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question re f=ma/weight


chrismohr
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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

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OK, OK, but can anyone give me an estimate of what this 10 pound bowling ball would weigh when dropped 13 feet onto a scale that didn't have much spring in it? I don't need an exact answer, just a close approximation.

 

Thanks,

Chris Mohr

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If you are not considering impulse, then the ball will be the same weight, since it is under the same acceleration. If you are considering the fact it is de-accelerating really fast (once it hits the scale), you would need the impulse, but you would need to know how fast it would stop, in order to calculate this.

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what if we assume that the scale has no "spring" to it? He could convert the weight of the 10lb ball into mass. Then calculate the velocity it reaches from 13 feet high (probably converted to meters/sec)....and then what resulting force it would put on the scale. I don't remember the equations from physics so I can't really help beyond this.

 

The answer to the question doesn't seem as simple as F=ma.

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The spring in the scale is used to measure the force. If there's no spring, there's no measurement. It's true that the spring of the scale will slow the ball down (otherwise the ball would crush the scale) but that's exactly how the scale itself measures the force.

 

If I take my bathroom scale (which operates on the 'spring' concept) and press it, I apply force on it. The spring is compressed in a certain way that states how much force I put on it. This is then displayed on the screen - either with a rotating disk or conversion to digital display.

 

If you want to know what the spring itself (of the measurement device) did to the movement, then you need another measurement-- like, for instance, under your scale, put another scale. So you have the measurement of the force applied on the ground, rather than on the scale.

 

Does this make sense?

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OK, wait, so you're asking "how much would the ball appear to weigh based on the force exerted on the scale" forgetting about the spring and all that. Basically, "how much force is the ball hitting the scale with?"

 

So, I'm still agreeing with my previous post, kind of... no matter what happens after it hits the surface of the scale.

 

I hope this is correct but what about velocity = 1/2 x acceleration due to gravity x time squared? Please somebody correct me here if I'm wrong!

 

I thing that F=ma can be used and then F (in Newtons) can be converted into pounds. But you first have to find what the speed of the mass is from a 13ft drop.

 

Basically, I'm converting my thoughts into words here even though it doesn't help. I could go pull out my physics book from my library but I don't want to do all the work for ya (apologies!).

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http://wiki.answers.com/Q/How_much_is_1_KN_in_Kg

 

Newtons

--------------

1 N = 100 g

10 N = 1 kg

--------------

 

KiloNewtons

--------------

1 kN = 1000 N = 100 kg

10 kN = 10000 N = 1000 kg

--------------

 

Kilograms are always for one 0 less than Newtons.

'Kilo' increases a value x1000 times.

If there are 12 N, then there are 1,2 kg.

 

___

 

That is only an estimation. A kiloNewton(kN) is a measure of force while a kilogram (kg) is a measure of mass.

F=ma. F=force, m=mass, a=acceleration

When talking about weight, weight is a force. So if something weighs 1kN then it will have a mass of 102kg. Get this by dividing this force by the gravitational constant.

The acceleration gravitational constant is around 9.81m/s2.

 

F=ma

F(kN)=10lbs(4 535.9237grams) * -9.81(m/s2)

 

4 535.9237 * 9.81 = 44497.411497 g*m/s2

 

F=44497.411497/1000

F=(approx.) 45 kN. (44.497411497 to be precise)

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There are many types of forces. F=ma is the total force on an object under acceleration.

 

In this case, the force on the scale comes from the weight of the ball (F=mg) and the impact of the ball with the scale (Ft=mv). You can figure out the velocity of the ball once it hits the scale, and you know its mass. The only question that remains is how much time (t) took for the scale to absorb the entire force. Or, in other words, how much time it took for the ball to come to complete rest after it hit the scale.

 

That depends on the spring inside the scale, mostly. Unless you have the spring constant (k), you need to have this piece of information in advance. If you have the spring constant you can calculate it by using the spring equation (F=-kx).

 

~moo

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There are many types of forces. F=ma is the total force on an object under acceleration.

 

In this case, the force on the scale comes from the weight of the ball (F=mg) and the impact of the ball with the scale (Ft=mv). You can figure out the velocity of the ball once it hits the scale, and you know its mass. The only question that remains is how much time (t) took for the scale to absorb the entire force. Or, in other words, how much time it took for the ball to come to complete rest after it hit the scale.

 

That depends on the spring inside the scale, mostly. Unless you have the spring constant (k), you need to have this piece of information in advance. If you have the spring constant you can calculate it by using the spring equation (F=-kx).

