Stuck on how to calculate the angle of lip on fishing lure

[mizisyafie]

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..

 

[member 553083]

For x component.F=ma=DragF cos 63 = Dragma cos 63 = Dragmass of water = density x (volume of water - volume of lure)so, mass of water = ( 998.2 )(0.050348 - 0.000018196 ) = 50.2392So, (50.2392)( 0.02) cos 63 = 0.4562 NSo, the drag force i got is 0.4562.Using this Cd = Drag / 0.5 p A V^2 = 0.58For y-componentLift + Fb - W + F sin 63 = 0W - Fb- F sin 63 = Lift0.1785 - 0.1782 - (998.2)(0.050348 - 0.000018196 ) sin 63 = Liftso, 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