Answer:
Approximately [tex]71.6\; {\rm N}[/tex] (assuming that [tex]g = 9.81\; {\rm N \cdot kg^{-1}}[/tex].)
Explanation:
Refer to the diagram attached. Forces on this object are:
Assuming that [tex]g = 9.81\; {\rm N \cdot kg^{-1}}[/tex], the magnitude of the weight of the sign would be:
[tex]\begin{aligned}(\text{weight}) &= m\, g \\ &= (7.30\; {\rm kg})\, (9.81\; {\rm N\cdot kg^{-1}}) \\ &\approx 71.6\; {\rm N}\end{aligned}[/tex].
Note that weight points downwards (negative) and is entirely in the vertical direction. As a result, the [tex]y[/tex]-component of weight would be equal to [tex](-71.6)\; {\rm N}[/tex].
Hence, the [tex]y[/tex]-component of these forces would be:
Since forces on the object to be balanced, forces need to be balanced in each component. For forces in the [tex]y[/tex]-component to be balanced, forces in the vertical direction need to add up to [tex]0\; {\rm N}[/tex]:
[tex]0\; {\rm N} + (-71.6)\; {\rm N} + (\text{$y$-component of tension on the right}) = 0\; {\rm N}[/tex].
Hence, the [tex]y[/tex]-component of the tension from the wire on the right end would be [tex]71.6\; {\rm N}[/tex].
