The magnitude of the force on the left-hand pole is mathematically given as
f'=0.167N
Q=45 degrees
What is the magnitude of the force on the left-hand pole.?
Generally, the equation for Force is mathematically given as
[tex]\frac{mg}{sin \theta}[/tex]
Therefore
[tex]F=\frac{17.1*10^{-3}*9.8}{sin45}[/tex]
F=0.237N
Considering horizontal axis or plane
f'-fcos=0
Therefore
f'=0.237*cos45
f'=0.167N
In conclusion, the slope
[tex]tan\theta= 30/30\\\\tan\theta=1\\\\\theta=45[/tex]
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