Fundamentals of Fluid Mechanics Questions and Answers  
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Question by AME536A Student
Is there any homework due on class 8/29/2019?
Question by AME536A Student
For Assignment #2 on question #1, I find that the surface area exposed to air pressure on the diagonal wall is related to the hypotenuse of the triangle, and thus is sqrt{L^2 + H^2}. The F=ma gravity component of the moment about the origin is related to L as well. As a result there is a big ugly equation with a relationship between L and L^2. Do I need to solve with a neat and tidy L=? Or can I leave the answer as something ugly like A*L^2 + BL + C = D?
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Question by AME536A Student
Are the answers shown on the bottom of the Homework #2 Assignment the correct answers? If they are the correct answers for the assignment, when do you hold office hours? I'm unsure of what I am doing wrong and would like to talk through it.
Sure you can come. Although my door is closed, I'm in my office almost all the time. Just knock. And yes, the answers are correct: work on the problems until you get the right answer.
Question by AME536A Student
On assignment #2 on question #4, what does the depth of the water have to do with the maximum buoyant force? If the river is infinitely deep, the maximum buoyant force will still only be related to the volume of the canoe. By depth, does the question mean "How deep the canoe's base is submerged into the water"?
Hint: check the definition of “floating” on Google.
Question by AME536A Student
I am getting stuck on problem #4 on the homework (the canoe question). I am able to do part A and it gives me the correct answer for part c, but I am having trouble with the integral on part B. The task is to integrate the forces instead of using Archimedes principle. The basic premise is pressure multiplied by area, so starting off, we can say the area integral for the submerged part of the boat is $$ \int_{\theta_0}^{\theta_1} \int_{0}^x rd \theta dx $$ (I couldn't get Latex to work with subscripts in the integral bounds. It should be $\theta^{}_{0}$ and $\theta^{}_{1}$). Next we can use Pascal's law to find the pressure: $$ P = P^{}_{atm} + \rho^{}_{w}g(H-y) $$ Here is where I am getting stuck. Originally I had the integral set up using Y and integrating from 0 to H multiplied by the area to get force: $$ \int_{\theta}^\theta \int_{0}^x \int_{0}^H r(P^{}_{atm} + \rho^{}_{w}g(H-y)) \sin(\theta) d \theta dx dy $$ This doesn't seem like the right way to integrate to find the force. There isn't really a y direction, it's just backward along the radius. So, my thought was to find Y as a function of $\theta$: $$ y = R(1-\sin(\theta)) $$ So the integral becomes $$ \int_{\theta}^\theta \int_{0}^x r(P^{}_{atm} + \rho^{}_{w}g(H-R(1-\sin(\theta))) \sin(\theta) d \theta dx $$ Neither of the ways I've done the problem thus far are resulting in the correct answer. I'm unsure of what is going wrong. Are either of the two ways I mentioned the correct way to find the force applied on the bottom of the canoe?
It's difficult to say where the problem is because you don't define $y$ and $\theta_0$ and $\theta_1$. Is $y$ pointing up or down? What is $\theta_0$? This should be made clear through text or figure. You seem to be on the right track thus. There's probably a small mistake in the setup of the integral. Have you tried integrating only half of the canoe and then multiplying by 2? This will get rid of one potential error related to the angles.. PS. I fixed your problem in your post. To see how I did it, right click on the equation and choose “Show Math As” -> “TEX Commands”.
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