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Equilibrio de un cuerpo rígido

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Solución<br />

DCL<br />

• BD es <strong>un</strong> miembro <strong>de</strong> 2 fuerzas<br />

• Palanca ABC miembro <strong>de</strong> 3<br />

Ecuaciones <strong>de</strong> <strong>Equilibrio</strong><br />

θ= tan −1 ( 0 .7<br />

)<br />

=60 . 3∘<br />

0 . 4<br />

+→∑ F x<br />

=0 ; F A<br />

cos60. 3 ∘ −F cos45 ∘ +400 N= 0<br />

+↑∑ F y<br />

=0 ; F A<br />

sin60 .3 ∘ − F sin45 ∘ =0<br />

Resolviendo,<br />

F A<br />

=1 . 07 kN<br />

F=1.32 kN

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