 

~moo

 

I think this over-complicating the equation. The original question mentions nothing about applying Ft=mv, I think it would be safe to assume they want to know the weight of the ball the instant it hits the scale and not worry about factoring any of that stuff in

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I think this over-complicating the equation. The original question mentions nothing about applying Ft=mv, I think it would be safe to assume they want to know the weight of the ball the instant it hits the scale and not worry about factoring any of that stuff in

No, but when a ball drops onto the scale, this force is implied. Otherwise, F=mg will answer the question if the ball was put down ("gently") on the scale.

 

The extra factor here is the fact the ball is DROPPING from a height onto the scale. This is physics. It's an extra force that you must take into account. If you want it less complicated, make the question less complicated ;)

 

~moo

 

 

 

PS: I didn't want to really complicate thing s(although in my view it would be simpler) but my initial thought was just to consider energies, solve, and transform to force. You begin with potential energy (U=mgh), and you finish with Kinetic (K=1/2 mv^2) and with Elastic (Ue=1/2 kx^2). Equate the 'starting' and 'ending' energies, and you can solve the question. I think it's easier to think about, but I guess that's a mattr of opinion. YOu can solve the question either way.

 

You can't ignore the impact, though. If you throw a ball at the scale, the impact is absolutely part of the forces you must factor in, whether it's comfortable or not, it exists. It was also mentioned, I believe, in the second or third reply on this thread.

 

~moo

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It depends, on how you define the question, really. The "weight" of the ball does not change, in either scenario, but the "weight, shown by the scale" will be drastically different if it's dropped from 4 meters (as mentioned above, how much different, depends on the internal mechanisms of the scale itself).

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It depends, on how you define the question, really. The "weight" of the ball does not change, in either scenario, but the "weight, shown by the scale" will be drastically different if it's dropped from 4 meters (as mentioned above, how much different, depends on the internal mechanisms of the scale itself).

True. But the question asked what the force on the scale was, which means you need to take into account all forces that apply on the scale.

 

 

~moo

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Hi all,

 

Wow this really is complicated. Thanks all for the answers so far. At first I thought the answer would be simple to create: what weight would a scale show if a 10-lb bowling ball were dropped from 13 feet? Is there a better way to frame this question? Let me try. What are the differences in force exerted between a ten pound bowling ball dropping at 1 mph, 22 mph, and 50 mph? Would the force be measured in joules or how? I just want to find some kind of force ratio here.

 

Thanks again all,

Chris Mohr

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Force is always measured in Newtons. But what a scale reads is the normal force, rather than the weight, i.e. it is measuring the force it is exerting on the object (the reaction force of what is being exerted on it), and it is assumed that you are in a static situation when you interpret that as the weight.

 

If you want to know what the maximum the scale will read, another way to look at it is the potential energy. For a spring, this is 1/2 kx^2, and ideally this will equal the kinetic energy the ball has when it strikes. If you have the calibration of the spring of the scale (i.e. you know k), then you can find what the scale will indicate. But without additional information about the scale, I don't think you have enough data to determine what the reading will be.

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Not really -

in physics "weight" refers to

The weight of an object, is the force exerted by that object on its support

(http://en.wikipedia.org/wiki/Weight)

 

Mass however refers to:

active gravitational mass and passive gravitational mass. In everyday usage, mass is often taken to mean weight, but in scientific use, they refer to different properties

(http://en.wikipedia.org/wiki/Mass)

 

One thing I always remember with physics was they always try to trick question you with the terms to test your level of understanding so be very careful to read the question and look at the key points.

 

"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?"

 

it would weigh ten pounds

what weight would the scale show

 

This tells you which formula to apply

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Not really -

in physics "weight" refers to

The weight of an object, is the force exerted by that object on its support

(http://en.wikipedia.org/wiki/Weight)

That's true, but the question is asking for the weight that appears on the scale. The scale measures the 'normal force' (which is the proportional opposite force exerted back at the object by the scale's spring), which is what swansont said.

 

Mass however refers to:

active gravitational mass and passive gravitational mass. In everyday usage, mass is often taken to mean weight, but in scientific use, they refer to different properties

(http://en.wikipedia.org/wiki/Mass)

 

One thing I always remember with physics was they always try to trick question you with the terms to test your level of understanding so be very careful to read the question and look at the key points.

 

In physics, mass is the amount of matter (density x volume) in an object. In SI units, the questions are usually easier, because you have total separation of mass and weigh units -- gram/kg for mass, and Newton for weight.

 

Strictly speaking, pound is a force while slug is the unit of mass, but convention changed by use. Pound is what you see on the scale, usually already converted, and people tend to call it mass.

 

"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?"

 

it would weigh ten pounds

what weight would the scale show

 

This tells you which formula to apply

 

I think we might do well if we translate this to SI units so we stop looking at the minor details and focus on the actual question ;)

 

If we look at the question physically, it's asking about the difference between measuring the weight (what the scale shows, which is the normal-force, translated usually to weight already, using wrong units) of a stationary ball vs. measuring the same for a falling ball.

