Simple physics. Need help

[Sokar408]

So I'm self studying physics at gymnasium level, so that I can study at the university to become an engineer. However as so happens with things you have to do on your own, you run into roadblocks that, in hindsight, will look retardedly easy. I think I'm in one of those situations. On the topic of "Force"

PLEASE keep in mind that I'm translating this on the fly, from Danish to English, so if something doesn't sound right, its probably me translating something wrong

I am asked to:

Surface area: 12 m^2

Wood tile:

5 cm high

1.9 g/cm^3

Concrete:

10 cm high

2.3 g/cm^3

Styrolit Isolation

25 cm high

75 kN/m^2 (Pascal)

a. Calculate the gravitational force of the concrete and wooden tiles. Now I'm assuming this basically means, to calculate the weight of the concrete and wooden tiles. So basically recalculate the density from g/cm^3, to kg/m^3, and multiply that, by the m^3 amount of the wooden tiles and concrete. Is this correct?

b. Hvor big is the maximum force, the isolation can withstand, when the applied force is evenly distributed? This would baffles me. I am already told that Styrolit can withstand 75 kN/m^2, which as far as I'm aware, is the maximum force it can stand. I'm not sure what I am suppose to do here.

c. How many kg of furniture and other goods, can the floor withstand (evenly distributed). Here I am assume I need to know how much of a maximum force the isolation is taken up by the wooden tiles and concrete, and the left over is the answer. Since the formula stands that the area isn't relevant (I'll show why below) when only dealing with force, I can simply calculate the Pascal for the tiles and concrete, which I get to:

Wooden tile: (1,9 g/cm^3 * 100 cm * 100 cm * 100 cm)/1000 g/kg = 1900 kg/m^3

Concrete: (1,9 g/cm^3 * 100 cm * 100 cm * 100 cm)/1000 g/kg = 2300 kg/m^3

The formula for force is P = P(gravitational force on mass) + P(Atmosphere = 101325 Pascal) = ÃÂahg/a = ÃÂgh (unit: N/m^2)

So I get the following

Gravitational force of Wooden tile: 1900 kg/m^3 * 9.82 m/s^2 * 0.05 m = 932,9 N/m^2 (Pascal)

Gravitational force of concrete tile: 2300 kg/m^3 * 9.82 m/s^2 * 0.10 m = 2258,6 N/m^2 (Pascal)

Total 3191,5 N/m^2 (Pascal) = 3,2 kN/m^2 (Pascal)

That means that of the 75 kN/m^2 (Pascal), the isolation can take, only 3.2 kN/m^2 (Pascal) is used by the concrete and wooden tiles? This seems completely insane, and I must have something wrong here, but I can't figure out what. And if I add the atmosphere here, it'll just be way above it, and that is even more ludicrous. Please help.

Edited March 5, 2015 by Sokar408

[-Velocity-]

So I'm self studying physics at gymnasium level, so that I can study at the university to become an engineer. However as so happens with things you have to do on your own, you run into roadblocks that, in hindsight, will look retardedly easy. I think I'm in one of those situations. On the topic of "Force"

PLEASE keep in mind that I'm translating this on the fly, from Danish to English, so if something doesn't sound right, its probably me translating something wrong

I am asked to:

Surface area: 12 m^2

Wood tile:

5 cm high

1.9 g/cm^3

Concrete:

10 cm high

2.3 g/cm^3

Styrolit Isolation

25 cm high

75 kN/m^2 (Pascal)

a. Calculate the gravitational force of the concrete and wooden tiles. Now I'm assuming this basically means, to calculate the weight of the concrete and wooden tiles. So basically recalculate the density from g/cm^3, to kg/m^3, and multiply that, by the m^3 amount of the wooden tiles and concrete. Is this correct?

Don't forget you have to multiply by 9.81 m/s^2 to get the force in Newtons... kg is a unit of mass.

b. Hvor big is the maximum force, the isolation can withstand, when the applied force is evenly distributed? This would baffles me. I am already told that Styrolit can withstand 75 kN/m^2, which as far as I'm aware, is the maximum force it can stand. I'm not sure what I am suppose to do here.

75 kN/m^2 is a unit of force per surface area- of pressure. So you multiply that by your surface area (12 m^2) to get the maximum force.

c. How many kg of furniture and other goods, can the floor withstand (evenly distributed).

Divide the previous answer you got (in Newtons) by 9.81 m/s^2 to convert from Newtons to kilograms.

The formula for force is P = P(gravitational force on mass) + P(Atmosphere = 101325 Pascal) = ÃÂahg/a = ÃÂgh (unit: N/m^2)

.

You should ignore atmospheric force, you don't build floors over a vacuum. There is going to be a roughly equal atmospheric pressure on the other side too. Unless you're supposed to factor in the buoyancy force... which would be ridiculous because it would fairly negligible.

