Den Hartog’s Mechanics

A web-based solutions manual for statics and dynamics

Problem 14

There are two ways to go about solving this problem:

  1. we can find the perpendicular distance from the line of action of force P to point O; or
  2. we can break P into its horizontal and vertical components and find their individual moments about O.

The second way seems a bit faster because the horizontal component of P runs right through O and does not contribute to the moment about O. Thus, the sum of the moments about O (clockwise positive) is:

\sum M_O = 1000 \cdot 3 - (P \: \sin 30^\circ) \cdot 12 = 0

We solve this equation for P to get P = 3000 / 6 = 500\: \rm{lb}.

Problem 15Problem 13

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