To print higher-resolution math symbols, click the
Hi-Res Fonts for Printing button on the jsMath control panel.

No jsMath TeX fonts found -- using image fonts instead.
These may be slow and might not print well.
Use the jsMath control panel to get additional information.
jsMath Control PanelHide this Message


jsMath

Den Hartog’s Mechanics

A web-based solutions manual for statics and dynamics

Problem 51

Following our usual procedure, we find statically equivalent joint loads for each of the non-joint loads. The free-body diagram looks like this:

Ignoring, for the moment, the loads applied directly to the supports, we get this FBD:

Now everything is a two-force member, which means

Starting with the upper right pin, we get an FBD of

Vertical equilibrium tells us V_B = 500\,\rm{lbs}; horizontal equilibrium tells us F = 250\,\rm{lbs}.

The upper left pin has this FBD:

Vertical equilibrium gives us Q = 500\,\rm{lbs}; horizontal equilibrium gives us X = -250\,\rm{lbs}, which means that X is actually pulling to the left, rather than pushing to the right.

Superposition is needed to get the horizontal reaction at the right support; the others have already been calculated.

H_A = V_A = 500\,\rm{lbs}
H_B = 250\,\rm{lbs}
V_B = 500\,\rm{lbs}

Den Hartog combines H_A and V_A to get a single reaction of 500 \sqrt{2} at the left support, but I don’t see any value in doing so (especially since he didn’t combine the horizontal and vertical components in any of the other reactions).


Problem 52Problem 50


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