Steele, Lyndsey asks:

```
CATAGORY: Mechanics
QUESTION: 30. An oar is held by a rower at .35m from the oarlock.  The
oar makes contact with the water at an average of 1.3 m from the oarlock.
If the force applied by the rower on the oar is 80N.  what is:

a)the force of the oar on the water, and
b)the force on the oarlock

the anwers are a) 22N and b) 100N
I need to know the method to getting the answer correct!!!
```

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```
For part (a) we look at torque. Consider the following diagram

^  Reaction force              ^ 80N
|  of water on oar             |
|                  oarlock     |
|                   |          |
*-------------------*----------*
|       1.3m        |  .35m    |
|
|
v  F = force of oar on water

The torque about the oarlock must be equal.  That is to say
the force times the lever arm on one side of the oarlock
must be the same as on the other side. Thus we have

(1)     F(1.3m) = 80N(.35m) or

or

(2)     F = 80N(.35m)/1.3m = 21.53 N ~ 22N

Note: the answer is rounded to 2 significant digits

This gives us the first answer.  :-)

Now the force on the oarlock is the sum of the push of the person on
the oar which is 80N AND the reaction of the water on the oar which is
~22N and is IN THE SAME DIRECTION AS THE 80N push. (See the
figure above) The total force on the oarlock is then

(3)    F(oarlock) = 80N + 22N = 102N ~ 100N

Note: the answer is rounded to 2 significant digits.

This gives us the second answer.  :-)

Hope that this helps.

```

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