I may not be interpreting your problem right though, because they give you the thickness of the floor, I think, and you don't use it. But they often give you things you don't need in physics problems to try to trick you into using them.

[PB666]

So I'm self studying physics at gymnasium level, so that I can study at the university to become an engineer. However as so happens with things you have to do on your own, you run into roadblocks that, in hindsight, will look retardedly easy. I think I'm in one of those situations. On the topic of "Force"

PLEASE keep in mind that I'm translating this on the fly, from Danish to English, so if something doesn't sound right, its probably me translating something wrong

I am asked to:

Surface area: 12 m^2

Wood tile:

5 cm high

1.9 g/cm^3

Concrete:

10 cm high

2.3 g/cm^3

Styrolit Isolation

25 cm high

75 kN/m^2 (Pascal)

a. Calculate the gravitational force of the concrete and wooden tiles. Now I'm assuming this basically means, to calculate the weight of the concrete and wooden tiles. So basically recalculate the density from g/cm^3, to kg/m^3, and multiply that, by the m^3 amount of the wooden tiles and concrete. Is this correct?

b. Hvor big is the maximum force, the isolation can withstand, when the applied force is evenly distributed? This would baffles me. I am already told that Styrolit can withstand 75 kN/m^2, which as far as I'm aware, is the maximum force it can stand. I'm not sure what I am suppose to do here.

c. How many kg of furniture and other goods, can the floor withstand (evenly distributed). Here I am assume I need to know how much of a maximum force the isolation is taken up by the wooden tiles and concrete, and the left over is the answer. Since the formula stands that the area isn't relevant (I'll show why below) when only dealing with force, I can simply calculate the Pascal for the tiles and concrete, which I get to:

Wooden tile: (1,9 g/cm^3 * 100 cm * 100 cm * 100 cm)/1000 g/kg = 1900 kg/m^3

Concrete: (1,9 g/cm^3 * 100 cm * 100 cm * 100 cm)/1000 g/kg = 2300 kg/m^3

The formula for force is P = P(gravitational force on mass) + P(Atmosphere = 101325 Pascal) = ÃÂahg/a = ÃÂgh (unit: N/m^2)

So I get the following

Gravitational force of Wooden tile: 1900 kg/m^3 * 9.82 m/s^2 * 0.05 m = 932,9 N/m^2 (Pascal)

Gravitational force of concrete tile: 2300 kg/m^3 * 9.82 m/s^2 * 0.10 m = 2258,6 N/m^2 (Pascal)

Total 3191,5 N/m^2 (Pascal) = 3,2 kN/m^2 (Pascal)

That means that of the 75 kN/m^2 (Pascal), the isolation can take, only 3.2 kN/m^2 (Pascal) is used by the concrete and wooden tiles? This seems completely insane, and I must have something wrong here, but I can't figure out what. And if I add the atmosphere here, it'll just be way above it, and that is even more ludicrous. Please help.

Funny story.

I have a garage, essentially what we call here a tear down, it had huge cracks in the floor and ~11" (~28 cm) of subsidence in the back, when it rained the garage became a pond, that's how much subsidence had taken place, there was no way to tear the slab out. So I jacked up the garage walls and poured lightweight concrete under the edge with rebar, and the poured light weight concrete on Styrofoam sections (IIRC 4'x1.5"x15"; stack as needed) with occasional beam. I was really conservative about concrete use because of the weight and expense, and so the top-slab tended to range 3 to 4 inches (7.5 to 10 cm). I did this in order to keep the weight of the material down, the soil in our area is more or less a spongy mud so you don't want to put alot of force on one side of the foundation. During the last 2 hurricane, my wheels were bone dry. Worked so well I jacked up the adjacent slab and slid Styrofoam 1/2" x 4' x 8' sheets underneath and compressed sand to fill and that hasn't lost level in 10 years. Given one had moved 12" in 45yr and the other 6" (3' of tilt) in 45yr I would say its holding quite well. The stability I attribute to the styrofoam, it absolutely stopped wicking and I believe the wicking between rains was causing the soil to deorganify faster and shrink. It also absorbs shifts better than concrete (which tends to crack). I was sure the floor would crack and subside where the Styrofoam was, but to date no cracks, got a fully loaded pleasure boat wheels and a car wheels setting on it so . . . .yeah styro holds. Smartest thing you can do if you have to build up on soft soils.

Do an experiment yourself. Take a long pipe and a flat piece of Styrofoam (shipper foam will do), seal a cap on one end of the pipe, hold it strait up on the styrofoam and add water to the other end (or get help) you will see that it will take alot of water. You can remove the pipe and see how much the stryofoam has given, your are probably going to have to buy another section of pipe before it seriously damages the stryofoam. Now imagine the weight if you had a room filled with that much water (and proportional weight of pipe). When they pour the concrete on the foam beads there will be some dimpling (small rocks, aeration damage, etc) but after that it will resist compression. Where you will see perturbations of the foam is on the bottom, where roots, critters, cracks and deformations of the soil occur. If that was your concrete slab, there would be a good chance it would have cracked, the stryo is a buffer between the hideous DoT forces that occur underground and your slabs desire not to be bent or twisted (it only likes to be compressed).

