Question

A 180-lb man carries a 15-lb can of paint up a helical staircase that encircles a silo with radius 20 ft. The silo is 80 ft high and the man makes exactly two complete revolutions. Suppose there is a hole in the can of paint and 12 lb of paint leaks steadily out of the can during the man's ascent. How much work is done by the man against gravity in climbing to the top?

Answer 1

To solve the problem it is necessary to transform the data given in the problem to mathematical expressions.

The weight of the man in magnitude of force is determined by 180lb and 30lb of the paint, in relation to the exchange rate for the hole (approximate circular) of the 6lb can.

Mathematically the Force would then be:

$F = (180+30-\frac{6t}{2(2\pi)})$

$F = (210 -\frac{3t}{2\pi})$

The path taken on the axes is determined by the radius of 15 feet horizontally, vertically and 80 feet high depending on the circumference circumference that travels, in other words the path in the components [x, y, z ,] would

$r= (15cost,15sint,\frac{80t}{2\pi(2)})$

$r= (15cost,15sint,\frac{20t}{\pi})$

Drifting,

$dr = (-15sint,15cost,\frac{20}{\pi})$

The work done is the Force traveled by the displacement made, from the floor to the top this would be:

$F\cdot dr = (210-\frac{3t}{2\pi})(\frac{20}{\pi})$

$F\cdot dr = \frac{4200}{\pi}-\frac{30t}{\pi^2}$

Therefore the work done from 0 to 4\pi is

$\int F\cdot dr = \int\limit_0^{4\pi} (\frac{4200}{\pi}-\frac{30t}{\pi^2})dt$

$\int F\cdot dr = \big[\frac{4200t}{\pi}-\frac{15t^2}{\pi^2}\big]\limit^{4\pi}_0$

$\int F\cdot dr = 16800-240$

$\int F\cdot dr = 16560ft-lb$

Therefore the total work done by the man against gravity in climbing to the top is 16560ft-lb