# Homework Help: Force exerted on a swimmer

1. Apr 19, 2013

### Joe123

1. The problem statement, all variables and given/known data
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=u1A=u1B= 3.5m/s
d1=d1A=d1B= 20*10^-3 m
Angle $\alpha$ = 25 degrees
d2A= 23*10^-3 m
d2B = 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=P1A=P1B
The flow pattern is effectively two-dimensional, i.e, you would see the same pattern near the fingertips as near the wrist.

2. Relevant equations

A1V1=A2V2

Bernoulli's Equation form of

1/2(V2^2-V1^2)=-1/ρ(P2-P1)

3. 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=V2AA2 (height will be the same so)
V2A= d1/d2A * V1

and i did the same for V2B

So the values for the velocities i got were
V2A=3.04 m/s
V2B= 4.67 m/s

Using these values i put them into Bernoulli's Equation

(V2A^2-V1^2)=-1/ρ(P2A-P1)
(V2B^2-V1^2)=-1/ρ(P2B-P1)
sub values and get:

(P2A-P1)= -1504.2 pascal
(P2B-P1)= 4779.45 pascal

I then subbed for P1 and got

(P2A-P2B)=-6283.65 pascal So this value is the Pressure difference between the flow streams.

I then used
(P2A-P2B)*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