Section 15.3 Measuring and Using PressureWhat is the

Step 1 of 2

We have to find the minimum hose diameter of an ideal vacuum cleaner that could lift a 10 kg (22 lb) dog off the floor.

If the vacuum cleaner could lift the dog off the floor then the gravitational force on the dog is balanced by the force resulting from the pressure difference between the atmosphere and the vacuum \(\left(p_{\text {hose }}=0\right)\) in the hose.

\(F_{\text {vacuum }}=F_{g}\)

\(A\left(p_{\text {atm }}-p_{\text {hose }}\right)=m g\)

Where,

\(A=\) Area of the hose in \(\mathrm{m}^{2}\)

\(m=\) mass of the dog \(=10 \mathrm{~kg}\)

\(g=9.80 \mathrm{~m} / \mathrm{s}^{2}\)

\(p_{\text {atm }}=1.013 \times 10^{5} \mathrm{~Pa}\)

\(p_{\text {hose }}=0 \mathrm{~Pa}\)

Thus,

\(A =\frac{m g}{\left(p_{\text {atm }}-p_{\text {hose }}\right)}\)

\(=\frac{(10)(9.80)}{1.013 \times 10^{5}-0}\)

\(=9.7 \times 10^{-4} \mathrm{~m}^{2}\)