Question

A weight of 5.00 x 10^3 N is suspended in equilibrium by two cables. Cable 1 applies a horizontal force to the right of the object and has a tension, FT1. Cable 2 applies a force upward and to the left at an angle of 37.0° to the negative x-axis and has a tension, FT2. What is FT2?

Answer 1

8310 N

Step-by-step explanation

Step 1

Free Body Diagram

Step 1

Newton's first law says that if the net force on an object is zero, then that object will have zero acceleration

so

`\begin{gathered} \sum ^(\square)_(\square)x=F_1-F_2\cos 37=0 \\ \text{replace} \\ F_1-F_2\cos 37=0\rightarrow equation\text{ (1)} \end{gathered}`

and

`\begin{gathered} \sum ^(\square)_(\square)y=F_2\sin 37-w=0 \\ \text{replace} \\ F_2\sin 37-w=0 \\ F_2\sin 37-5.0\cdot10^3=0 \\ so \\ F_2=\frac{5\cdot10^3}{\sin \text{ 37}} \\ F_2=8308.2\text{ N} \\ \text{rounded} \\ F_2=8310\text{ N} \end{gathered}`

therefore, the answer

8310 N

I hope this helps you