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COSMOS: Complete Online <strong>Sol</strong>utions Manual Organization System<br />
Chapter 2, <strong>Sol</strong>ution 61.<br />
Note: Refer to Note in Problem 2.60<br />
Free-Body Diagram At C:<br />
Force Triangle<br />
(a) Law of Cosines<br />
P<br />
2<br />
2 2<br />
( ) ( ) ( )( )<br />
= 1400 N + 700 N − 2 1400 N 700 N cos85°<br />
or<br />
P = 1510 N !<br />
(b) Law of Sines<br />
sin β sin85°<br />
=<br />
1400 N 1510 N<br />
sin β = 0.92362<br />
β = 67.461°<br />
α = 180° − 55° − 67.461°<br />
or α = 57.5°!<br />
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,<br />
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell<br />
© 2007 The McGraw-Hill Companies.
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