Calculus problem related to physics

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elcalculusman

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A flat plate of negligible thickness is in the shape of a right triangle with base 5 ft and height 10 ft.
The plate is submerged in a tank of water. Find the force on the face of the plate where the plate is submerged horizontally so that it rests flat on the bottom of the tank at a depth of 14 feet. (Use 62.4 lb/ft^3 as the density of water)

I'll be honest, I don't really know what to do. I have been working on this lesson all day and threw the towel on this one, decided to register for this forum. Can anyone offer some direction or show me the way towards a solution?
 
elcalculusman said:
A flat plate of negligible thickness is in the shape of a right triangle with base 5 ft and height 10 ft.
The plate is submerged in a tank of water. Find the force on the face of the plate where the plate is submerged horizontally so that it rests flat on the bottom of the tank at a depth of 14 feet. (Use 62.4 lb/ft^3 as the density of water)

I'll be honest, I don't really know what to do. I have been working on this lesson all day and threw the towel on this one, decided to register for this forum. Can anyone offer some direction or show me the way towards a solution?
Click to expand...

If the plate is lying horizontally, every point on this plate will experience equal amount of pressure (62.4*14 psft - corrected) now multiply by the area of the triangle to find the force.
 
Hold up.. wouldn't it be...

V = area of triangle * height = 25 * 14 = 350
then Mass= Volume * density = 350 * 62.4 = 21840

then Force = mass * accel of gravity = 32 * 21840 = 698880... ???

This is my introduction to physics. =X
 
elcalculusman said:
Hold up.. wouldn't it be...

V = area of triangle * height = 25 * 14 = 350
then Mass= Volume * density = 350 * 62.4 = 21840

then Force = mass * accel of gravity = 32 * 21840 = 698880... ???

This is my introduction to physics. =X
Click to expand...

62.4 lb is weight not mass - the 'g' is already factored in here. So the last step should be omitted.
 
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