Den Hartog’s Mechanics

A web-based solutions manual for statics and dynamics

Problem 25

The free-body-diagram of the ladder looks like this

Taking moments about the hinge, H, we get the equilibrium equation

\sum M_H = 1000\frac{L}{2} \sin 45^\circ + 200 \frac{4 L}{5} \sin 45^\circ - T \cos 15^\circ L = 0

where T cos 15° is the component of the rope tension perpendicular to the ladder. Because the unknown length of the ladder, L, is common to all terms, it drops out of the equation, and we can solve for the rope tension, T = 483 lbs.

We can get the two components of the hinge force from the two force equilibrium equations. For horizontal forces

\sum F_x = 483 \sin 30^\circ - F_{Hx} = 0

and F_{Hx} = 242\:\rm{lbs}. For vertical forces

\sum F_y = 483 \cos 30^\circ - 1000 - 200 + F_{Hy} = 0

and F_{Hy} = 782\:\rm{lbs}. Except for some roundoff—which I’m going to attribute to some sloppy slide rule work by one of Den Hartog’s grad students—these answer match those in the back of the book.

Problem 26Problem 24

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