Force on a Surface
Maths and statistics discussion, revision, exam and homework help.
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Force on a SurfaceOk so the example mentions a pipe strikes a nearby wall without rebounding.
The cross sectional area of the pipe is 6cm^2 and the water travels at 4 m/s.
I have to find the force acting on the wall given that the density of the water is 1000kg/m^3
it says
And I just don't get this maths, anyone care to explain?Last edited by I am Ace; 29-07-2012 at 14:14. -
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Re: Force on a SurfaceOK so how would I find the volume when the area = 0.015m^2?(Original post by Joinedup)
Think about the cylinder of water coming out of the pipe in one second. The volume of that cylinder is going to be given by length (4m) multiplied by it's crosssectional area isn't it?
I tried multiplying the velocity by 0.015^2 which didn't work out -
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Re: Force on a Surfaceif you're given a value for the area you don't square it, it's in area units which are length units squared.(Original post by I am Ace)
OK so how would I find the volume when the area = 0.015m^2?
I tried multiplying the velocity by 0.015^2 which didn't work out -
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Re: Force on a Surfaceas I understand it you've got a pipe squirting out an endless cylinder of water at a constant velocity.(Original post by I am Ace)
So how would I do it?
The length of the part of that endless water 'cylinder' coming out of the pipe in a fixed amount of time is the velocity (i.e. Distance/time) of water coming out of the pipe multiplied by the time. Do you agree?
The volume of the water cylinder will be the length multiplied by the area, if they've given you the area in cm^2 you need to convert to the same units as you're using for the length, meters. To convert an area in cm^2 to an area in m^2 you divide by 10000.
Once you've got your length in m and your area in m^2 you multiply the length by the area to get a volume in m^3. this is the volume that comes out of the pipe in the specified time.
