Three ropes are attached to the post at A in Fig. (a). The forces in the ropes are F1 = 260N, F2 = 75 N, and F3 = 60 N. 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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Accepted Answer

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

[latex]F_{1}=260\lambda _{AB}=260\frac{\overrightarrow{AB} }{\left|\overrightarrow{AB} \right| }=260(\frac{−3i − 12j + 4k}{13} ) =−60i − 240j + 80k N[/latex]

[latex]F_{2}=75\lambda _{AC}=75\frac{\overrightarrow{AC} }{\left|\overrightarrow{AC} \right| }=75(\frac{−3i + 4k}{5} ) =−45i + 60k N[/latex]

[latex]F_{3}=-60j N[/latex]

The resultant force is given by

[latex]R = ΣF=F_{1} + F_{2} + F_{3}[/latex]

= (−60i − 240j + 80k) + (−45i + 60k) + (−60j)= −105i − 300j + 140k N

The magnitude of R is

[latex]R=\sqrt{(-105)^{2}+(-300)^{2}+(140)^{2}}=347.3 N[/latex]

Part 2

The unit vector λ in the direction of R is

[latex]\lambda =\frac{R}{R}=\frac{−105i − 300j + 140k}{347.3}= −0.3023i − 0.8638j + 0.4031k[/latex]

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:

[latex]\frac{\left|\lambda _{x}\right| }{3}=\frac{\left|\lambda _{y}\right| }{12-y_{D}}=\frac{\left|\lambda _{z}\right| }{z_{D}}[/latex]

Substituting the components of λ, this becomes

[latex]\frac{0.3023 }{3}=\frac{0.8638 }{12-y_{D}}=\frac{0.4031 }{z_{D}}[/latex]

yielding

[latex]y_{D} = 3.43 m[/latex] [latex]z_{D} = 4.0 m[/latex]

Question 2.2: Three ropes are attached to the post at A in Fig. (a). The f...

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