A 10 kg mass placed on an infinitely long horizontal massless flat platform is to be supported by a steady vertical water jet as shown in the figure. The diameter of the jet is 5 cm. What minimum average velocity is required to hold the mass in place?
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To solve for the minimum velocity in fluid mechanics problems, ensure the balance of forces, using the relationship between mass flow rate and the velocity of the jet.
We need to find the minimum average velocity of the water jet required to hold the 10 kg mass in place. The force due to the water jet must balance the weight of the mass.
The weight of the mass is:
\[
W = mg = 10 \, \text{kg} \times 10 \, \text{m/s}^2 = 100 \, \text{N}
\]
The force exerted by the water jet can be expressed as:
\[
F = \dot{m} v
\]
where \( \dot{m} \) is the mass flow rate and \( v \) is the velocity of the water jet. The mass flow rate \( \dot{m} \) is related to the density and the velocity of the jet by the equation:
\[
\dot{m} = \rho A v
\]
where \( \rho = 1000 \, \text{kg/m}^3 \) is the density of water and \( A \) is the cross-sectional area of the jet.
The area of the jet is given by:
\[
A = \pi r^2 = \pi \left( \frac{d}{2} \right)^2 = \pi \left( \frac{5 \, \text{cm}}{2} \right)^2 = \pi \times (0.025)^2 = 1.9635 \times 10^{-3} \, \text{m}^2
\]
Now, using \( F = W \), we can solve for the velocity \( v \):
\[
100 = \rho A v^2
\]
\[
100 = 1000 \times 1.9635 \times 10^{-3} \times v^2
\]
\[
v^2 = \frac{100}{1000 \times 1.9635 \times 10^{-3}} = 50.943
\]
\[
v = \sqrt{50.943} = 7.14 \, \text{m/s}
\]
Thus, the minimum average velocity required to hold the mass in place is:
\[
\boxed{7.14 \, \text{m/s}}
\]
