Question

A triangular plate with base 4 m and height 5 m is submerged vertically in water so that the tip is even with the surface. Express the hydrostatic force against one side of the plate as an integral and evaluate it. (Round your answer to the nearest whole number. Use 9.8 m/s2 for the acceleration due to gravity. Recall that the weight density of water is 1000 kg/m3.)

Answer 1

hydrostatic force is 327000.00 N

Step-by-step explanation:

given data

base = 4 m

height = 5 m

density of water = 1000 kg/m3

to find out

hydrostatic force

solution

we know this is a triangle so we consider here a small strip PQ whoes area da in with length x and width dy

so area will be

da = base/H × height = base/H × y

so da = x dy = base/H × y dy

and we know pressure = ρ × g × h

here h = y

hydrostatic force = pressure × area

df = (ρ × g × h) × base/H × y dy

now integrate it from 0 to 5 height

f = ρ × g
`\int_(0)^(5) (h) (base/H)ydy`

f = ρ × g
`\int_(0)^(5) (y) (4/5)ydy`

f = ρ × g × 4/5 ×
`(y^(3) /3)^(5) _0`

now put value ρ = 1000 and g = 9.81

f = ρ × g × 4/5 ×
`(y^(3) /3)^(5) _0`

f = 1000 × 9.81 × 4/5 × (5³/3)

force = 7848 × 41.666667

force = 327000.00 N

hydrostatic force is 327000.00 N

Answer 2

The hydro-static force is 326666.67 N.

Step-by-step explanation:

Given that,

Base = 4 m

Height = 5 m

We need to calculate the area of strip

using similar triangle,

`(y)/(x)=(B)/(H)\Rightarrow y=(B)/(H)x`

Area of strip,
`dA =y\ dx`

`dA =(b)/(H)* x\ dx`

Where, b = base

dx = width of strip

H = height

Put the value of base into the formula

`dA=(4)/(5)* x dx`

We need to calculate the force

Using formula of force

`dF= P* dA`

Where, P = pressure

A = area

Put the value into the formula

`dF=\rho g h* dA`

`dF=\rho g x*(4)/(5)xdx`

On integration

`\int_(0)^(F)dF=\int_(0)^(5){\rho g x}*(4)/(5)* xdx`

`F = \rho g\int_(0)^(5)((4)/(5)x^2)dx`

`F=\rho g*(4)/(5)((x^3)/(3))_(0)^(5)`

`F=1000*9.8*(4)/(5)*(5^3)/(3)`

`F=326666.67\ N`

Hence, The hydro-static force is 326666.67 N.