364 Static Equilibrium and ElasticityAdditional Problems*P12.38(a)The beam is perpendicular to the wall, since 345222+=. Then sinθ=4 m5 m; =°53 1..(b)τhinge∑=0:+−=Tsin32500m N10mafafT=°=×250053 1104 103Nm3 mNsin..(c)xTk==××=01263..N8.25 10 N mm3The cable is 5.126 m long. From the law of cosines,45126323512635126423 512651 222212=+−=−coscos...afafafα4 m5.126 m3 mFIG. P12.38(d)From the law of sines, the angle the hinge makes with the wall satisfies sin.sin.512651 24m=°sin.sin...==+°−=∑0998580351 2 2500 9980107 103hingeN10m58NTTafaf(e)x=××=01293..Nm3=−cos...12129423 512951 1af(f)Now the answers are self-consistent:sin.sin..sin.....=°=°−==512951 1409983511 2500 99800129551 13mm51N10m51NmTTxafafafP12.39Let nAand nBbe the normal forces at the points of
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .