How Do Swimmers Calculate the Force Exerted by Arm Strokes?

[Joe123]

Homework Statement

So the question i have involves finding the force exerted on a swimmer using only arm strokes through the water.

A snapshot of the swimmer is taken by a camera underwater when the swimmer's arm is vertical and pointing at the camera.
Here is a diagram of the hand:

(forgot to label the top stream as A and bottom stream as B)

The question asks to calculate the Force component that drives the swimmer forward (in this case the x direction), at the time the image was taken.

Given values are u₁=u_1A=u_1B= 3.5m/s
d₁=d_1A=d_1B= 20*10^-3 m
Angle [itex]\alpha[/itex] = 25 degrees
d_2A= 23*10^-3 m
d_2B = 15*10^-3 m

Given Assumptions

The effective area of the hand is a rectangle, 0.15m *0.25m.
The density of water is 1000kg/m^3
The flow around the time the image was taken is approximately steady.
The pressure upstream where u1 was measured is the same on both sides of the hand, i.e. P₁=P_1A=P_1B
The flow pattern is effectively two-dimensional, i.e, you would see the same pattern near the fingertips as near the wrist.

Homework Equations

A₁V₁=A₂V₂

Bernoulli's Equation form of

1/2(V₂^2-V₁^2)=-1/ρ(P₂-P₁)

The Attempt at a Solution

To start off with i found the velocities of each of the flow streams near the end (point 2)

To do this i used the ratios of the d's to get

V₁A₁=V_2AA₂ (height will be the same so)
V₁d₁=V_2Ad_2A
V_2A= d₁/d_2A * V₁

and i did the same for V_2B

So the values for the velocities i got were
V_2A=3.04 m/s
V_2B= 4.67 m/s

Using these values i put them into Bernoulli's Equation

(V_2A^2-V₁^2)=-1/ρ(P_2A-P₁)
(V_2B^2-V₁^2)=-1/ρ(P_2B-P₁)
sub values and get:

(P_2A-P₁)= -1504.2 pascal
(P_2B-P₁)= 4779.45 pascal

I then subbed for P₁ and got

(P_2A-P_2B)=-6283.65 pascal So this value is the Pressure difference between the flow streams.

I then used
(P_2A-P_2B)*Area(hand) to get the force
=-235.63 N
then for the force in the x direction (since the force acts normal to the object)
-235.63*Cos(115degrees) = 99.58 N

The value sort of sounds reasonable but I'm not sure if I've made any errors going about this.

In particular is there anything in the y direction i need to include?

Any feedback on the working would be appreciated

 

[anathema]

.
Thank you for your question. Your approach seems to be on the right track, but there are a few things that need to be clarified and corrected.

Firstly, when using Bernoulli's equation, it is important to note that it is only valid for a steady, incompressible, and inviscid flow. In this case, the flow is assumed to be steady, but it is not incompressible (since the swimmer is moving through the water) and it is not inviscid (since there is friction between the swimmer's hand and the water). Therefore, Bernoulli's equation may not give accurate results in this case.

Secondly, in your calculation of the velocities at point 2, you have used the ratio of the distances, but you have not taken into account the angle of the hand (25 degrees). This will affect the velocity of the flow streams and will change your final result.

Thirdly, in your calculation of the pressure difference, you have used the wrong formula. The correct formula is:

(P2A-P2B) = 1/2 * ρ * (V2B^2 - V2A^2)

Finally, in order to calculate the force in the x-direction, you need to take into account the angle of the hand (25 degrees) and the pressure difference (P2A-P2B). The correct formula is:

Fx = (P2A-P2B) * A * cos(25 degrees)

where A is the area of the swimmer's hand.

I hope this helps and clarifies any doubts you may have had. Keep up the good work!