Calculus: Force..A dam is inclined at an angle of 30∘ from the vertical and has the shape of an isosceles trapezoid 100 ft wide at the top and 50 ft wide at the bottom and with a slant height of 70 ft. Find the hydrostatic force (in lb) on the dam when it is full of water.

Question

xapproachesinfinity · Answer

wanna how is this done

darkigloo · Answer

all i know is that the perpendicular height h is 70cos(30)= 60.6 ft

darkigloo · Answer

yes

xapproachesinfinity · Answer

so the force on the dam is directed where?

darkigloo · Answer

the triangle?

xapproachesinfinity · Answer

what triangle do you mean?

darkigloo · Answer

oh im sorry, i was thinking of another question. 
i don't know where its being directed.

xapproachesinfinity · Answer

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xapproachesinfinity · Answer

actually this seems much good to me

darkigloo · Answer

thats how i drew it.

xapproachesinfinity · Answer

ok so what's the problem

darkigloo · Answer

not sure how to approach the problem

xapproachesinfinity · Answer

what do you know about hydrostatic forces

xapproachesinfinity · Answer

that could be the link your looking for
the fact that you have all the given looks like this is related area

xapproachesinfinity · Answer

link to what you are looking for*

dumbcow · Answer

First you need formula for finding pressure and Force
$$P = \rho g d$$
$$\rho = 62.4 lb/ft^3, g = 32 ft/s^2$$
d = depth under water
$$F = PA$$
where A is surface Area
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The total force of trapezoid will be the sum of all the rectangle forces of height "dy"
$$A = 2x dy = 2(\frac{25}{70}y +25) dy = (\frac{5}{7} y + 50) dy$$
depth is the perpendicular height 
$$d = \cos 30 * y = \frac{\sqrt{3}}{2} y$$
Finally put it all together to get Force equation, then integrate over "y" from 0 to 70
$$F = \int\limits_0^{70}(62.4)(32)(\frac{\sqrt{3}}{2}y)(\frac{5}{7}y+50) dy$$
$$F = (62.4)(16 \sqrt{3}) \int\limits_0^{70} [\frac{5}{7} y^2 + 50y] dy$$
Integrate using power rule, apply limits and you are done