A triangular plate with height 6 ft and a base of 8 ft is submerged vertically in water so that the top is 3 ft below the surface. How do you express the hydrostatic force against one side of the plate as an integral and evaluate it?

Question

A triangular plate with height 6 ft and a base of 8 ft is submerged vertically in water so that the top is 3 ft below the surface. Express the hydrostatic force against one side of the plate as an integral and evaluate it. (Recall that the weight density of water is 62.5 lb/#ft^3.#)

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Answer 1 · 2017-10-03T20:00:13

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It's #int_3^9# #4/3# #delta# (9 - x)x dx. Solving it gives us:

= #4/3# #int_5^9# #delta# (9 - x)x dx

= #4/3# #delta# #int_3^9# 9x - #x^2# dx

= #4/3# #delta# [#9/2# #x^2# - #1/3# #x^3# #]_3^9# dx

= #4/3# #delta# [#9/2# #9^2# - #1/3# #9^3# - (#9/2# #3^2# - #1/3# #3^3#)]

= #4/3# #delta# [#9/2# #81# - #1/3# #729# - (#9/2# #9# - #1/3# #27#)]

= #4/3# #delta# [#729/2# - #729/3# - (#81/2# - #27/3#)]

= #4/3# #delta# [#729/2# - #729/3# - #81/2# + #27/3#]

= #4/3# #delta# [#729/2# - #81/2# - #729/3# + #27/3#

= #4/3# #delta# [#648/2# - #702/3#]

= #4/3# #delta# [324 - 234]

= #4/3# #delta# [90]

= #360/3# #delta#

= 120 #delta#

Note that #delta# is the weight of density of water. Now we must use #delta# = 62.5

= 120(62.5)

= 7500