Solving for Car Mass with Gauge Pressure and Surface Area | Physics Problems

[jrd007]

Fluid Problems

1) The gauge pressure in each of the four tires of a car is 240kPa. If each tire has a "footprint" of 220cm^2, estimate the mass of the car. answer: 2.2 x 10^3 kg pressure = 240 kPa, Area = 2.2 x 10^3 kg

The first thing I assume i need to do is convert kPa but I have not found that in my book... I know that pressure = Force/surface area = mg/A. I could use that formula to solve for mass?

240kPa = (m x 9.8 m/s^2)/2.2 m^2

2) A balloon has a radius of 7.35 m and is filled with helium. How large a cargo can it lift, assuming that the skin and structure of the balloon has a mass of 930 kg. (Neglect the buoyant force on the cargo volume itself).920 kg

r = 7.35 m, m = 930 kg, cargo mass = ? Again I have no idea how to attempt this. This chapter has been extremely hard for me.

3) What is the likely idenity of a metal if a sample has a mass of 63.5g when measured in air and an apparent mass of 55.4 g when submerged in water? (It also gives us a table and let's us know the densities of several metals but the important one is steel/iron which is 7.8 x 10^2 kg/m^3)

How do we use the two masses given to find out that it has that density. I have no ideas about this one.

The answer is iron/steel

 

[SoccaCrazy24]

for 1)
kPa = Kilo Pascal
Pascal = 1 Newton/meter^2
so you have your pressure in N/m^2 and your area is in cm^2 so you can easily change area from cm^2 to m^2

 

[jrd007]

Yes I know but I mean the kPa part. I know that 220 cm^2 = 2.2 m^2.

Would 240kPa = 240/1000 = .24 Pa = .24 N/m^2

 

[jrd007]

So would it be .24 N/m^2 x 2.2 m^2 = m(9.8) ?

because that does not work...

 

[SoccaCrazy24]

ok well... kilo means thousand so you don't divide bya 1000... you multiply by a thousand... and you also have to remember there are 4 tires supporting a car so you have to account for all 4 tires... but I am unsure of whether you converted your area correctly because its squared... so you can't just divide by 100 if i am not mistaken

 

[jrd007]

Actually there is an example in my book and it has m^2.

24,0000 Pa x 2.2 m^s = 4m x 9.8 m/s^s

so if I divide both sides by 9.8 and 4 m I get 13,469... that is not correct.

* 1 Pa = 1 N/m^2

 

[Fermat]

1 Pa = 1 N/m²
1 kPa = 1 kN/m² = 1000 N/m²
======================

1 cm = 1/100 of 1 m
1 cm² = 1/100² of 1 m²
220 cm² = 220/100² of 1 m²
220 cm² = 0.022 m²
===============

You know the pressure, and you can work out the total area from the 4 footprints, so using F = PA, you can get the total force provided by the pressure in the tyres, hence the weight of the car.

 

[mezarashi]

As Fermat's noted, understand the units conversion first. Then apply

Fgrav = 4PA (since there are 4 tires)

For your other problems, apply the concept of the buoyant force. The buoyant force acting on any object submerged in a fluid (including air) is the weight of the displaced fluid.

$$F_B = \rho V g$$
where rho indicates the fluid's density, and V is the volume of the object.

You can use the buoyant force in a force diagram like you would any other. For example in your third question, you have the condition:

$$F_{net} = 0 = F_{grav} - T - F_B$$

 

[Integral]

1.

You know that
$$P= \frac {m g} A$$
The first step is to solve for the quantity you need, that is mass. so:

$$m = \frac {PA} g$$

you have P= 240kPa

and A = 220 cm² for EACH TIRE. How many tires are on the average car?

How many cm² in 1 m²?

 

[jrd007]

So taking acrea of number 1 - m=PA/g = (4 x 240,000)(.022m)/(9.8)

= 2155 = 2.2 x 10^3 N!

Wow. I cannot believe it was that simple.

#2) I use $$F_B = \rho V g$$ some how.

I can figure out the volume by the equation (4/3)(pie)(r)^3
= 1662

But where do I go from there?

 

[Doc Al]

#2) I use $$F_B = \rho V g$$ some how.

I can figure out the volume by the equation (4/3)(pie)(r)^3
= 1662

But where do I go from there?

