Question

What is the minimum hose diameter of an ideal vacuum cleaner that could lift a $10kg$($22lb$) dog off the floor?

Short answer

The minimum hose diameter of an perfect vacuum cleaner is$d=3.51cm$

Step-by-step solution

Force.

A force may be a push or pull on an object caused by the interaction of the thing with another object.

Anytime two things interact, a force is exerted on each of them.

The $2$items do not feel the force after the interaction ends.

Find A to derive diameter.

calculating the force $F$,

localid="1650283275175" $F=mg$

localid="1650283280316" $F=(10\mathrm{kg})\left(9.8\frac{\mathrm{m}}{{\mathrm{s}}^2}\right)$

localid="1650283284371" $F=98N$

In the case of an perfect vacuum, the air pressure localid="1650283354042" $P$is localid="1650283288842" $1.013\times {10}^5\frac{N}{m^2}$

localid="1650283292792" $P=\frac{F}{A}$

localid="1650283296736" $A=\frac{98N}{1.013\times {10}^5{\displaystyle \frac{N}{m^2}}}$

localid="1650283300351" $A=9.674\times {10}^{-4}{m}^2$

Find the diameter of the hose.

Solving for the diameter of the opening of the hose,

$A=\pi r^2$

$9.674\times {10}^{-4}{\mathrm{m}}^2=\pi {\left(\frac{d}{2}\right)}^2$

$9.674\times {10}^{-4}{\mathrm{m}}^2=\pi \times {\frac{d}{4}}^2$

$d=\frac{4(9.674\times {10}^{-4}{\mathrm{m}}^2)}{\pi }$

$d=\sqrt{\frac{4\left(9.674\times {10}^{-4}{\mathrm{m}}^2\right)}{\pi }}$

$d=0.351m$