Answer to Question #234357 in Mechanics | Relativity for kobe

Question

Replace the three forces acting on the guy wires by a single, equivalent force acting on the flagpole, Use T1=200lb, t2=400 lb and T3=350 lb

Answer 1

Given Data:

The tension in cable AB is: T₁=200 lb

The tension in cable AC is: T₂=400 lb

The tension in cable AD is: T₃=350 lb

The unit tension in cable AB is:

$\vec{T_1}=(200 \;lb) \times \frac{(5i-6j-14k)\;ft}{\sqrt{(5 \;ft)^2 + (-6 \;ft)^2 + (-14 \;ft)^2}} \\ =(200\;lb) \times \frac{(5i-6j-14k) \;ft}{16.03 \;ft} \\ = (62.38i -74.86j -174.67k) \;lb$

The unit tension in cable AC is:

$\vec{T_2}=(400 \;lb) \times \frac{(-3i-3j-14k)\;ft}{\sqrt{(-3 \;ft)^2 + (-3 \;ft)^2 + (-14 \;ft)^2}} \\ =(400\;lb) \times \frac{(-3i-3j-14k) \;ft}{14.63 \;ft} \\ = (-82.02i -82.02j -382.78k) \;lb$

The unit tension in cable AD is:

$\vec{T_3}=(350 \;lb) \times \frac{(2i+6j-14k)\;ft}{\sqrt{(2 \;ft)^2 + (6 \;ft)^2 + (-14 \;ft)^2}} \\ =(350\;lb) \times \frac{(2i+6j-14k) \;ft}{15.36 \;ft} \\ = (45.57i -136.72j -319k) \;lb$

The resultant force acting on the flaqpole is:

$\vec{R}= [(62.38 -82.02+45.57)i +(-74.86-82.02+136.72)j +(-174.67-382.78-319)k] \\ \vec{R} = 25.93i -20.16j -876.45k \;lb$