- #1
MisterX
- 764
- 71
Let [itex]\lambda[/itex] be a linear density of a rope which is moving into a scale at velocity v. The additional force on the scale due to the collision is given as
[itex]\frac{d p}{d t} = v\frac{d m}{d t} = \lambda v^2[/itex]
Where as the stagnation pressure from stopping a column of water in excess of static pressure is
[itex]\frac{1}{2}\rho v^2[/itex]
We can easily compare the forms by, for example multiplying by the width of the column to obtain a linear density of the fluid, or consider hitting the scale with a continuum of infinitesimal ropes. It seems the [itex]\frac{1}{2}[/itex] factor would remain different.
So what is the explanation for this relative factor of [itex]\frac{1}{2}[/itex]? I have tossed around a few ideas but I'm curious what you may think.
[itex]\frac{d p}{d t} = v\frac{d m}{d t} = \lambda v^2[/itex]
Where as the stagnation pressure from stopping a column of water in excess of static pressure is
[itex]\frac{1}{2}\rho v^2[/itex]
We can easily compare the forms by, for example multiplying by the width of the column to obtain a linear density of the fluid, or consider hitting the scale with a continuum of infinitesimal ropes. It seems the [itex]\frac{1}{2}[/itex] factor would remain different.
So what is the explanation for this relative factor of [itex]\frac{1}{2}[/itex]? I have tossed around a few ideas but I'm curious what you may think.