Question

A flag in the shape of a right triangle is hung over the side of a building as shown below. The total weight of the flag is 250 pounds and it has uniform density. a = 15 and b = 39.

Answer 1

B hope this helps thanks

Answer 2

Answer:

(a) The density of flag is $\dfrac{25}{27}$ Pounds per square foot

(b) The weight of strip is $\dfrac{20}{9}(15-h)\Delta h$ Pounds

(c) The work is $\dfrac{20}{9}(15-h)h\Delta h$ foot-pounds

(d) The exact work by roof is 1250 foot-pounds

Step-by-step explanation:

(a) We are given a flag in the shape of a right triangle.

The total weight of flag is 250 pounds and Uniform density.

Base of the flag $=\sqrt{b^2-a^2}$

                          $=\sqrt{39^2-15^2}=36$

Area of the flag $=\dfrac{1}{2}\times 36\times 15 = 270$

Weight of flag = 250 pounds

$Density =\dfrac{Weight}{Area}=\dfrac{250}{270}=\dfrac{25}{27}$

Hence, The density of flag is $\dfrac{25}{27}$

(b) Weight of strip which is h feet below the roof.

Length of strip $=\dfrac{36}{15}(15-h)$

Width of strip $\Delta h$

Weight = Density x area

            $=\dfrac{25}{27}\times \dfrac{36}{15}(15-h)$

            $=\dfrac{20}{9}(15-h)\Delta h$

Hence, The weight of strip is $\dfrac{20}{9}(15-h)\Delta h$

(c) Work slice to move h feet above to the roof

work$=Weight\times displacement$

       $=\dfrac{20}{9}(15-h)\Delta h\times h$

       $=\dfrac{20}{9}(15-h)h\Delta h$

Hence, The work is $\dfrac{20}{9}(15-h)h\Delta h$ foot-pounds

(d) Exact work on the roof by hanging flag

$W=\int_0^{15}\dfrac{20}{9}(15-h)hd h$

$W=\dfrac{20}{9}(\dfrac{15h^2}{2}-\dfrac{h^3}{3})|_0^{15}$

$W=1250-0$

$W=1250$

Hence, The exact work by roof is 1250 foot-pounds