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Good, except lose the r^3 and the L^3joshmccraney said:$$F_n=\frac{4Fr^3}{3L^3}\cos{\theta}$$ Thank you both for the input!
Good, except lose the r^3 and the L^3joshmccraney said:$$F_n=\frac{4Fr^3}{3L^3}\cos{\theta}$$ Thank you both for the input!
Question: Why is the upper limit of integration L? The drawing attached to post #16 shows the fluid extending only partially along the full length of the rod. Or is it just a drawing?Chestermiller said:The moment of the drag force is ##M=FL\cos{\theta}=\int_0^L{k'r^3dr}=k'\frac{L^4}{4}##
That's not how I interpreted the diagram. I don't know what the correct interpretation is. Josh?kuruman said:Question: Why is the upper limit of integration L? The drawing attached to post #16 shows the fluid extending only partially along the full length of the rod. Or is it just a drawing?
There is no mention of the ground in the OP. Have you stated the problem word-for-word as given?joshmccraney said:The initial question posed was asking what force the hinge exerts on the ground.