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Replace the force and couple system acting on the frame by an equivalent resultant force and specify where the resultant’s line of action intersects member AB, measured from A.

Step-by-step

\underrightarrow { + } {F }_{ Rx }\sum { F}_{ x }  ;   {F }_{ Rx }150\frac {4 }{ 5 }  + 50 sin 30° = 145 lb

+ \uparrow  {F }_{ Ry }   =  \sum { F}_{ y }  ;      {F }_{ Ry }  =  50 cos 30° +  150\frac {3 }{ 5 }  =  133.3 lb

{ F }_{ R }  = \sqrt { ({ 145 )}^{ 2 }+({ 133.3)}^{ 2 }} = 197 lb

\theta = \tan ^{ -1 }{ } \frac {133.3}{ 145 } = 42.6°

\hookrightarrow + {M}_{ RA } = \sum { M}_{ A }  ;        145 d  = 150 \frac {4 }{ 5 } (2) –  50 cos 30° (3) + 50 sin 30° (6) + 500 d = 5.24 ft

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