research Engineer
Hospital for Special Surgery
Dave,
After the weight reaches terminal velocity, if it remains moving in a constant speed, the force applied on the weight from crane = the force applied from gravity (m*g). Newton's first law states "In the absence of force, a bosy either is at rest or moves in a straight line with constant speed."
I'm not sure if this answers your question,
Hsinyi Liu
Just like ridding in an elevator. You feel heavier as the elevator accelerates upwards. No change once constant velocity is reached, then lighter as the elevator decelerates to a stop at the higher floor.
During acceleration up, measurement of forceplatform in N is 200kg * G + 200*2m/s/s + 2000kg *G
No additional force to the 200kg*G + 2000kg*G is required once constant velocity is reached.
During decleration as mass nears the top of the crane the weight measured will be less. 200kg * G - 200*?m/s/s + 2000kg *G
Regards,
Matthew Brodie PhD
I think you may be confusing the weight force due to the acceleration
(g) and an acceleration (a) of a body when writing Newton's law F = ma
and/or not separating the crane and weight masses correctly.
For your example, if the weight is not moving OR at a constant velocity
v=4 m/s, then for both cases acc of the weight is zero. Thus, the force
plate would read:
F = (2000 + 200)kg * 9.81 m/s^2 = 21850 N, which is the sum of the crane
& weight masses multiplied by the acceleration of gravity.
If the crane accelerated the weight (and I think in your case, only the
weight is accelerated) with a=2 m/s^2, then the force plate would read F
= 21850 +/- (200 kg * 2 m/^2), depending on direction
(positive/negative, up/down, same as g / opposite of g) of the weight's
acceleration.
Hope that resolves it, so that concept & math are in agreement.
Warmest regards.
Jen
Hi Dave,
I often find free-body diagrams helpful in sorting through questions like the one you raised. But first, some helpful "ground rules:"
I will avoid calling the load lifted by the crane a "weight," because this is often used to refer to the force of gravity on such a load. Note also that, in this context, weight should not be measured in kg, as this is a unit of mass. The weight would better be expressed in Newtons, as this is a unit of force (mass x acceleration, or kg-m/s^2). This may seem like semantics, especially for colleagues in exercise physiology, but it adheres to appropriate concepts in physics.
FORCES
A free-body diagram of the load will have a force vector pointing downward (mg) that represents the force of gravity (i.e. the weight of the load). It will also have a force vector pointing upward equal in magnitude to (mg), that represents the crane's cable connection. Because these two forces are equal and opposite, there is no net force on the load, and it continues at zero acceleration (constant velocity) in the direction it was moving (upward) when the forces just canceled each other.
(Note that during the initial upward acceleration, the force at the cable connection would have been greater than (mg).)
A free-body diagram of the crane at the same time (i.e., the constant velocity period) will include a force vector pointing downward at the cable connection, equal in magnitude to (mg). It will also include a force vector pointing downward equal to the weight of the unloaded crane (Mg). Since neither the load nor the crane are accelerating, the reaction force from the ground on the crane/load system must be upward, and equal to (mg + Mg). Hence, the force plate will measure a force of (mg + Mg) while the load is moving with constant velocity.
MOMENTS
The load will rotate such that the upward cable force vector and the downward weight vector are co-linear. This will result in no net moment on the load, and therefore no additional rotation of the load.
It is unlikely that the crane/load system will be able to rotate. So, the ground reaction force (mg + Mg) will have a center-of-pressure beneath the center-of-mass for the combined crane/load system.
CENTER OF MASS (COM)
The location of the COM of the crane/load system will not be constant in this example, because the mass of the load is moving upward at constant velocity. I believe this means the change in location of the COM will also be at constant velocity, i.e., at zero acceleration, so this will not change the measured ground reaction force of (mg + Mg).
CAVEAT
All of this assumes that there are negligible accelerations of the mechanism that causes the load to move upward with a constant velocity. If there were large internal masses associated with this mechanism, and they were accelerating, this could cause the COM of the crane/load system to accelerate, with accompanying changes in the forces I noted above.
That's how I see it...
