Question

Each spring has an unstretched length of 2 m and a stiffness of k = 320 N/m.

Determine the stretch in OA and OB spring required to hold the 19-kg

Answer 1

Since the given system is in equilibrium, therefore the resultant force is;

$\small \sum F=0$

   $\small \Rightarrow \overrightarrow{T}_{OA}+\overrightarrow{T}_{OB}+\overrightarrow{T}_{OC}+\overrightarrow{W}=0$

   $\small \Rightarrow T_{OA}(-\hat{j})+T_{OB}(-\hat{i})+T_{OC}\left ( \frac{6\hat{i}+4\hat{j}+12\hat{k}}{\sqrt{6^{2}+4^{2}+12^{2}}} \right )+W(-\hat{k})=0$

   $\small \Rightarrow T_{OA}(-\hat{j})+T_{OB}(-\hat{i})+T_{OC}\left ( \frac{6\hat{i}+4\hat{j}+12\hat{k}}{14} \right )+W(-\hat{k})=0$

   $\small \Rightarrow \left (\left ( \frac{6}{14} \right )T_{OC}-T_{OB} \right )\hat{i}+\left (\left ( \frac{4}{14} \right )T_{OC}-T_{OA} \right )\hat{j}+\left (T_{OC}\left ( \frac{12}{14} \right )-W \right )\hat{k}=0$

Now compare the i, j, k coefficients. we get;

   $\small T_{OC}=\left ( \frac{14}{12} \right )W=\left ( \frac{14}{12} \right )mg$

   $\small \Rightarrow T_{OC}=\left ( \frac{14}{12} \right )(19)(9.8)=217.23\: N$

   $\small T_{OB}=\left ( \frac{6}{14} \right )T_{OC}=\left ( \frac{6}{14} \right )217.23=93.1\: N$

   $\small T_{OA}=\left ( \frac{4}{14} \right )T_{OC}=\left ( \frac{4}{14} \right )217.23=62.1\: N$

Thus the stretch in spring OA is;

   $\small S_{OA}=\frac{T_{OA}}{k}=\frac{62.1}{320}=0.194\: m$

and the stretch in spring OB is;

   $\small S_{OB}=\frac{T_{OB}}{k}=\frac{93.1}{320}=0.291\: m$