[KSK]

Don't forget you have to multiply by 9.81 m/s^2 to get the force in Newtons... kg is a unit of mass.

75 kN/m^2 is a unit of force per surface area- of pressure. So you multiply that by your surface area (12 m^2) to get the maximum force.

Divide the previous answer you got (in Newtons) by 9.81 m/s^2 to convert from Newtons to kilograms.

You should ignore atmospheric force, you don't build floors over a vacuum. There is going to be a roughly equal atmospheric pressure on the other side too. Unless you're supposed to factor in the buoyancy force... which would be ridiculous because it would fairly negligible.

I may not be interpreting your problem right though, because they give you the thickness of the floor, I think, and you don't use it. But they often give you things you don't need in physics problems to try to trick you into using them.

I think you needed the height of the wood and concrete layers to calculate their mass for the first question. But I'm not sure why the height of styrolit was needed.

Edited March 6, 2015 by KSK

[-Velocity-]

I think you needed the height of the wood and concrete layers to calculate their mass for the first question. But I'm not sure why the height of styrolit was needed.

Is styrolit the name of the material the floor was made out of? I didn't really understand the questions if not. I think the OP needs a basic physics book, like a high school textbook. If he/she going to hopefully go to college, then the college can take it the rest of the way. Also, one is going to need high school algebra skills to get started in calculus.

[Sokar408]

Is styrolit the name of the material the floor was made out of? I didn't really understand the questions if not. I think the OP needs a basic physics book, like a high school textbook. If he/she going to hopefully go to college, then the college can take it the rest of the way. Also, one is going to need high school algebra skills to get started in calculus.

Styrolit is the name of the isolation under the concrete. According to google translate, it should be a thing in english as well. As for algebra and calculus. I don't have any issues with either (Though don't really call it algebra and calculus, we just call it math, and they are individually taught without classification. Not to mention we don't have high schools, or colleges at all. We have Elementary School 1-9th grade. Then Gymnasium 11-13 grade, and then University for 3-7, or sometimes 10 years, if very specific cases)

It should be apparent that its the question I'm unclear on, not the actually math part.

[KSK]

Sounds like Styrofoam to me but I could be wrong.

I read the question as being about a two layered floor (concrete base, wood tile upper surface) resting on a base of styrofoam insulation. From there:

1. Calculate the weight of the floor given it's density and sufficient information to calculate the volume of concrete and wood tile.

2. Calculate the maximum weight (force) the base can withstand given how much pressure it can withstand.

3. Calculate the mass of extra furniture the floor can support (subtract result of 1) from result of 2) and convert weight to mass)

Don't know if that helps?

[Sokar408]

Sounds like Styrofoam to me but I could be wrong.

I read the question as being about a two layered floor (concrete base, wood tile upper surface) resting on a base of styrofoam insulation. From there:

1. Calculate the weight of the floor given it's density and sufficient information to calculate the volume of concrete and wood tile.

2. Calculate the maximum weight (force) the base can withstand given how much pressure it can withstand.

3. Calculate the mass of extra furniture the floor can support (subtract result of 1) from result of 2) and convert weight to mass)

Don't know if that helps?

If that is right, then yes. I'm still flabbergasted by the wording of question two in the book though, because it spend 15 pages talking about "force" as being a unit of Newtons, not weight. But assuming poor wording is the only way to get anywhere with it. Given that my physics teacher apparently died yesterday (only 45, poor woman :/ ), I can't even ask her.

[-Velocity-]

If that is right, then yes. I'm still flabbergasted by the wording of question two in the book though, because it spend 15 pages talking about "force" as being a unit of Newtons, not weight. But assuming poor wording is the only way to get anywhere with it. Given that my physics teacher apparently died yesterday (only 45, poor woman :/ ), I can't even ask her.

The unit of force is Newtons. Weight is force generated by the product of mass (kg) and gravitational acceleration (m/s^2), and is given in Newtons as well. People sometimes (incorrectly) give weight in kilograms, because we all feel about the same gravitational acceleration.

[Ten Key]

It's not an accepted SI unit, but kilograms can be used as a unit of force, typically abbreviated as kg_f.

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

[-Velocity-]

It's not an accepted SI unit, but kilograms can be used as a unit of force, typically abbreviated as kg_f.

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

I know, but it's not going to be used in physics or in a physics book. Let's not confuse the guy, please.

