Question

A banner in the shape of an isosceles triangle is hung over the side of a building. The banner has a base of 30ft (at the roof), a height of 20ft, and weighs 50lb. Write down an integral for the work required to lift the banner entirely onto the roof of the building.

Answer 1

Answer:

W = 9600000 joules

Step-by-step explanation:

From the given information:

The area of the banner = 1/2 × 30 × 20 = 300 ft²

The mass of the banner = 300 × 50 = 15000 pounds

Now; the integral for the required work is:

$W = \int^h_o F \ ds$

here;

h = height of the building

F = mg = 15000g

g = 32 ft/s²

Therefore;

$W = \int ^h_o 15000g \ ds$

$W = \int ^{20}_0 \ 15000 (32)(20)$

W = 9600000 joules