The rigid rod AB is attached to a hinge at A and to two springs, each of constant k. If h...

Calculate the distance displaced by the spring.

\(x=d \theta\)

Here, distance of the spring from \(A\) is \(d\).

Calculate the spring force when the body is rotated by an angle \(\theta\).

\(F=k x\)

Here, spring constant is \(k\) and distance displaced by the spring is \(x\).

Substitute \(d \sin \theta\) for \(x\) in the above equation.

\(F=k d \theta\)

Calculate the spring constant by taking moments about \(A\).

\(\sum M_{A}=0\)

\(2 F y-m g x^{\prime}=0\)

Here, spring force is \(F\), mass of the sleeve is \(m\), acceleration due to gravity is \(g\), perpendicular distance of force \(F\) from \(A\) is \(y\) and perpendicular distance of weight of the rod from \(A\) is \(x^{\prime}\).

Substitute \(d \cos \theta\) for \(y, k d \sin \theta\) for \(F\) and \(h \sin \theta\) for \(x^{\prime}\).

\((2 k d \sin \theta \times d \cos \theta)-m g h \sin \theta=0\)

\(k=\frac{m g h \sin \theta}{2 d^{2} \sin \theta \cos \theta} \quad \ldots \ldots\)

\(k=\frac{m g h}{2 d^{2} \cos \theta}\)

Here, angle turned is so small that the value of \(\cos \theta\) is approximately equal to 1 .

\(k=\frac{m g h}{2 d^{2}}\)

Substitute \(200 \mathrm{~kg}\) for \(m, 9.81 \mathrm{~m} / \mathrm{s}^{2}\) for \(g, 0.45 \mathrm{~m}\) for \(h, 0.3 \mathrm{~m}\) for \(d\) in equation (1).

$$ \begin{aligned} k &=\frac{200 \times 9.81 \times 0.45}{2 \times 0.3^{2}} \\ &=4905 \mathrm{~N} / \mathrm{m} \\ &=4.905 \mathrm{kN} / \mathrm{m} \end{aligned} $$

Therefore, the range of values of spring constant is \(k>4.91 \mathrm{kN} / \mathrm{m}\).