question re f=ma/weight

[Double K]

Double K, swansont and myself offered two ways to solve this question, using net-force or using energies.

 

I'm not sure I understand what you're disagreeing with

~moo

 

Just for the record, I'm not disagreeing

 

I was trying to uncomplicate the calculation for someone who stated at the start they weren't a student or physics major etc.

[mooeypoo]

Just for the record, I'm not disagreeing

 

I was trying to uncomplicate the calculation for someone who stated at the start they weren't a student or physics major etc.

Noble attempt, but I don't think it worked

 

This isn't a "simple" question. It's a combination of forces (one depends on time) *or* a question of preservation of energies. Either way, those are usually things that are a bit different than the basic highschool physics.

 

They're far from being hard, though, they're just not immediately intuitive.

 

~moo

[chrismohr]

I have a few rough figures I'm pretty confident of with this poor bowling ball. In one second the ball would drop about 30 feet and be going 22 miles per hour. By the next second it would be going 45 mph and would have dropped about 90 feet. So with this information, can anyone actually calculate the force in Newtons of a 10 lb bowling ball dropping at 1 mph, 22 mph and 45 mph? Let's assume the spring-action scale has long ago been crushed and now we are just calculating force in Newtons. This would give me an actual answer I could use. Thanks again everyone, I can't believe all the responses this has generated so far.

[D H]

In general, no. You need to know the spring constant.

 

Think of it this way: A block of granite is a spring, albeit a very stiff one. A pile of air mattresses is also a spring, albeit a very non-stiff one. Dropping a bowling ball from the same height onto a block of granite versus dropping one from the same height but onto a pile of air mattresses will result in forces of significantly different magnitudes acting on the bowling ball.

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In general, no. You need to know the spring constant.

And the damping coefficient as well. Scales are obviously fairly heavily damped; ideally, they are critically damped. Suppose your mass is 160 lb. Step on an undamped scale and the weight shown on the scale will oscillate between 0 and 320 pounds and never stop. Even a lousy scale will settle to 160 lb, and do so rather quickly.

Edited May 10, 2010 by D H
Consecutive posts merged.

[mooeypoo]

And the damping coefficient as well. Scales are obviously fairly heavily damped; ideally, they are critically damped. Suppose your mass is 160 lb. Step on an undamped scale and the weight shown on the scale will oscillate between 0 and 320 pounds and never stop. Even a lousy scale will settle to 160 lb, and do so rather quickly.

This is a good exmaple, btw. If we want to see what the *highest* number the scale records, then "320" is it. If we want to see what the stable force is recorded, then it's 160lb.

 

When you have a ball falling from a distance, though, eventually it will "settle" on the spring (after its motion is damped). But when it settles, it will do so according to the ball's weight regardless of the falling part. If you want to measure the force INCLUDING the impact from the fall, then the "max" readout is probably what you're looking for.

[Physicsfan]

since the displacement caused by the weighing machine is negligible,the force exerted by the machine on the ball(newtons third law) would be almost equal to the momentum of the ball when it hit the wighing machine, so that means the force exerted by the ball on the machine(weight) would be almost equal to its momentum.

 

am i correct?

[swansont]

since the displacement caused by the weighing machine is negligible,the force exerted by the machine on the ball(newtons third law) would be almost equal to the momentum of the ball when it hit the wighing machine, so that means the force exerted by the ball on the machine(weight) would be almost equal to its momentum.

 

am i correct?

 

Momentum is not a force. You would need to know over what time the momentum was transferred, because F = dP/dt. If the reaction takes longer, there is a smaller force.

[insane_alien]

since the displacement caused by the weighing machine is negligible,the force exerted by the machine on the ball(newtons third law) would be almost equal to the momentum of the ball when it hit the wighing machine, so that means the force exerted by the ball on the machine(weight) would be almost equal to its momentum.

 

am i correct?

 

no.

 

what you said is that the force will be equal to the momentum therefore the force is equal to the momentum.

 

not only wrong, but circular reasoning.

 

you cannot assume the displacement to be negligble. negligble pretty much means zero. in order to stop something from any velocity to zero in zero distance then you need an infinite acceleration. to get an infinite acceleration you need an infinite force.

 

as infinite forces clearly don't exist in this situation, the displacement of the scale is NON negligble and must be taken into consideration.

 

Now, so the thread can progress, can we assume the scale displaces by 1cm?

 

bugger you all, i'm going for it anyway.

