Answer :
Answer:
The answer to the question is
[tex]93.75\int\limits^7_3 {y^2-11y+28} \, dy[/tex] = 250 lb force
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 = [tex]2(\frac{(4-y)3}{4} )[/tex]×Δy
Force on a strip is given by Force = Pressure × Area
= 62.5×(7-y)× [tex]2(\frac{(4-y)3}{4} )[/tex]×Δy Which gives 93.75 ×(7-y)(4-y)Δy
Hence hydrostatic force = [tex]93.75\int\limits^7_3 {(7-y)(4-y)} \, dy[/tex] = [tex]93.75\int\limits^7_3 {y^2-11y+28} \, dy[/tex]
=[tex]93.75[{\frac{y^3}{3} -11\frac{y^2}{2} +28y} ]\limits^7_3[/tex] = 250 lb force
Answer:
3281.25 lb
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
[tex]\frac{p}{6}=\frac{4-x_i}{4}[/tex]
[tex]p=6(\frac{4-x_i}{4})[/tex]
hence, the Area of the strip =
[tex]P_2=p \delta x[/tex]
[tex]P_2=6(\frac{4-xi}{4}) \delta x[/tex]
The pressure on the strip is :
[tex]\delta d_i=62.5(3+xi)[/tex]
And the force on the strip is:
[tex]\delta d_iA_i=62.5(3+xi^*)6(\frac{4-xi}{4}) \delta x[/tex]
The total force is gotten by the addition of the forces on all the strips and taking the limit
[tex]F= \lim_{x \to 0} E^*_{i=1}62.5(3+xi)6(\frac{4-xi}{4}) \delta x[/tex]
[tex]=\int\limits^4_0 62.5(3+x)(6)(\frac{4-x}{4}) \, dx[/tex]
[tex]=\frac{6(62.5)}{4} \int\limits^4_0 (3+x)(4-x) \, dx[/tex]
[tex]=\frac{375}{4}\int\limits^4_0(12-3x+4x-x^2) \, dx[/tex]
[tex]=\frac{375}{4}\int\limits^4_0(12+x-x^2) \, dx[/tex]
[tex]=\frac{375}{4}[(12x+ \frac{x^2}{2}- \frac{x^3}{3} ]^4_0[/tex]
[tex]=\frac{375}{4}[(12(4)+ \frac{(4)^2}{2}- \frac{(4)^3}{3} ][/tex]
[tex]=\frac{375}{4}[48+8-21 ][/tex]
[tex]=\frac{375}{4}[35 ][/tex]
= 3281.25 lb
Therefore, the hydrostatic force = 3281.25 lb
