The triangular plate ABC weighs 225 lb and its center of gravity is at point G. The bar CE is weightless and connected to the plate by the cable BD. Find the reaction forces in pins A and C, the tension in cable BD, and the support reaction at E.

Ay=166.7 lb

Ax=21.2 lb

Cy=108.2 lb

Cx=2.9 lb

NE=62 lb

TBD=-55.5 lb

Explanation:

from diagram a) we have to:

∑MA=0

NE*cos70(46-30tan45)+NE*sin70(30+26)-(225*16)=0

NE(0.342*16)+NE(0.939*56)-3600=0

NE=62 lb

from the force balancing, we have:

∑Fy=0

Ay+NE*sin70-225=0

Ay+62*sin70-225=0

Ay=166.7 lb

∑Fx=0

Ax-NE*cos70=0

Ax-62*cos70=0

Ax=21.2 lb

From the b) diagram we have:

θ=tan^-1(15/(15+x)=tan^-1(15/(15+16))=25.8°

∑MC=0

-TBD*cosθ(15)-TBD*sinθ(15)-NE*cos70(30)+NE*sin70(30)=0

-TBD*cos(25.8)*(15)-TBD*sin(25.8)(15)-62*cos70(30)+62*sin70(30)=0

TBD=-55.5 lb

∑Fy=0

-Cy+NE*sin70-TBD*cos25.8=0

-Cy+62*sin70-TBD*cos25.8=0

Cy=108.2 lb

∑Fx=0

-Cx-NE*cos70-TBDsin25.8=0

Cx=2.9 lb

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