 

first off, we'll have to change all those nasty imperial units into SI because I refuse to use imperial. (i used wolfram alpha for the conversions if you want to check)

 

so: mass = 4.536kg

height = 3.962m

 

using mgh this gives us an energy of 176.3J

 

then using 1/2kx^2 for elastic energy to find k we get

 

k=3.526*10^6 N/m

 

so force equals k*x which gives us 35260N or in earth gravity 3594kg or if you want it in silly imperial units 7924 lb

[charles langley]

Hi all,

 

This is my first post. I'm a 56 year old guy (not a student) and I don't know how to apply the f=ma formula to solve this question.

 

If a 10-pound bowling ball is sitting on a scale at sea level it would weigh ten pounds. What if that same bowling ball were dropped from a height of 13 feet onto the scale? If it were a bathroom scale it would probably crush it, but if the scale were strong enough, what weight would the scale show if it were able to adjust instantly to the ball hitting it? I'm assuming no wind resistance so the ball is dropping at 9.8 meters/second squared. Could you demonstrate how you came up with the answer too?

 

Thanks!

Chris Mohr

Casual Science hobbyist

 

Chris

 

I read through this thread and was wondering if you had found the answers you were looking for , so I thought it might be best to just post my findings.

 

F=ma

 

F (44.45205226 N) = M(4.5359237 kg) * A(9.8 m/s^2)

 

the previous post had the answer as follows

 

so: mass = 4.536kg

height = 3.962m

 

using mgh this gives us an energy of 176.3J

I couldnt understand the following

 

then using 1/2kx^2 for elastic energy to find k we get

 

k=3.526*10^6 N/mso force equals k*x which gives us 35260N or in earth gravity 3594kg or if you want it in silly imperial units 7924 lb

 

35,260 newton = 7,926.763 pound-force

which is not hardly a correct answer to your problem.

 

so I wanted to prevent a possible accident in case you were building something.

 

below is a web site that gives several examples to help you learn how to

do these calculations , we might just be seeing what you have in a exam. LOL

 

so to avoid a later accident it may be a good idea for you to learn this instead of just being provided with answers.

 

http://www.jupiterscientific.org/sciinfo/examplesfeq.html

 

 

what weight would the scale show if it were able to adjust instantly to the ball hitting it?

 

44.45205226 kilogram-force = 98.00 pound-force

 

the weight scale would read 98.00 pounds the first time it hits the scale.

 

I suppose that was what you were looking for.

 

here is a good web site to perform conversion also that might come in handy.

 

http://www.onlineconversion.com/force.htm

 

Hope this answers your question.

Edited June 27, 2010 by charles langley

[insane_alien]

35,260 newton = 7,926.763 pound-force

which is not hardly a correct answer to your problem.

 

can you explain why this is not correct? assuming an elastic collision where all the energy is absorbed by the spring then this answer is perfectly correct.

 

i admit that the collision will not be perfectly elastic in reality and a fair amount of energy will be converted to heat. but to do a more precise calculation, we'd need more details to work with.

 

a quick bit of math shows that with your numbers (38.8m/s^2 acceleration) the scales have to deform by 1 meter to stop the bowling ball, i have yet to see a bathroom type scale that deforms this much.

 

but never mind that, where did you get the value of 38.8m/s^2 from?

 

at least with my answer i gave the reasoning behnd it.

Edited June 27, 2010 by insane_alien

[swansont]

44.45205226 kilogram-force = 98.00 pound-force

 

the weight scale would read 98.00 pounds the first time it hits the scale.

 

That's what the scale would show if the ball were placed on the scale, i.e. it is at rest, which is not the question that was asked.

[charles langley]

can you explain why this is not correct? assuming an elastic collision where all the energy is absorbed by the spring then this answer is perfectly correct.

 

i admit that the collision will not be perfectly elastic in reality and a fair amount of energy will be converted to heat. but to do a more precise calculation, we'd need more details to work with.

 

a quick bit of math shows that with your numbers (38.8m/s^2 acceleration) the scales have to deform by 1 meter to stop the bowling ball, i have yet to see a bathroom type scale that deforms this much.

 

but never mind that, where did you get the value of 38.8m/s^2 from?

 

at least with my answer i gave the reasoning behnd it.

 

hello insane alien

 

Im sorry , my first post was a calculation error , Im not sure how I did that but it seemed

a good bit too high so I recalculated , and edited my post.

 

he gives the mass and the distance in the question.

 

Mass = 10 pounds = 4.5359237 kilograms

 

Height = 13 feet = 3.9624 meters

 

so its a matter of a = f/m to find the acceleration of 9.8 m/s^2

 

and F=ma delivers the force.

 

4.5359237 kilograms * 9.8 m/s^2 = 44.45205226 N

 

and he gave the distance that the ball would travel of 3.9624 meters

[swansont]

hello insane alien

 

Im sorry , my first post was a calculation error , Im not sure how I did that but it seemed

a good bit too high so I recalculated , and edited my post.

