Den Hartog’s Mechanics

A web-based solutions manual for statics and dynamics

Problem 64

I’m not sure of the practical applications of this problem, but it is cute. The c.g. of the semicircular arc is 2r/\pi from the center of the arc (which you can confirm here), so the free-body diagram will look like this

and the moment equilibrium about the point of contact with the ground gives us

w_1 \frac{2 r}{\pi} \sin\alpha = w_2 r \cos\alpha

A bit of algebra gives us

\frac{w_1}{w_2} = \frac{\pi}{2 \tan\alpha}

which is equivalent to the answer given in the back of the book. I prefer giving answers in terms of tangent because they can be entered in a calculator more directly—calculators don’t have a cotangent key.

Problem 65Problem 63

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