Question

2) A rope is attached to a wall in three-dimensional space, the other end of the rope is attached to a metal pole that is supported by the rope. The tension force in the rope (Fr) and the position vector () that is directed along the metal pole are given below. What is the magnitude of the force along the metal pole (Fu)? F = (+600 N)<+ (+200 N)ỹ + (-200 Nữ 1 = (+3 m)X + (-1 m)9+(+2m)2 rope Metal pole 321N 280 N 663N 217 N 92 N 426 N

Answer 1

So you need to find the force in the direction of the metal pole. So you need to find the direction vector of $\textbf{r}.$ This is nothing but, $\hat{r} = \frac{\textbf{r}}{|\textbf{r}|} = \frac{3\hat{x}-\hat{y}+2\hat{z}}{\sqrt{3^2 + 1^2 + 2^2}}.$ Now $\hat{r}\cdot{\textbf{F}},$ gives the force along the metal pole, which is,

600 x3 – 200 x 1 200 x 2 1200 F.F= 320.713 N 321.0 N. V14 V14