# Thread: Need help setting up hydrostatic force problem

1. ## Need help setting up hydrostatic force problem

I need help with number 24.

Here is my attempt at trying to figure out the width of this triangle:

And here is my attempt at trying to set this problem up:

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.

2. 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.

3. Wow thank you so much for clearing that up for me!