# On levers and balance of forces

Here’s a picture of the oar at two points in time. On the left-hand side of the picture, the frame of reference is the boat. This is the usual perspective of the rower moving down the slide and pulling the oar. On the right-hand side, the same oar, at the same points in time, in the reference frame of the public standing on the bank. What’s not clear from the perceptive of the rower but should be noticed by an accurate observer on the bank, is that the blade slips through the water. The oar is a lever, but the effective outboard length of the lever is given by $l_2$ rather than by $l_{\rm out}$.

To calculate the forces on the boat, let’s however first consider an idealized case without blade slip. Imagine the rower to catch poles standing in the canal, as depicted in the following figure:

The rower pulls the handle with a force $F_{\rm handle}$. Through the lever principle, this results in forces $F_{\rm oarlock}$ and $F_{\rm blade}$ at the oarlock and the blade, respectively. As the blade doesn’t slip, no work is done at the blade. Thus, the work at the handle being equal to the work at the oarlock, the oarlock force is given by:

$F_{\rm oarlock} = \frac{L}{l_{\rm out}} F_{\rm handle}$

where $L$ is the total length of the oar (measured from the handle to the center of the blade). On the right hand side of the figure, the forces on the hull are depicted. The oarlock force accelerates the boat, while the drag force $W$ and the rowers footboard force $F_{\rm foot}$ act in the opposite direction. Using Marinus van Holst’s notation, $\dot{x}$ is the velocity of the hull, $\dot{z}$ is the velocity of the center of mass, and $\dot{y}$ is the velocity of the rower with respect to the hull. The equations of motion are thus:

$F_{\rm prop} = (m_c + m_b) \ddot{z} = m_b \ddot{x} + m_c (\ddot{x}+\ddot{y})$

$m_b \ddot{x} = \frac{L}{l_{\rm out}} F_{\rm handle} - F_{\rm foot} - W$

$m_c (\ddot{x}+\ddot{y}) = F_{\rm foot} - F_{\rm handle}$

Inserting the last two equations into the one for the net propulsive force on the system, $F_{\rm prop}$, we obtain:

$F_{\rm prop} = \frac{L}{l_{\rm out}} F_{\rm handle} - F_{\rm foot} - W + F_{\rm foot} - F_{\rm handle}$

Thus, finally we obtain:

$F_{\rm prop}= \frac{l_{\rm in}}{l_{\rm out}} F_{\rm handle} - W = F_{\rm blade} - W$

Voilà, the net force on the system equals the blade force minus the drag force. This is logical, as the blade and the hull area are the only points of contact between the boat and the rest of the world.

Now, for the slipping blade situation, the balance of forces is slightly different. From work considerations, we have

$F_{\rm handle} (\dot{x}+\dot{y}) - F_{\rm oarlock} \dot{x} - F_{\rm blade} (\dot{\phi} l_{\rm out} - \dot{x}) = 0$

Where the blade force is still* given by:

$F_{\rm blade} = \frac{l_{\rm in}}{l_{\rm out}} F_{\rm handle}$

The system net propulsive force is given by $F_{\rm blade} - W$.

Still nothing new here, dear readers. In the next post, I will finally address the energy balance question. This will be an interesting one, as there are differences between the findings of the two giants (on whose shoulders I stand) about the amount of energy dissipated in the body of the rower.

I have removed the references to a virtual outboard $l_2$ from this text. The use of this virtual outboard is not necessary and leads to erroneous results. I have updated my model accordingly.