Den Hartog’s Mechanics
A web-based solutions manual for statics and dynamics
The book
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Problems solved:
1–.
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Problem 53
Because the weight of the door is uniformly distributed, its center of gravity is at its geometric center. The free-body diagram looks like this:
Taking moments about the lower hinge, point B, we get the equilibrium equation
and the solution is H_A = 11.25\,\rm{lb}. Horizontal equilibrium tells us that H_B = H_A = 11.25\,\rm{lb}. Vertical equilibrium gives V_A + V_B = 30\,\rm{lb}, but there isn’t enough information to determine how that 30 lbs is split between the upper and lower hinges. Yes, it’s tempting to say that each hinge carries 15 lbs, but there’s no mechanical or geometric justification for it. The situation is statically indeterminant.
The answer in the back of the book has the directions opposite to those we’ve shown, but that’s because the question asks for the forces the door exerts on the hinges and we’ve calculated the forces the hinges exert on the door. Newton’s Third Law accounts for the direction difference.
Last modified: January 22, 2009 at 8:32 PM.