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Category > Engineering Posted 28 May 2017 My Price 8.00

Determine the internal forces at point J knowing that θ=30°.

A semicircular rod is loaded as shown. Determine the internal forces at point J knowing that θ=30°.

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Alpha Geek
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Status NEW Posted 28 May 2017 10:05 AM My Price 8.00

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file 1495967051-1586701_1_636314773287546965_reaction-on-rod.docx preview (49 words )
----------- FB-----------D o-----------f t-----------he -----------sem-----------ici-----------rcu-----------lar----------- ro-----------d ----------- F-----------or -----------the----------- eq-----------uil-----------ibr-----------ium----------- of----------- th-----------e p-----------ort-----------ion----------- sh-----------own----------- M-----------A=0-----------(Rc-----------*35-----------*16-----------0)+-----------(Rc-----------*45-----------*16-----------0)------------(28-----------0*3-----------20)-----------=0R-----------c=4-----------00N----------- (i-----------n t-----------he -----------dir-----------ect-----------ion----------- of----------- CD-----------)-----------Fx-----------=0R-----------h+4-----------00*-----------45=-----------0Rh-----------=-3-----------20N-----------Rh=-----------320-----------N (-----------lef-----------war-----------d)-----------F-----------y=0-----------400-----------*35------------RV------------28-----------0=0-----------RV=------------40-----------RV=-----------40N----------- (u-----------pwa-----------rd)-----------↑-----------For----------- ca-----------lcu-----------lat-----------ing----------- fo-----------rce----------- at----------- po-----------int----------- J -----------FBD----------- of----------- th-----------e f-----------igu-----------re -----------wil-----------l b-----------e b-----------elo-----------w ----------- F-----------or -----------the----------- eq-----------uil-----------ibr-----------ium----------- of----------- po-----------rti-----------on -----------AJ ----------- -----------Ft-----------=0F-----------t-3-----------20*-----------sin-----------30------------40*-----------sin-----------60=-----------0Ft-----------=19-----------4.6-----------4 N----------- -----------F-----------n=0-----------Fn------------(32-----------0*c-----------os3-----------0)+-----------40c-----------os6-----------0=0-----------Fn=-----------257-----------.13-----------N -----------M-----------O=0-----------194-----------.64-----------*16-----------0-4-----------0*1-----------60+-----------Mj=-----------0Mj-----------=-2-----------474-----------2.4-----------
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