Calculating Force and Pressure: Vacuum Cleaner and Octopus Example"

[Izmad]

Homework Statement

a) A very powerful vacuum cleaner has a hose 2.91 cm in diameter. With no nozzle on the hose, what is the weight of the heaviest brick that the cleaner can lift?

(b) A very powerful octopus uses one sucker of diameter 2.91 cm on each of the two shells of a clam in an attempt to pull the shells apart. Find the greatest force that the octopus can exert in salt water 32.3 m depth.

Homework Equations

p=F/A
P=P(atm)+pgh
A=pi*r^2
density of salt water=1.025kg/m^3
P(atm)=1.013*10^5 Pa

The Attempt at a Solution

I'm a little confused. This is what I did with question b.
Pressure=P(atm)+(1.025)(9.8)(32.3)=1.016*10^5
A=pi*(.0146)^2=.0007
Area*Pressure=Force .0007*1.016*10^5=71.12 N
THIS IS WRONG. I'm not sure what I'm doing wrong. Can someone please explain

And i don't know what to do with question a. I found the area of the vacuum. But I'm not sure how to figure out a mass. Does P(atm) play a role in this? Could i get some direction please?

Thanks

 

[learningphysics]

Your units aren't right... what units are you using for pressure?

Your method for part b looks right to me...

use the same method for part a... what is the pressure in part a?

 

[Izmad]

No pressure is given for part a. 71.12 is wrong. So, i must be missing something.

both problems are in Newtons.

 

[Izmad]

also, maybe pressure= mg/A plays a role.

I'm still confused. :(

 

[learningphysics]

also, maybe pressure= mg/A plays a role.

I still confused. :(

what units of pressure are you using? 1.013*10^5 implies you're using Pa. That's fine.

But then you need to use density in kg/m^3. what units is 1.025? I think you need 1025kg/m^3

Use 1025 instead of 1.025.

 

[Izmad]

this is what i did.

1.013x10^5+ (1025)(9.8)(32.3)=4.261x10^5 Pa

4.261x10^5*.0007=298.2N

This answer was close but no cigar. It was within 10% of the correct value.
However, I'm still not sure what that value is. :(

Further help would be very much appreciated.

thank you

 

[learningphysics]

this is what i did.

1.013x10^5+ (1025)(9.8)(32.3)=4.261x10^5 Pa

4.261x10^5*.0007=298.2N

This answer was close but no cigar. It was within 10% of the correct value.
However, I'm still not sure what that value is. :(

Further help would be very much appreciated.

thank you

you are rounding too much. especially the area 0.0007... and use the exact value for radius 0.01455m.

you need to keep more decimal places.

don't round till the very end, when you get your final answer.

 

[Izmad]

Ahh, well I just changed the float of my calculator to 7. This has helped. I got my final answer to be 283.2N. Thanks for the advice learningphysics.

Also, are there any ideas in regards to problem a? I'm still unsure how to arrive at a mass, and what pressures to use.

thank you

 

[learningphysics]

Ahh, well I just changed the float of my calculator to 7. This has helped. I got my final answer to be 283.2N. Thanks for the advice learningphysics.

Also, are there any ideas in regards to problem a? I'm still unsure how to arrive at a mass, and what pressures to use.

thank you

The pressure would just be atmospheric pressure ie: 1.013*10^5 Pa. other than that everything is just like part b.

You get the force using pressure... then

we need:

Fpressure - mg = 0

 

[Izmad]

area*pressure=mass*gravity

wow, i don't know what I'm doing wrong, i can get the correct answer. :(

but thank you very much for your help. You are wonderful. :)

 

[learningphysics]

area*pressure=mass*gravity

wow, i don't know what I'm doing wrong, i can get the correct answer. :(

but thank you very much for your help. You are wonderful. :)

thanks. you're welcome. did you mean you can't get the correct answer? what number are you getting?