How to apply momentum equation to wedge with angle θ

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Saladsamurai
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Homework Statement



Picture1-35.png


Homework Equations


[tex]\sum(\dot{m}\vec{v})_{exit}-\sum(\dot{m}\vec{v})_{inlet}=\sum F_{ext}[/tex]


I only know that this is cons of momentum because my prof told us. I am having a hard time visualizing how to incorporate THETA into the above equation since it is not a nice right triangle.

I am thinking that since the exit velocities are equal, than the inlet flow must be being split about the wedges horizontal axis of symmetry.

Thus, I think I can divide the wedge into an "upper" and "lower" right triangle whose angle wrt to the horizontal is [itex]\frac{\theta}{2}[/itex].

Then I can resolve the momentum eq into Cartesian coordinates.

Sound good?
 
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Nice! I guess that what is confusing me now, is why they gave me
force/unit length into page

But I am just going to start plugging in and see what happens.
 
Saladsamurai said:
I guess that what is confusing me now, is why they gave me
force/unit length into page
Consider it to be an arbitrarily wide sheet of water. Use momentum per unit length as well.
 
Doc Al said:
Consider it to be an arbitrarily wide sheet of water. Use momentum per unit length as well.

Okay, so the m-1 just drops out anyway, and this problem reduces to regular cons of mom
 
Thanks! 83.4 degrees sounds reasonable I think :smile:
 
Saladsamurai said:
83.4 degrees sounds reasonable I think
Show how you got that answer.
 
Rethink your result for m, the mass flow rate. You want it to be mass flow per unit width (or depth).
 
I am not sure that I follow. Errr... Okay. So that inlet with the 4 cm dimension is NOT a PIPE...right?

It is a "sheet" with area 4cm*WIDTH. Thus my, mass flow rate should be, with h=4cm:

[tex]\dot{m}=\rho V (h*W)\Rightarrow \frac{\dot{m}}{W}=\rho Vh[/tex]

Am I with you now?
 
Typical 'not-paying-attention' mistake. Perhaps I should turn off Band of Brothers while I do my studies? :smile:

Thanks Doc!