# Homework Help: Equation of motion canoe stroke

1. Jan 13, 2012

### The-alexandra

1. The problem statement, all variables and given/known data

What is the equation of motion of canoe stroke in canoeing sport

2. Relevant equations

F=ma
Momentum=mV

3. The attempt at a solution

My

P - DT = MT (dVcrew/dt + dVboat/dt)

the Book response,

P - DT = (m dVboat/dt ) + MT (dVcrew/dt + dVboat/dt)

P=Propulsive force ; Vboat= Boat velocity
DT= Total drag force ; Vcrew= Crew velocity
m=Boat mass ; MT=Total mass ( boat + crew + paddle)

Alexandra

2. Jan 13, 2012

### I like Serena

Welcome to PF, The-alexandra!

I'm afraid neither your nor the book's response is right.

For the proper response it is necessary to know whether the crew velocity is relative to the boat, or whether it is an absolute velocity.
Can you tell?

3. Jan 13, 2012

### The-alexandra

Hi, I cannot find the answer

(Copy and paste)Book;

According to Newton’s second law, a linear force, F, generated by a system can be given by
ƩF=ma (1)

where m is the mass of the system and a is its acceleration. If input forces and masses of a system are known, then the velocity, V, of that system can be calculated as the integral of the system acceleration, such that
V=∫a dt (2)

In the case of the canoe stroke, equation (1) can be further developed, giving

P - DT = (m dVboat/dt ) + MT (dVcrew/dt + dVboat/dt)

P=Propulsive force ; Vboat= Boat velocity
DT= Total drag force ; Vcrew= Crew velocity
m=Boat mass ; MT=Total mass ( boat + crew + paddle)

where P is the total propulsive force, D is the total drag force, m is the mass of the boat , M is the combined mass of the boat, paddlers and paddles, vboat is the horizontal velocity of the boat relative to the earth and vcrew is the horizontal velocity of the paddler’s centre of mass relative to the boat.

Someone can help me, Please I m lost

Last edited: Jan 13, 2012
4. Jan 13, 2012

### I like Serena

Good. So indeed the crew velocity is relative to the boat.

We want to do everything in the same frame of reference, for which we typically use the earth frame.
So the actual crew velocity relative to the earth is $V_{crew} + V_{boat}$.

The paddles are not consistently mentioned, but I'll assume they move together with the paddlers.

The forces need to be charted separately, taking each object, its mass, and its velocity into account separately.

The force on the boat is $F_{boat} = m_{boat} {dV_{boat} \over dt}$ and the force on the crew+paddles is $F_{crew+paddles} = m_{crew+paddles} {d(V_{crew} + V_{boat}) \over dt}$.

Since the mass $m_{crew+paddles}$ is not given, we will replace it by $M_T - m_{boat}$, which is the same.

Can you add those 2 forces together and simplify the resulting expression?

5. Jan 13, 2012

### The-alexandra

thanks so much for your help, "I like Serena "
I'm really sorry, but i can't find the same answer:
for the right side of the equation

=(mboat dVboat/dt ) + (MT−mboat)((dVcrew/dt + dVboat/dt))

=MT(dVcrew/dt + dVboat/dt) - mboat dVcrew/dt

could you tell, please where is my mistake?

thanks so much, I really appreciate your help

Alexandra

6. Jan 13, 2012

### I like Serena

No mistake.

I said before that the Book's response was wrong.

7. Jan 13, 2012

### The-alexandra

Last edited: Jan 13, 2012
8. Jan 13, 2012

### The-alexandra

I'm sorry, I found the same equation in two scientific paper, and we are wrong,

you have 5 more minutes for help me ??

thanks again

Last edited by a moderator: May 5, 2017
9. Jan 13, 2012

### I like Serena

Well, I claim the paper is wrong.

Actually, I think they made a typo.

With this correction the formula would be right.

EDIT: Am I still in time?

10. Jan 13, 2012

### The-alexandra

yes, you are in time.. for save the science

thanks again and have a nice day =)