question re f=ma/weight

[charles langley]

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.

[John Cuthber]

mass * distance does not solve for force.

 

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?

[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.

Edited June 28, 2010 by charles langley

[John Cuthber]

Where in the name of all that's holy have you got 13 ft/s from?

130 foot pound would be the correct energy- of course

130 foot pounds over one twentyfourth of a foot gives 3120 Lbf

Like I gave before.

 

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

 

No, we don't. Just the distances will give the right answer- it's just that you refuse to accept it.

They will give the average force as I calculated; doubling that gives the peak force.

Edited June 29, 2010 by John Cuthber
Consecutive posts merged.

[AlphaSheeppig]

You don't even use F=ma in this problem...

 

It depends on the scale. Measure the height of the scale with nothing on, then measure the height of the scale with the ten pound bowling ball. Call the difference in height in ft x. The spring constant k is then (10 lb/x). Then calculate the total kinetic energy the ball will have when it hits. This is E=mgh where m is the mass in pounds, h is the height in ft, and g is your gravitational constant (32.2 ft/s/s). The distance the scale will "bounce" through is d = sqrt(2E/k), and the force you read off is then F=kd.