Answer:
7 deg
Explanation:
[tex]m[/tex] = mass of the rod = [tex]1.4 kg[/tex]
[tex]W[/tex] = weight of the rod = [tex]mg = (1.4) (9.8) = 13.72 N[/tex]
[tex]k_{L}[/tex] = spring constant for left spring = [tex]59 Nm^{-1}[/tex]
[tex]k_{R}[/tex] = spring constant for right spring = [tex]33 Nm^{-1}[/tex]
[tex]x_{L}[/tex] = stretch in the left spring
[tex]x_{R}[/tex] = stretch in the right spring
[tex]L[/tex] = length of the rod = 0.75 m
[tex]\theta[/tex] = Angle the rod makes with the horizontal
Using equilibrium of force in vertical direction for left spring
[tex]k_{L} x_{L} = (0.5) W\\(59) x_{L} = (0.5) (13.72)\\x_{L} = 0.116 m[/tex]
Using equilibrium of force in vertical direction for right spring
[tex]k_{R} x_{R} = (0.5) W\\(33) x_{R} = (0.5) (13.72)\\x_{R} = 0.208 m[/tex]
Angle made with the horizontal is given as
[tex]\theta = tan^{-1}(\frac{(x_{R} - x_{L})}{L} )\\\theta = tan^{-1}(\frac{(0.208 - 0.116)}{0.75} )\\\theta = 7 deg[/tex]