Answer:
the answer would be 78.4N worth of tension each.
Explanation:
we know that the sign is motionless and all of the tensions are the same, so let us represent the system with newtons second law with the y-axis pointing upwards
[tex]\sum F_y=4t-mg[/tex]
where t is the tension of one of the strings and m is the mass of the signpost. We know that the net force in the y direction is zero, so[tex]4t-mg=0\Rightarrow{t=\frac{mg}{4}=\frac{32.0*9.8}{4}=78.4}[/tex]
