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?

Short answer

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

Step-by-step solution

Given

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

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,

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

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$282.8\mathrm{lbs}$.