Answer to Question #278848 in Physics for Simmy

Question #278848

The uniform bar of length l 40 N and is subjected to four forces shown find the magnitude location and direction of a fith force F needed to keep the bar in equilibrium



1
Expert's answer
2021-12-14T09:28:45-0500

Fx=80cos30°=69.282 (N)F_x=80\cdot\cos30°=69.282\ (N)


Fy=60+70+405080sin30°=80 (N)F_y=60+70+40-50-80\cdot\sin30°=80\ (N)


F=69.2822+802=105.83 (N)F=\sqrt{69.282^2+80^2}=105.83\ (N)



τa=0700.2L+400.5L+60L500.8L\sum\tau_a=0\to70\cdot0.2L+40\cdot0.5L+60\cdot L-50\cdot0.8L-


80xL=0x=0.675-80\cdot xL=0\to x=0.675


α=tan1(80/69.282)=49.1°\alpha=\tan^{-1}(80/69.282)=49.1°


Answer:


F=105.83 (N)F=105.83\ (N)


α=49.1°\alpha=49.1°


x=0.675Lx=0.675L




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