At least he doesn't have to do some problems in U.S. customary units, like physics books here in the U.S. will do. They'll throw in a problem where everything is in feet, inches, gallons (or, God forbid, tablespoons), pounds force, pounds mass, PSI, horsepower, etc. Those problems are most easily solved by just converting the initial problem statement into SI, solve the problem, then convert the answer back to U.S. customary to make the book/teacher happy.

Edited March 6, 2015 by |Velocity|

[Ten Key]

Don't forget foot-pounds, slugs and BTUs.

I only mention it because I can remember being confused by this back in high school physics. As you say, the textbook will always treat kilograms as units of mass, and newtons as units of weight. But it is so common for people to use "kilograms" as a unit of weight outside of science and engineering that it's easy to get confused. I've never seen newtons used for weight outside of a STEM textbook or paper.

So when your friend says, "I weigh 70 kilograms", they are using kg_f. Your physics textbook would say, "Your friend weighs 686.5 newtons".

[-Velocity-]

Don't forget foot-pounds, slugs and BTUs.

I only mention it because I can remember being confused by this back in high school physics. As you say, the textbook will always treat kilograms as units of mass, and newtons as units of weight. But it is so common for people to use "kilograms" as a unit of weight outside of science and engineering that it's easy to get confused. I've never seen newtons used for weight outside of a STEM textbook or paper.

Newtons are used by engineers and scientists constantly, for any unit that involves force.

So when your friend says, "I weigh 70 kilograms", they are using kg_f. Your physics textbook would say, "Your friend weighs 686.5 newtons".

I look at it differently. When your friend says, "I weigh 70 kilograms", what he is really meaning is "I mass 70 kilograms", because we know he lives on Earth. If he were to say "I weigh 70 kilograms", and he lived on Mars, then you'd have to ask him what he meant. But it is true that scales that give weights in kilograms are actually measuring kilograms force, unless it's a balance scale.

Edited March 6, 2015 by |Velocity|

[PB666]

I know, but it's not going to be used in physics or in a physics book. Let's not confuse the guy, please.

At least he doesn't have to do some problems in U.S. customary units, like physics books here in the U.S. will do. They'll throw in a problem where everything is in feet, inches, gallons (or, God forbid, tablespoons), pounds force, pounds mass, PSI, horsepower, etc. Those problems are most easily solved by just converting the initial problem statement into SI, solve the problem, then convert the answer back to U.S. customary to make the book/teacher happy.

Yes but when I say I weigh 150 lbs, its factually correct (unless you are in Britian, in which a pound is a piece of paper with an ole lady's face illustrated on it, whose value appears to dive when any number of bank scandals happen, go figure that one).

this is an answer to multiple posts not just this one.

In the lab we 'weigh' out grams. There is a reason for that usage.

A grams weight is subject to change, but for the sake of an experiment if I want a 1 molar solution that is:

MW in g/mole * desire M in mole/liter * desired final volume in liter(s) = and so that answer is not in Newtons, the answer is in grams. To state it correctly would be to measure 1 gram on a balance and that was true 40 years ago when we used balances to weigh bigger things, and still true if you are measuring say 0.1 milligram on an analytical balance. But since we don't use balances anymore and the devise is electronic you really are weighing something and converting the weight to mass electronically.

When you go to the doctors office and he puts you on a scale (the tall thing with the sliding weights on it), that is actually a balance, he is measuring your mass and the scale units assume that gravity is 9.8m/s. So guess what both are wrong terms. Technically a weight can only come from a calibrated tension measuring device, and a mass can only directly come from a balance, and a pound is not a piece of paper with a lady's face on it.

But these restrictions appear to have no meaning in modern lexicon........

The real answer here is that unit conversion is real sucky, we should all convert to the kg m sec newt watt joule system and universally use scientific notation, just like everyone should use IAPUC nomenclature to describe chemicals (unlike some pharmaceuticals that have 2 dozen different names), but we don't and so space-ship on occasion has been known to crash into martian surface, with embarrassed red faces somewhere saying 'doh'. Live with it.

You live in a society where they trust garbagemen and oil derrick-trained CEO's opinion on global warming more than that of 99% of scientist and their preacher/reverend/priest more than evolutionary biologist on the topic of evolution, and where they believe the US constitution was written by god's reincarnation on earth . . . . . . . . . then the unit conversion issue is the least of your problems.

[YNM]

The difference of gravitational acceleration on various Earth surface is too small to be cared for general use - hence why the balance we use gives us the original reading (kilograms force ? Never actually used it) divided with a pretty much stable coefficient, the (mean ?) gravitational acceleration at Earth's surface.

Had it mattered, even the question where the Moon and Sun is and their current distance would matter so much, for every time we measure mass. (so the difference due to elevation and reliefs is in a similar magnitude to tidal force I guess)

Edited March 8, 2015 by YNM