 

There's no need to overcomplicate matters. A stationary ball on the scale will show the normal force to be F=mg. A Falling ball will add an extra effect of an impact force, for a limited time, F=mv/t. After the impact is 'delivered', the scale will rest back on only showing F=mg, assuming the impact didn't break it.

 

If the goal is to see the maximum 'reading' of the scale, you need to take the maximum mv/t and add that to the 'mg' force.

 

It's all about framing the question, but seeing as the OP admitted he's not extremely well versed in physics, I think it might be better to try and not confuse him further by getting into the confusion of weight/mass/force/normal-force and just look at the question from a principle point of view.

 

~moo

 

~moo

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That's true, but the question is asking for the weight that appears on the scale. The scale measures the 'normal force' (which is the proportional opposite force exerted back at the object by the scale's spring), which is what swansont said.

 

 

 

In physics, mass is the amount of matter (density x volume) in an object. In SI units, the questions are usually easier, because you have total separation of mass and weigh units -- gram/kg for mass, and Newton for weight.

 

Strictly speaking, pound is a force while slug is the unit of mass, but convention changed by use. Pound is what you see on the scale, usually already converted, and people tend to call it mass.

 

 

 

I think we might do well if we translate this to SI units so we stop looking at the minor details and focus on the actual question ;)

 

If we look at the question physically, it's asking about the difference between measuring the weight (what the scale shows, which is the normal-force, translated usually to weight already, using wrong units) of a stationary ball vs. measuring the same for a falling ball.

 

There's no need to overcomplicate matters. A stationary ball on the scale will show the normal force to be F=mg. A Falling ball will add an extra effect of an impact force, for a limited time, F=mv/t. After the impact is 'delivered', the scale will rest back on only showing F=mg, assuming the impact didn't break it.

 

If the goal is to see the maximum 'reading' of the scale, you need to take the maximum mv/t and add that to the 'mg' force.

 

It's all about framing the question, but seeing as the OP admitted he's not extremely well versed in physics, I think it might be better to try and not confuse him further by getting into the confusion of weight/mass/force/normal-force and just look at the question from a principle point of view.

 

~moo

 

~moo

 

Just for the record, I'm not disagreeing that all these factors are points to consider, but your last paragraph really sums up my point in that the OP was from a guy not a student and more than likely not so complicated a question, wouldnt the required answer be a simple vector force calculation? Just seems this is what the question probes for taking into consideration the units it gives for calculation. Otherwise they would give a resistance of the spring etc. Granted that a (t) unit could be derived from the units given in the question however.

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Not really -

Not really, not really what? The original poster was inquiring about the maximum value that would be shown on the scale. That number depends on the spring coefficient of the scale.

 

in physics "weight" refers to

The weight of an object, is the force exerted by that object on its support

(http://en.wikipedia.org/wiki/Weight)

Terrible article, and yet another example of why wikipedia is (or can be) an awful source.

 

There are (at least) four technical definitions of the term weight.

  1. Legally (and colloquially), weight is a synonym for mass. It is not a unit of force in this sense. This sense predates the use of the term in physics, so in a sense this is the definition that should prevail. Look at it this way: We scientists do not like it when the lay community usurps/abuses scientific terminology (e.g., "Evolution is only a theory"). We similarly should not usurp/abuse common terms.
     
     
  2. In most introductory physics texts (up to sophomore year in college) and many engineering texts, weight is a synonym for gravitational force. There is one problem with this definition: It is an immeasurable quantity.
     
     
  3. In a few introductory physics texts and some aerospace engineering texts, weight is the net sum of all forces acting on a body except the gravitational force.
     
     
  4. Most general relativity texts use a definition that is closely allied with definition #3: weight is a synonym for proper acceleration, which is what ideal accelerometers measure. Multiply this by rest mass and you get something very similar to what scales measure.

 

Scales measure the net sum of all forces acting on a body except the gravitational force. In other words, it is definitions #3/#4 that are in play here, not the freshman physics definition of weight (which, once again is immeasurable), and not the legal/colloquial definition, either.

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Just for the record, I'm not disagreeing that all these factors are points to consider, but your last paragraph really sums up my point in that the OP was from a guy not a student and more than likely not so complicated a question, wouldnt the required answer be a simple vector force calculation? Just seems this is what the question probes for taking into consideration the units it gives for calculation. Otherwise they would give a resistance of the spring etc. Granted that a (t) unit could be derived from the units given in the question however.

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, why these aren't "simple vector calculations" or what your caviat point was meant to say.

 

Physics is clear, but not necessarily simple, at least not for everyone. The two solutions we posted are simple to people who know physics. They might not be as simple to people who aren't as well versed in it.

 

 

~moo

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