Respuesta :
Answer:
The appropriate response is "[tex]\tau=\frac{1}{6} PgLh^3[/tex]". A further explanation is described below.
Explanation:
The torque ([tex]\tau[/tex]) produced by the force on the dam will be:
⇒ [tex]d \tau=XdF[/tex]
On applying integration both sides, we get
⇒ [tex]\tau = \int_{0}^{a}x pgL(h-x)dx[/tex]
⇒ [tex]= pgL\int_{0}^{h}(h-x)dx[/tex]
⇒ [tex]=pgL[\frac{h^3}{2} -\frac{h^3}{3} ][/tex]
⇒ [tex]=\frac{1}{6} PgLh^3[/tex]
The torque acting on the dam wall is due to the force of water pressure
which depends on the density and depth of the water and on gravity.
- [tex]\displaystyle The \ \mathbf{torque} \ at \mathbf{ \ point \ P}\ due \ to \ the \ reservoir \ water \ is, \, \underline{\tau = \frac{1}{6} \cdot \rho \cdot g \cdot w \cdot h^3 \right)}[/tex]
Reasons:
The question parameters are;
Width of the dam = w
Height of the dam = h
Required:
The torque due to the water about point P at the base of the dam
Solution:
The pressure of the water vary with depth
Force, F = Pressure, P × Area, A
Therefore, for an infinitesimal area, dA, where the pressure can be taken as constant, we have;
dF = P × dA
dA = w × dy
P = ρ·g·y
∴ dF = ρ·g·y·w·dy
Torque, τ = Force, F × Distance, d
The distance of an infinitesimal point below the surface of the water in the dam = h - y
Where;
h = The height of the water in the dam
The infinitesimal torque at an infinitesimal point where the pressure is constant is therefore;
dτ = dF × (h - y)
Which gives;
dτ = (h - y)·ρ·g·y·w·dy = ρ·g·w·(h·y - y²)·y
The torque is therefore;
[tex]\displaystyle \int\limits d\tau = \mathbf{ \int\limits^h_0 {\rho \cdot g \cdot w \cdot (h \cdot y - y^2)} \, dy}[/tex]
Which gives;
[tex]\displaystyle \tau = \int\limits^h_0 {\rho \cdot g \cdot w \cdot (h \cdot y - y^2)} \, dy = \mathbf{\rho \cdot g \cdot w \cdot \left[\frac{h \cdot y^2}{2} - \frac{y^3}{3}\right]^h_0} = \rho \cdot g \cdot w \cdot \left(\frac{h ^3}{2} - \frac{h^3}{3}\right)[/tex]
[tex]\displaystyle \tau = \rho \cdot g \cdot w \cdot \left(\frac{h ^3}{2} - \frac{h^3}{3}\right) = \rho \cdot g \cdot w \cdot \left(\frac{3 \cdot h ^3 - 2\cdot h^3}{6} \right) = \mathbf{ \rho \cdot g \cdot w \cdot \left(\frac{h ^3 }{6} \right)}[/tex]
[tex]\displaystyle The \ torque \ at \ the \ base \ of \ dam \ is, \, \tau = \frac{1}{6} \cdot \rho \cdot g \cdot w \cdot h^3 \right)[/tex]
The force of the torque at the base of the dam is on the same line of action as the point P, therefore, the torque at the base is equal to the torque at point P, which gives;
[tex]\displaystyle The \ torque \ at \ the \ base \ of \ dam \ is, \, \tau = \mathbf{\frac{1}{6} \cdot \rho \cdot g \cdot w \cdot h^3 \right)} = Torque \ at \ point \ P[/tex]
[tex]\displaystyle The \ torque \ at \ point \ P\ due \ to \ the \ reservoir \ water \ is, \, \underline{\tau = \frac{1}{6} \cdot \rho \cdot g \cdot w \cdot h^3 \right)}[/tex]
The parameters used are based on a similar question obtained online.
Learn more about the application of integration here:
https://brainly.com/question/19578741
https://brainly.com/question/18568839