FB
Frank L Buczek Jr, PhD
Branch Chief, HELD/ECTB
Coordinator, MSD Cross Sector Program
National Institute for Occupational
Safety and Health (NIOSH)
1095 Willowdale Road MS 2027
Morgantown, WV 26505
304-285-5966 voice, 304-285-6265 fax
fbuczek@cdc.gov
Good one Frank
Dear Dave Smith,
As for your main question I can tell you that the force you ought to be
measuring at a constant lifting speed is indeed 2200kg. At constant
speed the lifting force (force on the cable) is equal to the gravity force.
What you might not have considered yet is that the lifting acting by the
crane can cause some low frequency (probably From the standpoint of the weight, you are quite right that there is an
external force applied by the crane even when the weight is moving at
constant velocity; this external force is equal and opposite to the weight
and keeps the weight in its state of constant velocity. In other words,
the sum of the forces on the weight equal its mass times
acceleration (equal to zero in this state of constant velocity) and thus
the (negative) gravity force of the weight is balanced by the equal and
opposite (positive) force applied by the crane.
>From the standpoint of the crane, it has the weight pressing down on it,
just as it would if the weight were stationary. Contrast this to the
situation in which the weight is free-falling along its track (or
whatever connection it has to the crane). In this free-fall scenario, the
crane exerts no force on the weight, and the GRF reflects only the crane.
Anyhow, back to the constant-velocity scenario: the crane has the weight
pressing down on it (i.e. the crane is supporting the weight), just as it
would in a static situation. Thus in the constant-velocity state, the GRF
equals the sum of the crane+weight, just as in a static state.
An alternative way of approaching this problem is to consider the center
of mass of the combined crane+weight system, since the GRF equals the sum
of the combined weight and the product of the combined mass times the
acceleration of the COM of the combined system. One could plot the COM
throughout the entire scenario using the following equation:
COM position = [(Crane position)*(2000kg) + (Weight
position)*(200kg)]/2200kg
where Crane position = constant
Weight Position = initial position + (1/2)(2 m/s/s)*t^2 from t=0 to t=2
sec; and
Weight position = (Height at 2 sec) + (4 m/s)*(t-2 sec) after t=2 sec
Taking the first and second derivative of the COM position would show that
the COM of the crane+weight system is accelerating only when the weight is
accelerating, and the COM is moving at a constant velocity (i.e. zero
accel) when the weight is at terminal velocity. Thus, with no
acceleration of the COM, the GRF must equal the combined weight.
I hope this helps (and I hope that it's correct, too!). Thanks again for
the post.
-Dave ( another good one thanks Dave G)
Dear Dave,
according to Newton's first law, you need only external forces to accelerate something. The 2200kg (N) your force plate form registered is based on the gravitation of the earth. There is no additional need of forces, because there is no acceleration.
If you look at your experiment with horizontal forces it should be more clear.
On the other hand, you may also compare people with different weight. They all should have different "acceleration forces" if they stand on a force plate form. They do not have it. What your force plate register is the different force of gravity that exists between bodies with different mass.
Greetings from
Ferdi
___________________________________
Dr. Ferdinand Tusker
TU-München
Fakultät für Sportwissenschaft
Tel.: 089 289 24575
Interesting question - I think it's easier to imagine that the ground that
the crane sits on is flat, and it's accelerating upwards at 9.81 m/s per
second. Easier to intuitively picture than the Earth's mass warping space
and time so that it *seems* like the flat ground is accelerating upward.
When the 200kg mass is accelerating at 2 m/s/s, then for those two seconds,
the force measured by the force plate will be 2200kg*9.81m/s/s +
200kg*2m/s/s, because both accelerations are upward. Once 4m/s velocity is
reached, then the weight will go down to 2200kg*9.81m/s/s because velocity
requires no force to maintain. (The crane will only need force,
200kg*9.81m/s/s, to counter the continued 9.81m/s/s acceleration of the flat
Earth).
Would you forward other answers to me? I look forward to hearing if I'm
totally wrong.
Andy Lammers
Superb thanks Andy
Dave, Just do free body diagrams of the box (the
"weight" being lifted) and of the crane. The vertical
ground reaction force will always be total weight of
crane and the box plus the mass of the box times its
acceleration (positive acceleration defined as
upward).
Peter