Question

A triangular plate with height 4 ft and a base of 6 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/ft3.)

Answer 1

3281.25 lb

Step-by-step explanation:

Given that:

A triangular plate is submerged vertically in water 3 ft depth from the surface of the water.

A diagrammatic representation is shown below illustrating what we have in the question.

From the diagram; we divide the region D into horizontal strips of equal width [0,4]

Let's take a look at the triangle OPS and RSQ

`(p)/(6)=(4-x_i)/(4)`

`p=6((4-x_i)/(4))`

hence, the Area of the strip =

`P_2=p \delta x`

`P_2=6((4-xi)/(4)) \delta x`

The pressure on the strip is :

`\delta d_i=62.5(3+xi)`

And the force on the strip is:

`\delta d_iA_i=62.5(3+xi^*)6((4-xi)/(4)) \delta x`

The total force is gotten by the addition of the forces on all the strips and taking the limit

`F= \lim_(x \to 0) E^*_(i=1)62.5(3+xi)6((4-xi)/(4)) \delta x`

`=\int\limits^4_0 62.5(3+x)(6)((4-x)/(4)) \, dx`

`=(6(62.5))/(4) \int\limits^4_0 (3+x)(4-x) \, dx`

`=(375)/(4)\int\limits^4_0(12-3x+4x-x^2) \, dx`

`=(375)/(4)\int\limits^4_0(12+x-x^2) \, dx`

`=(375)/(4)[(12x+ (x^2)/(2)- (x^3)/(3) ]^4_0`

`=(375)/(4)[(12(4)+ ((4)^2)/(2)- ((4)^3)/(3) ]`

`=(375)/(4)[48+8-21 ]`

`=(375)/(4)[35 ]`

= 3281.25 lb

Therefore, the hydrostatic force = 3281.25 lb

Answer 2

The answer to the question is

`93.75\int\limits^7_3 {y^2-11y+28} \, dy` = 250 lb force

Step-by-step explanation:

The given variables are

Height of triangular plate = 4 ft

Base of triangular plate = 6 ft

The hydro-static force on the triangle is the product of pressure at a particular depth and the area of the surface of the triangle at that depth

the width of the triangle is given by

slope =

Here we have y = -4/3x +4 and the depth from the top at any y is 7 -y

Therefore pressure = 62.5×(7-y)

And the area of a strip is is given by 2 times the x coordinate and Δy

That is area =
`2(((4-y)3)/(4) )`×Δy

Force on a strip is given by Force = Pressure × Area

= 62.5×(7-y)×
`2(((4-y)3)/(4) )`×Δy Which gives 93.75 ×(7-y)(4-y)Δy

Hence hydrostatic force =
`93.75\int\limits^7_3 {(7-y)(4-y)} \, dy` =
`93.75\int\limits^7_3 {y^2-11y+28} \, dy`

=
`93.75[{(y^3)/(3) -11(y^2)/(2) +28y} ]\limits^7_3` = 250 lb force