Den Hartog’s Mechanics

A web-based solutions manual for statics and dynamics

Problem 22

Here’s another problem where the drawing in the book is good enough to use as a free-body-diagram.

There are two unknowns in this problem: the force, R, that the water exerts on the oar, and the force, G, that the oarlock exerts on the oar. To determine R without first (or simultaneously) determining G, we take the moments about the oarlock. The equilibrium equation is

\sum M_G = 50\:\cdot\:15 - 60 R = 0

which leads to R = 12.5 lbs.

To solve for G, we can either

Since we did Problem 21 the first way, let’s do this one the second way. Taking moments about the tip of the oar, we get this equilibrium equation:

\sum M_R = 50 (15+60) - 60 G = 0

The solution is G = 62.5 lbs.

Addendum
Now you may be asking yourself: Is it kosher to use two moment equilibrium equations in a planar statics problem? Aren’t the three equations of planar statics \sum F_x = 0, \sum F_y = 0, and \sum M = 0? As it turns out, the answer to both these questions is “Yes,” and there is no contradiction.

The three equations of planar statics are indeed \sum F_x = 0, \sum F_y = 0, and \sum M = 0, but mathematics allows us to use linear combinations of these equations to solve them. So, for this problem

\sum F_y = 50 - G + R = 0

and if we multiply this equation by 60 and add it to the \sum M_G equation we used to solve for R, we get

50\:\cdot\:60 - 60 G + 60 R + 50\:\cdot\:15 - 60 R = 0

which simplifies to

50 (15+60) - 60 G = 0

the same as the \sum M_R equation.

This is not magic. The equilibrium equation for moments about some point will always be a linear combination of the equilibrium equation for moments about another point and one or more of the force equilibrium equations. So it is kosher to use two moment equilibrium equations, and it will often be advantageous for us to do so.

Problem 23Problem 21

Last modified: January 22, 2009 at 8:32 PM.