You calculate the buoyant force, using that formula. (You'll need the density of air.) That buoyant force must support the weight of the balloon structure, the helium inside the balloon (what's the density of helium?), plus any added cargo. Set up the equation to solve for the maximum cargo that the balloon can lift.

 

[jrd007]

$$F_B = \rho V g$$

airs denisty = 1.29, helium's = 0.179

so I add those denisties together and then solve for Fb?

 

[Doc Al]

Why would you add the densities? The buoyant force is the weight of the displaced fluid, which, in this case, is air.

 

[jrd007]

Okay so...

(1.29)(1662)(9.8) = Fb
= 21011 N for a boyancy force.

But how does thi shelp me with the mass of the cargo?

 

[Doc Al]

The buoyant force is the upward force on the balloon. When maximally loaded, that upward force will be balanced by the downward forces. One of those downward forces is the weight of the cargo. (See post #11.)

 

[jrd007]

So I do something w/ the denisty of helium? Correct?

Find it's force? but how? And then once I have heliums I can minus the two forces to get the cargo mass?

 

[Doc Al]

There are three "parts" to the balloon/cargo system: the helium, the balloon structure, and the cargo. It is their weight that pulls the system down. You'll need the density of helium to find the weight of the helium.

 

[jrd007]

As I previously stated the denisty of helium is 0.179

weight of balloon = (930 kg)(9.8) = 9114 N

F(boyancy) = 21011 N

How do I change helium, by using denisty, into a mass?

 

[Doc Al]

What's the definition of density?

 

[jrd007]

Mass/volume.

 

[Doc Al]

Right. Since you know the volume and density, you can calculate the mass of the helium.

 

[jrd007]

Okay so .179 = mass/1662
mass = 297 kg

Then to get the Force in which that mass exerts I take that times 9.8 = 2915 N

Now I know Fb = 21011 N, Heliums mass = 297 (F = 2915N) and the ballon's structure = 930 kg (F = 9114 N)

But how will I use that to get the weight of the cargo, which is 920 kg?
If I saw 21011 - 2915 - 9114 = mass of cargo x 9.8
I get a mass of 916... four digits off of 920...

 

[jrd007]

Did I miss calculate something?

 

[Doc Al]

I didn't check your arithmetic, but 916 rounds off to 920 very nicely.

 

[jrd007]

I see. So how do I relate what you are saying to # 3?

 

[Doc Al]

Just some arithmetic "housekeeping" advice to minimize calculation errors: Never grab for the calculator until the last step. Here's how I'd set up the problem:
$$\rho_a V g = \rho_h V g + 920 g + m g$$

Thus the "g" drops out:
$$m = (\rho_a - \rho_h) V - 920 = (\rho_a - \rho_h) (\frac{4}{3}\pi r^3) - 920$$

Only when I reached that last step would I plug in numbers and reach for the calculator. (And, using your numbers, you'll probably get a slightly different answer. Don't worry about it.)

 

[Doc Al]

So how do I relate what you are saying to # 3?

The goal of problem #3 is to figure out the density of the unknown metal. You have its mass. Use the given data (from which you can deduce the buoyant force) to figure out the volume of the sample. Give it a shot.

 

[jrd007]

Isn't bouyancy force = pVg? - I know neither p(denisty) or volume or the Fb so how can I solve for it?

 

[jrd007]

And I guess we could not use the true mass + apparent mass = volume?

= 118.9 cm3

 

[jrd007]

Also, just wanted to let you know when I did it your way with the

$$(\rho_a - \rho_h) (\frac{4}{3}\pi r^3) - 930$$ I get 917.8

 

[Doc Al]

Isn't bouyancy force = pVg? - I know neither p(denisty) or volume or the Fb so how can I solve for it?

The fluid is water, so you should know the density. Also: true weight - buoyant force = apparent weight.

 

[jrd007]

denisty of water is 1000 kg/m^3

So since mass is 63.5 I just divide that by the volume and tha give me density?

 

[Doc Al]

Yes. Once you figure out the volume you can get the density by $\rho = m/V$. (Be careful with units.)

 

[jrd007]

yes but is it not jus 63.5/1000 ? I thought that was why I was trying to find the volume of water?

 

[Doc Al]

1000 kg/m^3 is the density of water, not the volume of water displaced. (You'll need the density and the buoyant force to solve for the volume.)