Q. 54 - Calculus

Question

Each bolt attached in a certain ceiling is known to be able to withstand a force of 200 pounds before it pulls out. What is the maximum weight of an object that can be suspended in the following system? 

Step-by-Step Solution

Answer

The maximum weight of the object is $282.8 \mathrm{lbs}$.

Step 1: Given

Let us consider that the maximum weight of an object that can be suspended in the following system is $m$ lbs.

Step 2: Find maximum weight

Since the weight is stationary, the sum of the three vectors is $u+v+w=0$. We decompose each vector into its horizontal and vertical components, using the standard convention that motion to the right is positive on the x-axis and motion upward is positive on the y-axis.

Therefore, 

$$\begin{gathered} u=-\left|u\right|\cos 45 i+\left|u\right|\sin 45 j=-\frac{\sqrt{2}}{2}\left|u\right|i+\frac{\sqrt{2}}{2}\left|u\right| j \\ v=\left|v\right|\cos 45 i+\left|v\right|\sin 45 j=\frac{1}{\sqrt{2}}\left|v\right|i+\frac{1}{\sqrt{2}}\left|v\right| j \\ w=-m j \end{gathered}$$

Since, $u+v+w=0$

$$\begin{gathered} \left(-\frac{\sqrt{2}}{2}\left|u\right|+\frac{1}{\sqrt{2}}\left|v\right|\right)i+\left(\frac{\sqrt{2}}{2}\left|u\right|+\frac{1}{\sqrt{2}}\left|v\right|-m\right)j=0 \\ \frac{\sqrt{2}}{2}\left|u\right|=\frac{1}{\sqrt{2}}\left|v\right| \mathrm{and} \frac{\sqrt{2}}{2}\left|u\right|+\frac{1}{\sqrt{2}}\left|v\right|=m \\ \mathrm{Given} \mathrm{that}, \left|\mathrm{u}\right|=\left|\mathrm{v}\right|=200 \mathrm{pounds} \\ \frac{\sqrt{2}}{2}\left(200\right)+\frac{1}{\sqrt{2}}\left(200\right)=m \\ m=100\sqrt{2}+100\sqrt{2}=200\sqrt{2}=282.8 \mathrm{lbs} \end{gathered}$$

Hence, the maximum weight of the object is $282.8 \mathrm{lbs}$.