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COSMOS: Complete Online <strong>Sol</strong>utions Manual Organization System<br />
Chapter 2, <strong>Sol</strong>ution 20.<br />
Using the force triangle and the Laws of Cosines and Sines<br />
We have: γ = 180° − ( 45° + 25° ) = 110°<br />
2<br />
2 2<br />
Then: ( ) ( ) ( )( )<br />
and<br />
R = 30 kN + 20 kN − 2 30 kN 20 kN cos110°<br />
2<br />
= 1710.42 kN<br />
R = 41.357 kN<br />
30 kN 41.357 kN<br />
sinα = sin110°<br />
⎛ 30 kN ⎞<br />
sinα<br />
= ⎜<br />
sin110°<br />
41.357 kN<br />
⎟<br />
⎝ ⎠<br />
= 0.68164<br />
α = 42.972°<br />
Finally: φ = α + 45° = 87.972°<br />
R = 41.4 kN 88.0° !<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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