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Photo from Wikimedia Commons |
When you spray water from a garden hose you feel a backward force as you hold the nozzle. Similarly, anything you spray the water at feels a force in the direction away from the spray. What is the origin of this force, and what determines its magnitude?
Solution
System:
Interactions:
We will ignore the vertical direction, so that the only interaction is the force due to the changes in momentum of the water and the item hit..
Model:
Approach:
Diagrammatic Representation
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Imagine a stream of water as a cylinder of uniform cross-sectional area A and density ρ. We consider an elemental unit of this that is Δx long. It travels, as does the rest of the stream, horizontally at velocity v (We will ignore the downward force of gravity here).
Mathematical Representation
Consider the element of length Δx and area A and density ρ. Its mass m must therefore be
\begin
[ m = \rho A \Delta x ] \end
Since it travels with velocity v, its momentum is thus
\begin
[ \vec
= \rho A \Delta x \vec
] \end
Let's assume that this element of the stream strikes an object (a wall, say), and breaks up, dissipating in all directions. the stream element loses all of its momentum in the process. The change in momentum of the stream element is thus
\begin
[ \Delta \vec
= \rho A \Delta x \vec
] \end
If this happens in a time Δt, then the change in momentum with time (which is just the Average Force) is
\begin
[ \vec{F_{avg}} = \frac{\Delta \vec{p}}
= \rho A \vec
\frac
] \end
But Δx/Δt = v , so the magnitude of the Average Force is thus
\begin
[ F_
= \rho A v^
]\end
\begin
[ m^
= \rho^
V^
= (\mbox
^
)(\mbox
\times\mbox
\times\mbox
) = \mbox
]\end
where ρwater is the density of water and Vrain is the volume of the accumulated rain.
We can now solve to find:
\begin
[ v_
= \frac{m^
v^
_{x,i}}{m^
+ m^{\rm rain}} = \mbox
]\end