Physics: Pressure vs depth in liquid; total force on rectangular and triangular surfaces.
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See attachment #1 for a diagram showing parameters.
A solid prism has a vertical rectangular face of height H= .90 m by width W=.5 m and a horizontal rectangular face of length L= 1.2 m, with W=.5 m also. The prism is submerged in liquid whose density is D= 1600 kg/m^3 with the horizontal face at depth B= .35 m.
Find the total force exerted by the liquid on the horizontal rectangular surface of length L by width W at depth B.
Find the total force exerted by the liquid on the vertical rectangular surface, height H, width W, top edge at depth B.
Find the total force exerted by the liquid on the vertical triangular face with edge H vertical and edge L horizontal.
See attachment #2
The general relationship required to find pressure at some depth h in a liquid of density D, is (1) P= D g h. For finding force F, exerted by the liquid on a surface area A we need (2) F = P A or the element dF = P dA.
Note that since the horizontal surface is at constant depth B, the pressure is also constant over this surface, so for the total force on area LW we can write:
(3) F = D g B L W. Substituting all knowns should give you F= 3293 nt. To give an example of how the units involved work out, the substitution of knowns, is shown with units for proper physics, on attachment #2.
The solution provides an excellent explanation of how to solve the problem and calculates the answers.