Three ropes are attached to the post at A in Fig. (a). The forces in the ropes are F1 = 260 lb, F2 = 75 lb, and F3 = 60 lb. Determine (1) the magnitude of the force R that is equivalent to the three forces shown, and (2) the coordinates of the point where the line of action of R intersects

Chapter 2 Q. 2.2

Engineering Mechanics Statics [EXP-90698]

Three ropes are attached to the post at A in Fig. (a). The forces in the ropes are $F_{1}=260\:lb$ , $F_{2} = 75\:lb$ , and $F_{3} = 60\:lb$ . Determine (1) the magnitude of the force $R$ that is equivalent to the three forces shown, and (2) the coordinates of the point where the line of action of $R$ intersects the $yz$ -plane.

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Part 1
The forces are concurrent at point A and thus may be added immediately. Because the forces do not lie in a coordinate plane, it is convenient to use vector notation. One method for expressing each of the forces in vector notation is to use the form F=Fλ, where λ is the unit vector in the direction of the force F. Thus

F_{1}=260λ_{AB}=260\frac{\overrightarrow{AB}}{|\overrightarrow{AB|}}=260\left(\frac{−3i−12j+4k}{13} \right) =−60i−240j+80k\:lb

F_{2}=75λ_{AC}=75\frac{\overrightarrow{AC}}{|\overrightarrow{AC|}}=75\left(\frac{−3i+4k}{5} \right) =−45i+60k\:lb

F_{3}=-60j\:lb

The resultant force is given by
$R =\Sigma{F}=F_{1}+F_{2}+F_{3}=(−60i−240j+80k)+(−45i+60k)+(−60j)=−105i−300j+140k\:lb$

The magnitude of $R$ is
$R =\sqrt{(−105)^2+(−300)^2+(140)^2}=347.3\:lb$

Part 2
The unit vector $λ$ in the direction of $R$ is
$λ=\frac{R}{R}=\frac{−105i−300j+140k}{347.3}=−0.3023i−0.8638j+0.4031k$

Let $D$ be the point where $λ$ intersects the $yz$ -plane, as shown in $Fig.\:(b)$ . The coordinates of $D$ can be determined by proportions:
$\frac{|λ_{x}|}{3}=\frac{|λ_{y}|}{12-y_{D}}=\frac{|λ_{z}|}{z_{D}}$

Substituting the components of $λ$ , this becomes
$\frac{0.3023}{3}=\frac{0.8638}{12-y_{D}}=\frac{0.4031}{z_{D}}$

yielding
$y_{D} =3.43\:ft$ $z_{D}=4.0\:ft$