Answer:
Explanation:
Given
Weight of strut [tex]W_1=400 N[/tex]
Weight of sign Board [tex]W_2=200 N[/tex]
In this question angle which cable makes with horizontal is not given so
assuming [tex]\theta =30^{\circ} [/tex]
From Diagram
Moment about hinge point
[tex]T\sin 30\times L=W_1\times \frac{L}{2}+W_2\times L[/tex]
[tex]T\sin 30=\frac{W_1}{2}+W_2[/tex]
[tex]T\sin 30=\frac{400}{2}+200[/tex]
[tex]T=800 N[/tex]
[tex]\sum F[/tex] in x direction is zero
[tex]F_x=T\cos 30[/tex]
[tex]F_x=800\cos 30=346.41 N[/tex]
[tex]\sum F[/tex] in Y direction is zero therefore
[tex]F_y+T\sin 30=W_1+W_2[/tex]
[tex]F_y=400+200-800\cdot \sin 30[/tex]
[tex]F_y=200 N[/tex]
[tex]F_{net}\ at\ hinge=\sqrt{F_x^2+F_y^2}[/tex]
[tex]F_{net}=399.99\approx 400 N[/tex]