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

(a) The density of flag is
`(25)/(27)` Pounds per square foot

(b) The weight of strip is
`(20)/(9)(15-h)\Delta h` Pounds

(c) The work is
`(20)/(9)(15-h)h\Delta h` foot-pounds

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

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
`=√(b^2-a^2)`

`=√(39^2-15^2)=36`

Area of the flag
`=(1)/(2)* 36* 15 = 270`

Weight of flag = 250 pounds

`Density =(Weight)/(Area)=(250)/(270)=(25)/(27)`

Hence, The density of flag is
`(25)/(27)`

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

Length of strip
`=(36)/(15)(15-h)`

Width of strip
`\Delta h`

Weight = Density x area

`=(25)/(27)* (36)/(15)(15-h)`

`=(20)/(9)(15-h)\Delta h`

Hence, The weight of strip is
`(20)/(9)(15-h)\Delta h`

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

work
`=Weight* displacement`

`=(20)/(9)(15-h)\Delta h* h`

`=(20)/(9)(15-h)h\Delta h`

Hence, The work is
`(20)/(9)(15-h)h\Delta h` foot-pounds

(d) Exact work on the roof by hanging flag

`W=\int_0^(15)(20)/(9)(15-h)hd h`

`W=(20)/(9)((15h^2)/(2)-(h^3)/(3))|_0^(15)`

`W=1250-0`

`W=1250`

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