# Need help setting up hydrostatic force problem

• May 20th 2011, 12:21 AM
florx
Need help setting up hydrostatic force problem
http://i55.tinypic.com/21cb3vl.jpg

I need help with number 24.

Here is my attempt at trying to figure out the width of this triangle:
http://i54.tinypic.com/59slmt.png

And here is my attempt at trying to set this problem up:
http://i51.tinypic.com/35kiiwo.jpg

Please tell me if this is wrong or right. I cannot check my answer in the back of the book. Thank you for your time.
• May 20th 2011, 02:41 AM
HallsofIvy
Well, you have the width correct. You are applying "similar triangles" to the two triangles- they have the same angles so they are similar. The larger triangle has top side 4 and the other side 3- their ratio is 3/4. The corresponding values in the small triangle are top= w and other side= 3- x. Another way to do that would be to write the equation of the straight line (hypotenuse of the right triangle) itself. Because it is a straight line, its equation must be of the form w= ax+ b. When x= 0 (at top of triangle) w= 4: 4= a(0)+ b so b= 4. When x= 3 (bottom of triangle) w= 0: 0= a(3)+ 4 so a= -4/3. w= -(4/3)x+ 3 is, of course, exactly the same as w= (4/3)(3- x).

However, It is a very good idea to actually write down what your variables represent. It took me a moment to realize that your x is distance down from the top edge- and I think you fooled yourself with that. You then want "weight of water times width times depth". Since you are measuring the depth downward from the top edge, which is itself 1 foot below the water, the depth is x+ 1, not 4- x.
• May 20th 2011, 06:30 AM
florx
Wow thank you so much for clearing that up for me!