Calculating Pressure on a Roof Due to a Sudden Change in Atmospheric Pressure

[vaxop]

Pressure on a roof (help!)

A house has a roof (colored gray) with the dimensions shown in the drawing. Determine the magnitude and direction of the net force that the atmosphere applies to the roof when the outside pressure rises suddenly by 10.0 mm of mercury, before the pressure in the attic can adjust.

ive been on this for hours and i can't figure it out

 

[Sirus]

Please show us the work that you have done for these hours.

 

[vaxopy]

its on paper lol (and a mess :P)

what program do u use to make equations look properly online?

 

[Sirus]

It's a typesetting language supported by the forums called Latex. See here for an introduction.

 

[vaxopy]

darn anyone have any ideas?

 

[Sirus]

Begin by converting the mercury reading to a more convenient unit of pressure. Think about pressure, surface area, and force; don't forget about direction...

You need to show us an effort at doing the problem...at least your reasoning or something.

 

[vaxopy]

I got the basics figured out..
its the angle that's bothersome :/

P = (13600kg/m^3 * 10mm * 9.81m/s^2) / 10^4 = 133.416N/m^2 (change in pressure)

The surface area is 14.5m * 4.21m = 61.045m, or 122.09m for the whole roof.

Id think that the pressure would just be directed straight down, and its magnitude would be cos30 * 133.416 = 115.5N/m^2

I know that the pressure will be directed (somewhat) down, because a change in pressure outside would be similar to going further down in water

but this seems waaaaay too easy.. the question is supposed to be very difficult :/

 

[michaelw]

I don't think that the two magnitudes would cross each other out..

But i don't know ><

 

[vaxopy]

am i on the right track?

 

[Sirus]

I think the pressure is the same regardless of the inclination of the surface. To find the magnitude of the net force on the roof, find the force vector for each of the two parts of the roof, then add vectorally.

 

[learningphysics]

I got the basics figured out..
its the angle that's bothersome :/

P = (13600kg/m^3 * 10mm * 9.81m/s^2) / 10^4 = 133.416N/m^2 (change in pressure)

The surface area is 14.5m * 4.21m = 61.045m, or 122.09m for the whole roof.

Id think that the pressure would just be directed straight down, and its magnitude would be cos30 * 133.416 = 115.5N/m^2

I know that the pressure will be directed (somewhat) down, because a change in pressure outside would be similar to going further down in water

but this seems waaaaay too easy.. the question is supposed to be very difficult :/

Double check your units conversion to N/m^2... I get 1333N/m^2. I believe your 10^4 in the denominator should be 10^3.

The way you've worked the problem is fine so far (except the conversion I believe). Just find the two forces on the two parts of the roof and add vectorially as Sirius said.