- #1
member 428835
I'm trying to just find the hydrostatic force on the hinge of the gate in the image attached, and while not labeled, the thickness out of the page is ##w=5##. My thoughts are to set up this integral $$\vec{F}_p = \iint \rho g y \hat{n}\, dA = w \rho g \int_0^8 y \left( \frac{-dx\hat{i}+dy\hat{j}}{\sqrt{dx^2+dy^2}} \right)\, \sqrt{dx^2+dy^2}\\
= w \rho g \int_0^8 y \left(-dx\hat{i}+dy\hat{j} \right)\\
= w \rho g \int_0^8 y \left(y'(x)\hat{j}-1\hat{i} \right) \, dx\\
=5\cdot 64 \int_0^8 \left(9 + \frac{3 x}{4}\right)(\frac{3}{4}\hat{j}-\hat{i})\, dx\\
=-30720 \hat{i} + 23040 \hat{j}$$
where the magnitude of the last expression gives me the correct answer of the magnitude of ##\vec{F}_p=38,400## lbf. So when finding the force on the hinge I would say it is simply ##-30720 \hat{i} + 23040 \hat{j}## but the book says something else. Where am I going wrong?
= w \rho g \int_0^8 y \left(-dx\hat{i}+dy\hat{j} \right)\\
= w \rho g \int_0^8 y \left(y'(x)\hat{j}-1\hat{i} \right) \, dx\\
=5\cdot 64 \int_0^8 \left(9 + \frac{3 x}{4}\right)(\frac{3}{4}\hat{j}-\hat{i})\, dx\\
=-30720 \hat{i} + 23040 \hat{j}$$
where the magnitude of the last expression gives me the correct answer of the magnitude of ##\vec{F}_p=38,400## lbf. So when finding the force on the hinge I would say it is simply ##-30720 \hat{i} + 23040 \hat{j}## but the book says something else. Where am I going wrong?