fluid force on vertical side of tank the weight density of water is 62.4 trapazoid

† † margin: y y x - 2 - 1 1 2 - 2 - 1 1 2 50 water line not to scale d ⁢ ( y ) = 50 - y Figure 6.5.8: Measuring the fluid force on an underwater porthole in Example 6.5.4. The truth is that it is not, hence the survival tips mentioned at the beginning of this section. This is counter-intuitive as most assume that the door would be relatively easy to open. Most adults would find it very difficult to apply over 500 lb of force to a car door while seated inside, making the door effectively impossible to open. Using the weight-density of water of 62.4 lb/ft 3, we have the total force as

We adopt the convention that the top of the door is at the surface of the water, both of which are at y = 0. Net vertical force weight of water contained in semicircular volume RBP + Weight of oil contained in arc portion OQP of the cylinder. Its length is 10 / 3 ft and its height is 2.25 ft. To find the area, we will multiply all of that by the change in height, which is Δx.SolutionThe car door, as a rectangle, is drawn in Figure 6.5.7. Now we can find the area of the trapezoid, which is the base added to double the width of the triangle (a), since the triangle is on both sides of the trapezoid. This is equal to the line from the base to the surface of the water (2 - x) divided by a, which is the width at 2-x.

Set up the equation so that you divide the total height (4m) by the maximum width of this section (2m since (8m - 4m)/2, there's 2m on each side).

First you need to find an area equation for the triangle section of the trapezoid where the width is increasing from 4m to 8m, keeping in mind that we're only interested in the area that is submerged in water (2m).

Fluid force on vertical side of tank the weight density of water is 62.4 trapazoid