Answer to Question #156190 in Analytic Geometry for Keleko

Question

A system of forces F1, F2 and F3 acting through points with position vectors r1, r2 and r3 respectively, where F1 = (3i - 3j + 4k) N, F2 = (3i + 4j + 3k)N, F3 = (-4i - 2j + mk) N,
r1 = (i + j + k)m , r2 = ( 3i + 2j + k)m , r3 = (2i - j)m reduces to a couple G and a single force F = (2i - j + 8k)N acting through the point with position vector r0 = (2i - j + 2k)
Find
(i) the value of m,
(ii) the vector moment of F2 about the point with position vector r0,
(iii) the Cartesian equation of the line of action of F,
(iv) the magnitude of the couple G.

Answer 1

$\vec{r_1} = \hat{i} + \hat{j}+\hat{k}$ , $\vec{r_2} =3 \hat{i} +2 \hat{j}+\hat{k}$ and $\vec{r_3} = 2\hat{i} - \hat{j}$

$\vec{F_1} = 3\hat{i} -3 \hat{j}+4\hat{k}$ , $\vec{F_2} = 3\hat{i} + 4\hat{j}+3\hat{k}$ and $\vec{F_3} =-4 \hat{i}-2 \hat{j}+m\hat{k}$

$\vec{F_0} = 2\hat{i}- \hat{j}+8\hat{k}$ and $\vec{r_0} = 2\hat{i} - \hat{j}+2\hat{k}$

(i) $\vec{F_0}=\vec{F_1}+\vec{F_2}+\vec{F_3}$

taking only z component.

$8 = 4+3+m \implies m=1$

(ii) Moment of force $F_2$ , $\vec{r} \times \vec{F_2} =(\vec{r_2}-\vec{r_0} )\times \vec{F_2} = ( \hat{i} +3 \hat{j}-\hat{k}) \times (3 \hat{i} +4 \hat{j}+3\hat{k}) = 13 \hat{i} -6 \hat{j}-5\hat{k}$

(iii) Line of action is the line along which force act. Since force is acting along the line $\vec{r} = 2\hat{i}-\hat{j}+2\hat{k}$ . So unit vector in that direction is $\hat{r}=\frac{( 2\hat{i}- \hat{j}+2\hat{k})}{3}$

(iv) Couple is formed when two parallel force act in opposite direction. Here there no two forces are parallel. So couple is zero.