- #1
mizisyafie
- 7
- 0
Homework Statement
Hello friend, i really stuck on how to calculate this thing. I have to find the drag force and lift force acting on fishing lure. By the way, the fishing lure has lip and there an angle on the lip.I am sorry my english is bad but i hope you can understand what i am trying to tell you guys. I am also attached a free body diagram so guys you can take a look to make better understand.
I simulated the fishing lure on Ansys and run in fluid of water. The data as below.
Water velocity = 0.59 m/s
Acceleration, a = 0.02 m/s^2
Volume of water = 0.050348 m^3
Water Density = 998.2 kg/m^3
mass of fishing lure =0.01820 kg
Weight of fishing lure = mg = 0.1785 N
Volume of fishing lure = 0.000018196 m^3
wetted area = 0.00452120 m^2
Buoyancy Force = 0.1782 N
Lip angle (as in free body diagram ) = 63°
Homework Equations
Ok friend, here is my calculation. I use mass of water that flow and hit the fishing lure.
For x component.
F=ma=Drag
F cos 63 = Drag
ma cos 63 = Drag
mass of water = density x (volume of water - volume of lure)
so, mass of water = ( 998.2 )(0.050348 - 0.000018196 ) = 50.2392
So, (50.2392)( 0.02) cos 63 = 0.4562 N
So, the drag force i got is 0.4562.
Using this Cd = Drag / 0.5 p A V^2 = 0.58
For y-component
Lift + Fb - W + F sin 63 = 0
W - Fb- F sin 63 = Lift
0.1785 - 0.1782 - (998.2)(0.050348 - 0.000018196 ) sin 63 = Lift
so, the lift force = - 0.895 ( - sign show the lift acting downward as the lure will dive )
Using this CL = lift / 0.5 p A V^2 = -1.14
The Attempt at a Solution
My problem is, i think something wrong on my equation in x component and y component especially on cos and sin...guys, can you correct me pleasezzz..