 

he gives the mass and the distance in the question.

 

Mass = 10 pounds = 4.5359237 kilograms

 

Height = 13 feet = 3.9624 meters

 

so its a matter of a = f/m to find the acceleration of 9.8 m/s^2

 

and F=ma delivers the force.

 

4.5359237 kilograms * 9.8 m/s^2 = 44.45205226 N

 

and he gave the distance that the ball would travel of 3.9624 meters

 

Take a bowling ball. Place it on your foot. Make note of what it feels like.

 

Now, lift it up to your head, and drop it on your foot.

 

Does it feel the same?

 

 

A dropped ball will have momentum, and the scale must exert a force to bring it to rest. The scale will indicate the force being exerted in doing so.

[charles langley]

Take a bowling ball. Place it on your foot. Make note of what it feels like.

 

Now, lift it up to your head, and drop it on your foot.

 

Does it feel the same?

 

 

A dropped ball will have momentum, and the scale must exert a force to bring it to rest. The scale will indicate the force being exerted in doing so.

 

Hello swansont

 

a ball with a mass of 4.5359237 kg at a acceleration of

9.8 meters a second / second will fall a distance of 4.9 meters in one FULL second , thats 1 meter further than the ball could fall the

3.9624 meters distance allowed in the question.

 

so the ball travels less than 1 second.

in fact it travels for a time of 0.899 seconds

 

its final velocity is 8.8102 m/s

and it has a average velocity of 4.4051 m/s durring the .899 seconds

 

the distance the ball travels is its average velocity * time of 0.899 seconds = 3.9601849 meters

 

you are welcome to check my numbers if you like

 

but the 35260N force could accelerate the 4.5359237 kilogram ball to 7773.499 m/s^2

 

a=f/m = 35260N / 4.5359237 kilograms =7773.499 m/s^2

 

do you believe me to be wrong?

 

If that were possible I would go buy up every bowling ball I could get my hands on

and I wouldnt pay any more electricity biolls , LOL

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Originally Posted by charles langley

44.45205226 kilogram-force = 98.00 pound-force

 

the weight scale would read 98.00 pounds the first time it hits the scale.

 

That's what the scale would show if the ball were placed on the scale, i.e. it is at rest, which is not the question that was asked.

 

swansonts

 

how could a 10 lb ball show 98 pounds weight if its just placed on the scale?

 

Im not looking for a argument , I was only trying to help one of the members to solve his problem it seemd as though he never did

get a correct answer from those who posted replies to his question.

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since the displacement caused by the weighing machine is negligible,the force exerted by the machine on the ball(newtons third law) would be almost equal to the momentum of the ball when it hit the wighing machine, so that means the force exerted by the ball on the machine(weight) would be almost equal to its momentum.

 

am i correct?

 

That is a good and clear explanation of what would "actually" happen.

Edited June 28, 2010 by charles langley
Consecutive posts merged.

[swansont]

 

how could a 10 lb ball show 98 pounds weight if its just placed on the scale?

 

I was looking at your whole post and only quoted part of it; you had 44.45 N, which is not what the scale would read if the ball were dropped on it. I admit, I stopped paying attention at that point. Otherwise I would have noticed the units of kilogram-force. What is a kilogram-force?

 

The real question is how you went from 10 lb to 98 lb. Magic?

[charles langley]

I was looking at your whole post and only quoted part of it; you had 44.45 N, which is not what the scale would read if the ball were dropped on it. I admit, I stopped paying attention at that point. Otherwise I would have noticed the units of kilogram-force. What is a kilogram-force?

 

The real question is how you went from 10 lb to 98 lb. Magic?

 

swansont

 

a kilogram-force is the magnitude of the force exerted on one kilogram of mass by a 9.80665 m/s^2 gravitational field

 

http:// http://en.wikipedia.org/wiki/Kilogram-force

 

The real question is how you went from 10 lb to 98 lb. Magic?

 

f=ma

 

physics math , the part of physics we use to determine if the wording of physics is right or not.

physics is self checking because of the math , so the math can prove if the words are right or wrong.

 

http:// http://en.wikipedia.org/wiki/Physics

 

Relation to mathematics and the other sciences

In the Assayer (1622), Galileo noted that mathematics is the language in which Nature expresses its laws. Most experimental results in physics are numerical measurements, and theories in physics use mathematics to give numerical results to match these measurements.

 

Physics relies upon mathematics to provide the logical framework in which physical laws may be precisely formulated and predictions quantified. Whenever analytic solutions of equations are not feasible, numerical analysis and simulations may be utilized. Thus, scientific computation is an integral part of physics, and the field of computational physics is an active area of research.

