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Den Hartog’s Mechanics

A web-based solutions manual for statics and dynamics

Problem 4

Same pulley system as in Problems 1—-3, but this time we have W_1= 20\:\rm{lbs}, W_2= 10\:\rm{lbs}, and W_3= 25\:\rm{lbs}. This is a bit annoying, because our polygon of forces will no longer be a right triangle. What to do?

Well, the book says to use “graphical construction” and use a protractor to measure the angles of the string. Remember, this book was written almost 60 years ago, when engineering students were expected to know how to draw. A student back then would pull out his (yes, almost always his) trusty compass and scale, lay off a vertical line of 25 units, then scribe a 20-unit-radius arc from the bottom of the vertical line and a 10-unit-radius arc from the top. Where the arcs cross is the third corner of the triangle. Draw in the other two sides of the triangle and measure the angles. Easy.

A method more comfortable to today’s student would be to use the law of cosines, the generalization of the Pythagorean theorem to non-right triangles.

10^2 = 20^2 + 25^2 - 2\cdot20\cdot25 \cos \alpha

which gives us \alpha = \cos^{-1} 0.925 = 22.3^\circ, which is the angle between the left part of the string and the vertical. Similarly,

20^2 = 10^2 + 25^2 - 2\cdot10\cdot25 \cos \gamma

which gives us \gamma = \cos^{-1} 0.65 = 49.5^\circ, which is the angle between the right part of the string and the vertical.

Problem 5Problem 3

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