Question

a uniform 1.2 - kg rod that is 0.90 m long is suspended at rest from the ceiling by two springs, one at each end. both springs hang straight down from the ceiling. the springs have identical lengths when they are unstretched. their spring constants are 33 n/m and 53 n/m. find the angle that the rod makes with the horizontal.

Explanation:

Step1: Calculate the weight of the rod

The weight of the rod \(W = mg\), where \(m = 1.2\space kg\) and \(g=9.8\space m/s^{2}\).
\(W=(1.2)(9.8)=11.76\space N\)
Since the rod is in equilibrium, \(F_{1}+F_{2}=W\), where \(F_{1}=k_{1}x_{1}\) and \(F_{2}=k_{2}x_{2}\) (\(k_{1} = 33\space N/m\), \(k_{2}=53\space N/m\))

Taking torques about the end with spring - 1. The torque due to the weight of the rod \(\tau_{W}=W\times\frac{L}{2}\) (where \(L = 0.90\space m\)), and the torque due to spring - 2 is \(\tau_{2}=F_{2}\times L\). In rotational equilibrium \(\tau_{W}=\tau_{2}\)

\(W\times\frac{L}{2}=F_{2}\times L\), so \(F_{2}=\frac{W}{2}\) (this is a wrong approach. Let's use the correct torque equation. Let the elongation of spring 1 be \(x_{1}\) and spring 2 be \(x_{2}\). Taking torques about the center of the rod. The torque due to spring 1 \(\tau_{1}=F_{1}\times\frac{L}{2}\) and the torque due to spring 2 \(\tau_{2}=F_{2}\times\frac{L}{2}\). In equilibrium \(\tau_{1}=\tau_{2}\) (no, wrong. Let's start over.

Let the elongation of spring 1 be \(x_{1}\) and spring 2 be \(x_{2}\). The forces \(F_{1}=k_{1}x_{1}\) and \(F_{2}=k_{2}x_{2}\). Since the rod is in translational equilibrium \(F_{1}+F_{2}=mg\). Taking torques about the end of the rod attached to spring 1. The torque due to the weight of the rod \(\tau_{w}=mg\times\frac{L}{2}\) and the torque due to spring 2 is \(\tau_{2}=F_{2}\times L\). In rotational equilibrium \(mg\times\frac{L}{2}=F_{2}\times L\), so \(F_{2}=\frac{mg}{2}\) (again wrong. Correct:

Let’s use the equations of equilibrium. For translational equilibrium \(F_{1}+F_{2}=mg\). For rotational equilibrium (taking torques about the center of the rod), \(F_{1}\times\frac{L}{2}=F_{2}\times\frac{L}{2}\) (no. Let's use the general torque formula. Let the angle of the rod with the horizontal be \(\theta\). The vertical displacements of the ends of the rod are related to the elongations of the springs. If the rod makes an angle \(\theta\) with the horizontal, and the length of the rod is \(L\), then \(x_{2}-x_{1}=L\sin\theta\).

From translational equilibrium \(k_{1}x_{1}+k_{2}x_{2}=mg\). From rotational equilibrium (taking torques about the center of the rod), \(k_{1}x_{1}\times\frac{L}{2}=k_{2}x_{2}\times\frac{L}{2}\), so \(k_{1}x_{1}=k_{2}x_{2}\)

From \(k_{1}x_{1}=k_{2}x_{2}\), we have \(x_{1}=\frac{k_{2}}{k_{1}}x_{2}\). Substitute into \(k_{1}x_{1}+k_{2}x_{2}=mg\)

\(k_{2}x_{2}+k_{2}x_{2}=mg\) (no, \(k_{1}\times\frac{k_{2}}{k_{1}}x_{2}+k_{2}x_{2}=mg\), \(2k_{2}x_{2}=mg\), \(x_{2}=\frac{mg}{2k_{2}}\) and \(x_{1}=\frac{mg}{2k_{1}}\)

Then \(x_{2}-x_{1}=L\sin\theta\)

\(\sin\theta=\frac{\frac{mg}{2k_{2}}-\frac{mg}{2k_{1}}}{L}\)

Substitute \(m = 1.2\space kg\), \(g = 9.8\space m/s^{2}\), \(k_{1}=33\space N/m\), \(k_{2}=53\space N/m\), \(L = 0.90\space m\)

\(\sin\theta=\frac{mg}{2L}(\frac{1}{k_{2}}-\frac{1}{k_{1}})\)

\(\sin\theta=\frac{(1.2)(9.8)}{2\times0.90}(\frac{1}{53}-\frac{1}{33})\)

First calculate \(\frac{(1.2)(9.8)}{2\times0.90}=\frac{11.76}{1.8}\approx6.533\)

\(\frac{1}{53}-\frac{1}{33}=\frac{33 - 53}{53\times33}=\frac{- 20}{1749}\approx - 0.0114\)

\(\sin\theta=6.533\times(- 0.0114)\approx - 0.0745\) (magnitude, \(\sin\theta=\frac{mg}{2L}(\frac{1}{k_{1}}-\frac{1}{k_{2}})\))

\(\sin\theta=\frac{(1.2)(9.8)}{2\times0.90}(\frac{1}{33}-\frac{1}{53})\)

\(\frac{(1.2)(9.8)}{2\times0.90}=\frac{11.76}{1.8}\approx6.533\)

\(\frac{1}{33}-\frac{1}{53}=\frac{53 - 33}{33\times53}=\frac{20}{1749}\approx0.0114\)

\(\sin\theta=6.533\times0.0114\approx0.0745\)

Step2: Calculate the angle \(\theta\)

Since \(\sin\theta\approx0.0745\), then \(\theta=\sin^{- 1}(0.0745)\)

\(\theta\approx4.3^{\circ}\)

Answer:

\(4.3^{\circ}\)