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 69
In outline, this problem is the same as the last one. The free-body diagram of the right-hand trap door looks like this:
where we’ve used a unit weight of 124.8 pcf to calculate the pressures and resultant forces associated with the trapezoidal pressure distribution. Taking moments about point A, we get this equilibrium equation:
\sum M_A = F_B \cdot 4 - 11,981 \cdot 2 - 3994 \cdot \frac{2 \cdot 4}{3} = 0
The solution is F_B = 7987\,\rm{lbs}, which, apart from round-off, is the answer given in the back of the book.
Last modified: January 22, 2009 at 8:32 PM.