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Den Hartog’s Mechanics

A web-based solutions manual for statics and dynamics

Problem 73

We’re going to reuse the solution Den Hartog graciously provided us on page 45:

\delta_1 = \frac{R}{kL} \left( 6 \frac{c}{L} - 2 \right)

Our plan of attack is to express R and c in terms of the values given in the problem, then set \delta_1 = 0 and solve for b.

With w_1 = 20, L = 12, and P = 1000, R = 20 \cdot 12 + 1000 = 1240\,\rm{lb}. The position of this resultant from the right end of the beam is

c = \frac{240 \cdot 6 + 1000 b}{1240} = \frac{36 + 25b}{31}

Plugging these and the given values into the expression for \delta_1 and setting it to zero gives

\delta_1 = \frac{1240}{(100/12) \cdot 12} \left( \frac{6}{12} \frac{36 + 25b}{31} - 2 \right) = 0

where we’ve divided by 12 to express k in lb/ft/ft. The solution is b = 88/25 = 3.52\,\rm{ft}. If P gets any closer to the right end of the beam, the left end will lift up off the ground.

Problem 74Problem 72

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