Chapter 3.3, Problem 19P

Determine the unstretched length of DB to hold the 40-kg crate in the position shown. Take k = 180 N/m.

To determine

The unstretched length of spring DB.

Answer to Problem 19P

The unstretched length $\left(l_0\right)$ of spring DB is $\underset{\_}{2.03\text{m}}$.

Explanation of Solution

Given information:

The stiffness of the spring is $180\text{N}/\text{m}$.

The mass of a crate is 40 kg.

Assumption:

Consider the point E in between the points B and C.

Show the free body diagram of the spring as in Figure 1.

Determine the length of the member CD using the formula.

$l_{CD}=\sqrt{l_{DE}^2+l_{CE}^2}$ (I)

Here, the length of the point DE is $l_{DE}$ and the length of the point CE is $l_{CE}$.

Determine the length of the spring BD using the formula.

$l_{BD}=\sqrt{l_{DE}^2+l_{BE}^2}$ (II)

Here, the length of the point BE is $l_{BE}$.

Determine the weight of the crate.

$W=mg$ (III)

Here, the mass of a crate is m and the acceleration due to gravity is g.

Determine the tension force of the member BD and CD by applying the equation of equilibrium.

Along the horizontal direction:

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ T_{BD}\cos {\theta }_{BD}-T_{CD}\cos {\theta }_{CD}=0\end{array}$ (IV)

Here, the tension force of the member CD is $T_{CD}$ and the angle for the member CD is ${\theta }_{CD}$.

Along the vertical direction:

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ T_{BD}\sin {\theta }_{BD}+T_{CD}\sin {\theta }_{CD}-W=0\end{array}$ (V)

Determine the unstretched length of the spring BD.

$\begin{array}{c}{F}_{BD}=k_{BD}{x}_{BD}\\ =k_{BD}\left(l_{BD}-l_0\right)\end{array}$ (VI)

Here, the tension force of the spring member BD is $F_{BD}$ and the un-stretched length of the spring BD is $l_0$.

Conclusion:

Substitute 2 m for $l_{DE}$ and 2 m for $l_{CE}$ in Equation (I).

$\begin{array}{c}{l}_{CD}=\sqrt{2^2+2^2}\\ =2\sqrt{2}\text{m}\end{array}$

Substitute 2 m for $l_{DE}$ and 3 m for $l_{BE}$ in Equation (II).

$\begin{array}{c}{l}_{BD}=\sqrt{2^2+3^2}\\ =\sqrt{13}\text{m}\end{array}$

Determine the $\cos {\theta }_{BD}$ value.

$\cos {\theta }_{BD}=\frac{l_{BE}}{l_{BD}}$

Substitute 3 m for $l_{BE}$ and $\sqrt{13}\text{m}$ for $l_{BD}$.

$\cos {\theta }_{BD}=\frac{3}{\sqrt{13}}$

Determine the $\cos {\theta }_{CD}$ value.

$\cos {\theta }_{CD}=\frac{l_{CE}}{l_{CD}}$

Substitute 2m for $l_{CE}$ and $2\sqrt{2}$ m for $l_{CD}$.

$\begin{array}{c}\cos {\theta }_{CD}=\frac{2}{2\sqrt{2}}\\ =\frac{1}{\sqrt{2}}\end{array}$

Substitute $\frac{3}{\sqrt{13}}$ for $\cos {\theta }_{BD}$ and $\frac{1}{\sqrt{2}}$ for $\cos {\theta }_{CD}$ in Equation (IV).

$\begin{array}{l}{T}_{BD}\cos {\theta }_{BD}-T_{CD}\cos {\theta }_{CD}=0\\ T_{BD}\left(\frac{3}{\sqrt{13}}\right)-T_{CD}\left(\frac{1}{\sqrt{2}}\right)=0\\ T_{BD}\left(\frac{3}{\sqrt{13}}\right)=T_{CD}\left(\frac{1}{\sqrt{2}}\right)\\ T_{BD}=T_{CD}\left(\frac{\sqrt{13}}{3\sqrt{2}}\right)\end{array}$ (VII)

Substitute 40 kg for m and $9.81\text{m}/{\text{s}}^{\text{2}}$ for g in Equation (III).

$\begin{array}{c}W=40\times 9.81\\ =392.4\frac{\text{kg}\cdot \text{m}}{{\text{s}}^{\text{2}}}\times \frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\\ =392.4\text{N}\end{array}$

Determine the $\sin {\theta }_{BD}$ value.

$\sin {\theta }_{BD}=\frac{l_{DE}}{l_{BD}}$

Substitute 2 m for $l_{DE}$ and $\sqrt{13}\text{m}$ for $l_{BD}$.

$\sin {\theta }_{BD}=\frac{2}{\sqrt{13}}$

Determine the $\sin {\theta }_{CD}$ value.

$\sin {\theta }_{CD}=\frac{l_{DE}}{l_{CD}}$

Substitute 2 m for $l_{DE}$ and $2\sqrt{2}\text{m}$ for $l_{CD}$.

$\begin{array}{c}\sin {\theta }_{CD}=\frac{2}{2\sqrt{2}}\\ =\frac{1}{\sqrt{2}}\end{array}$

Substitute $T_{CD}\left(\frac{\sqrt{13}}{3\sqrt{2}}\right)$ for $T_{BD}$, $\frac{2}{\sqrt{13}}$ for $\sin {\theta }_{BD}$, $\frac{1}{\sqrt{2}}$ for $\sin {\theta }_{CD}$, and 392.4 N for W in Equation (V).

$\begin{array}{l}{T}_{CD}\left(\frac{\sqrt{13}}{3\sqrt{2}}\right)\frac{2}{\sqrt{13}}+T_{CD}\frac{1}{\sqrt{2}}-392.4=0\\ T_{CD}\left(\frac{2}{3\sqrt{2}}\right)+T_{CD}\left(\frac{1}{\sqrt{2}}\right)=392.4\\ \left(\frac{1}{\sqrt{2}}\right)T_{CD}\left(\frac{2}{3}+1\right)=392.4\\ \frac{5}{3\sqrt{2}}{T}_{CD}=392.4\end{array}$

$\begin{array}{l}{T}_{CD}=\frac{392.4\times 3\sqrt{2}}{5}\\ T_{CD}=332.96\text{N}\end{array}$

Substitute 332.96 N for $T_{CD}$ in Equation (VII).

$\begin{array}{c}{T}_{BD}=332.96\left(\frac{\sqrt{13}}{3\sqrt{2}}\right)\\ =282.96\text{N}\end{array}$

Substitute 282.96 N for $F_{BD}$, $\sqrt{13}\text{m}$ for $l_{BD}$, and $180\text{N}/\text{m}$ for $k_{BD}$ in Equation (VI).

$\begin{array}{l}282.96=180\left(\sqrt{13}-l_0\right)\\ \frac{282.96}{180}=\left(\sqrt{13}-l_0\right)\\ l_0=\sqrt{13}-1.572\\ l_0=2.03\text{m}\end{array}$

Thus, the unstretched length of spring DB is $\underset{\_}{2.03\text{m}}$.