d3uk61

12-11-2009, 04:08 AM

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 :o)

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 :o)

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