91Ó°ÊÓ

• Consider a 100-pound weight hanging by two ropes.
• The goal is to determine the force magnitude in each rope.

The below given figure shows the weight of 100 pounds suspended by two ropes.

• Because the weight is stationary, the total of the three vectors is zero.
• As a result, $\mathbf{u}+\mathbf{v}+\mathbf{w}=0$.
• Consider that motion to the right is positive on the $x$-axis, and upward motion is positive on the $y$-axis.
• Decompose each vector into horizontal and vertical components at this point.

• Horizontal and vertical components are separated.
• $\mathbf{u}=-\cos 45°‖\mathbf{u}‖\mathbf{i}+\sin 45°‖\mathbf{u}‖\mathbf{j}$
  
• $\mathbf{v}=\cos 30°‖\mathbf{v}‖\mathbf{i}+\sin 30°‖\mathbf{v}‖\mathbf{j}$
  
• $\mathbf{w}=-100\mathbf{j}$
  

• $\mathbf{u}=-\frac{1}{\sqrt{2}}‖\mathbf{u}‖\mathbf{i}+\frac{1}{\sqrt{2}}‖\mathbf{u}‖\mathbf{j}$
  
• $\mathbf{v}=\frac{\sqrt{3}}{2}‖\mathbf{v}‖\mathbf{i}+\frac{1}{2}‖\mathbf{v}‖\mathbf{j}……\left(3\right)$
  
• $\mathbf{w}=-100\mathbf{j}……\left(4\right)$
  

Using (1), $\mathbf{u}+\mathbf{v}+\mathbf{w}=0$

As a result

$\mathbf{u}+\mathbf{v}+\mathbf{w}=\left(-\frac{1}{\sqrt{2}}‖\mathbf{u}‖\mathbf{i}+\frac{1}{\sqrt{2}}‖\mathbf{u}‖\mathbf{j}\right)+\left(\frac{\sqrt{3}}{2}‖\mathbf{v}‖\mathbf{i}+\frac{1}{2}‖\mathbf{v}‖\mathbf{j}\right)-100\mathbf{j}$