charles langley

Good thing that nobody multiplied a mass by a distance then, isn't it?

 

Lbf is a rather old fashioned unit, the pound force. It's the weight of an object with a mass of 1 pound.

 

So, what I multiplied was a force by a distance.

That gives you an energy

The energy concerned is the gravitational energy that the ball starts with- it's also (neglecting air resistance) the energy that ball has just before it hits the scales.

 

It is also, due to the conservation of energy, the energy that ends up stored in the spring of the scales.

 

That spring energy is there because a force- the force by which the scales retard the ball- moves through a distance ( the "give" in the scales).

 

Strictly speaking I think it's the mean force.

 

Now, if I watched carefully I could see how far the scales got squashed when the ball hit them, at least to a reasonable accuracy.

That's the only unknown in the system.

 

As I asked before, how overcomplicated an answer can you get?

As I asked before, how overcomplicated an answer can you get?

I have to admit that the standard units had me twisted up in a knot for awhile.

but hopefull we will end up with a correct answer to his question.

you had posted 3120 Lbf , so I used a potential energy equation as you were talking energy.

So, what I multiplied was a force by a distance.

That gives you an energy

PE = mgh

PE = 4.5359237 kg * 9.807 gc * 3.9624 m = 176.26 j

176.26 joule = 130.002 foot pound

for comparison

10 lbf * 13 ft/s = 130 lbf/s

thats a large difference from your 3120 Lbf and may be the correct answer to your energy equation.

I think that some are considering that the weight scale will read the maximum weight that the mass

can possibly supply differently depending on the time involved in stopping the mass.

like jumping on your bathroon scales will cause the needle to jump way past your weight.

and we would need the time involved to get an exact number here.

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.

force * distance does not solve for force.

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"

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.

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.

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.

Hi!

 

I just wanted to inform you about a recent video on Youtube that replicates the Thornson principle.

 

A student have made a device in Lego and used a balance weight to prove the Thornson inertial drive force. He also explains the simple concept....

 

And wow, the thing moves....The only explanation would of course be movement of air by the rotor....But, but... the rotor moves the air equally in all directions, so I don't see that as the explanation...

 

I'll have to build one from my son's Lego and test this wrapped in plastic...

 

Here is the link;

http://www.youtube.com/watch?v=q-f6-6GSyZ4

hey , just get a large sphere and cut it in half.

put a battery driven toy car inside the sphere.

turn the toy on so that the wheels turn.

then seal up the two halves of the sphere.

its now a closed system.

place the sphere on your floor.

the toy pushes against the INSIDE of the sphere with a force from the INSIDE ( not external ).

you now have a reactionless propulsion device , post it on youtube.

any device that does not have a force applied to its outside for propulsion ( external force ) is considered a reactionless propulsion device , and its not supposed to work.

the sphere rolls on the floor , because the toy car is providing a INTERNAL FORCE...

or as Wikipedia describes it

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

A reactionless drive or inertial propulsion engine (also reactionless thruster, reactionless engine, bootstrap drive, and inertia drive) is any form of propulsion not based around expulsion of fuel or reaction mass.

 

The name comes from Newton's Third Law of Motion, usually expressed as: "For every action, there is an equal and opposite reaction." Such a drive would use a hypothetical form of thrust that does not require any outside force or net momentum exchange to produce linear motion, and therefore necessarily violates the conservation of momentum, a fundamental principle of all current understandings of physics. In addition it can be shown that conservation of energy is violated.

Hi!

 

I just wanted to inform you about a recent video on Youtube that replicates the Thornson principle.

 

A student have made a device in Lego and used a balance weight to prove the Thornson inertial drive force. He also explains the simple concept....

 

And wow, the thing moves....The only explanation would of course be movement of air by the rotor....But, but... the rotor moves the air equally in all directions, so I don't see that as the explanation...

 

I'll have to build one from my son's Lego and test this wrapped in plastic...

 

Here is the link;

http://www.youtube.com/watch?v=q-f6-6GSyZ4

hey , just get a large sphere and cut it in half.

put a battery driven toy car inside the sphere.

turn the toy on so that the wheels turn.

then seal up the two halves of the sphere.

its now a closed system.

place the sphere on your floor.

the toy pushes against the INSIDE on the sphere.

the sphere beging to roll on the floor , because the toy car is providing a INTERNAL FORCE...

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

Merged post follows:

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.

Merged post follows:

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.

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

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.

hello everyone , I'm new to the forum.

 

and I have a theory of a new type of propulsion for space travel.

 

consider a rail gun that fires a 100 kg mass along a 500 meter rail.

 

then the mass enters a turnaround at the end of the 500 meter rail.

having accelerated the 500 meters , it will present a force to the turnaround.

 

the mass turns 180 degrees through the turnaround and begins traveling the opposite direction and free floats in zero g for 500 meters in the

opposite direction.

 

it then enters a second 180 degree turnaround and presents a force to the second turnaround.

the mass turns 180 degrees again in the second turnaround and now enters the rail gun a second time

and it is accelerated even faster the second time.

 

the two forces that are presented to the turnarounds by the 100 kg mass will cancel each other out and the force that was used to accelerate the 100 kg mass will propel the spacecraft.

 

 

so you end up with

2 positive forces for propulsion ;

1) the force used to accelerate the mass

2) the force presented to the second turnaround by the mass

and

1 negative force for propulsion ;

1) the force presented to the first turnaround by the mass

 

this process repeats giving the space ship the ability to accelerate as long

as power is used to accelerate the mass.

 

as the mass inside is following a path inside the space ship the momentum

of the mass in reference to the space ship is conserved.

 

and since force is not momentum the conservation of momentum

is not violated.

 

it is a closed system but matter is not crossing the boundary.

the forces that are applied to the space ship from the inside , push

the space craft the same way a mass ejection propulsion system works

by ejecting or throwing mass out of the spaceship.

 

 

the force that accelerates the mass inside the space ship is transfered to

the space ship which gives the space ship acceleration.

 

nothing needs to be ejected or thrown from the ship for propulsion.

you should have a look at the following web site.

http://technology.arc.nasa.gov/partnering/sbirsttr.cfm