Edited June 28, 2010 by charles langley

[swansont]

44.45 N is not the same as 44.45 kilogram-force.

[charles langley]

44.45 N is not the same as 44.45 kilogram-force.

 

your right 44.45205226 N = 4.532 847 839 kilogram-force

 

got to go for now though , so have a nice night.

[darkenlighten]

This is has gone on for way longer than it needed to imo, so I'm gonna settle this.

 

There can only be two scenarios:

 

1) You consider the impulse of the impact, in which the question does not have enough information for it to be solved.

 

2) You don't consider the impulse and the fact the the acceleration will be the same from either height when the bowling-ball hits the scale (ignoring change from the distance from the center of the earth, edit: which then still doesn't matter since the scale is at the same height ).

 

There isn't much more to this problem than this.

Edited June 28, 2010 by darkenlighten

[charles langley]

This is has gone on for way longer than it needed to imo, so I'm gonna settle this.

 

There can only be two scenarios:

 

1) You consider the impulse of the impact, in which the question does not have enough information for it to be solved.

 

2) You don't consider the impulse and the fact the the acceleration will be the same from either height when the bowling-ball hits the scale (ignoring change from the distance from the center of the earth, edit: which then still doesn't matter since the scale is at the same height ).

 

There isn't much more to this problem than this.

 

I agree.

 

I had to do something real quick , but when I got back and checked the

numbers I realized that the force of

44.45205226 newtons = 9.993218 pound-force

 

so in the 0.8 seconds it really didnt aquire much acceleration at all that would

really mean anything.

 

in fact , LOL , it looks like it lost some of the impact.

 

it originally weighed 10 lbs , but by the time all the conversions were done it

impacted with less force than it weighs. 0.006782 lbs less.

[darkenlighten]

And to add, your answer is most likely a conversion error, because it will be exactly 10-lbs (if impulse is not considered). There will be no loss, given the simplistic scenario.

[John Cuthber]

How overcomplicated an answer can you get?

The force neede to bring it to a halt is the force gravity exerts on it (10 Lbf) multiplied by the ratio of the distance it accelerated to the distance over which it's brought to a halt.

 

It falls 156 inches

If the spring on the scales brings it to rest in, for example, half an inch the force is

10 X156/0.5

 

=3120 Lbf

If it came to rest over just 0.1 of an inch the force would be 5 times greater; 1560Lbf

Without knowing more about the scales it's impossible to know, but you can get a reasonable estimate this way.

[insane_alien]

but the 35260N force could accelerate the 4.5359237 kilogram ball to 7773.499 m/s^2

 

a=f/m = 35260N / 4.5359237 kilograms =7773.499 m/s^2

 

do you believe me to be wrong?

 

your calculation of the acceleration is correct(ish, it will actually be non-linear and that is the peak acceleration)

 

you seem to be forgetting that this acceleration only lasts a tiny fraction of a second just as it brings the bowling ball to a stop.

 

i assumed the scales deformed 1 centimeter. work out the average acceleration for a 4.44kg ball going from 8.8m/s to 0m/s in a distance of 1cm.

 

it should be less than the value above but only because it is the average acceleration, not the peak acceleration as was asked.

 

i haven't seen you calculate anything other than its resting weight.

 

and obviously you must realise that an impact force is greater than the weight of the object, otherwise hammers wouldn't work.

[charles langley]

your calculation of the acceleration is correct(ish, it will actually be non-linear and that is the peak acceleration)

 

you seem to be forgetting that this acceleration only lasts a tiny fraction of a second just as it brings the bowling ball to a stop.

 

i assumed the scales deformed 1 centimeter. work out the average acceleration for a 4.44kg ball going from 8.8m/s to 0m/s in a distance of 1cm.

 

it should be less than the value above but only because it is the average acceleration, not the peak acceleration as was asked.

 

i haven't seen you calculate anything other than its resting weight.

 

and obviously you must realise that an impact force is greater than the weight of the object, otherwise hammers wouldn't work.

 

I was calculating the acceleration using your 35260N

 

not my 44.45205226 N

 

I was showing the difference between our answers.

 

I dont find that the force is 35,260 N

 

I was refering to the below that you posted

 

 

then using 1/2kx^2 for elastic energy to find k we get

 

k=3.526*10^6 N/m

 

so force equals k*x which gives us 35260N or in earth gravity 3594kg or if you want it in silly imperial units 7924 lb

 

Im currious about the above equation , could you elaborate?

 

what is "x"

what is "k"

[darkenlighten]

I still don't understand why you guys are trying to solve this using impulse. It almost seems like a battle of egos at this point lol.