Question

Write the position of point A, B and C as Cartesian vectors r_A, r_B, r_c. Express the position vectors from A to Band A to C as unit vectors, url, Ur2. Express F_b and F_c as Cartesian Vectors. Determine the magnitude of the resultant force at A.

Answer 1

(1) From the given diagram we can write the cartesian vectors as;

$\small r_{A}=6\hat{k}$

   $\small r_{B}=3\hat{i}-2\hat{j}$

$\small r_{C}=2\hat{i}+3\hat{j}$

(2) The position vector from A to B will be;

   $\small \overrightarrow{AB}=(3-0)\hat{i}+(-2-0)\hat{j}+(0-6)\hat{k}=3\hat{i}-2\hat{j}-6\hat{k}$

Therefore the unit vector will be;

$\small \hat{AB}=\frac{\overrightarrow{AB}}{\left | \overrightarrow{AB} \right |}=\frac{3\hat{i}-2\hat{j}-6\hat{k}}{\left | 3\hat{i}-2\hat{j}-6\hat{k} \right |}=\frac{3}{7}\hat{i}-\frac{2}{7}\hat{j}-\frac{6}{7}\hat{k}$

The position vector from A to B will be;

   $\small \overrightarrow{AC}=(2-0)\hat{i}+(3-0)\hat{j}+(0-6)\hat{k}=2\hat{i}+3\hat{j}-6\hat{k}$

Therefore the unit vector will be

   $\small \hat{AC}=\frac{\overrightarrow{AC}}{\left | \overrightarrow{AC} \right |}=\frac{2\hat{i}+3\hat{j}-6\hat{k}}{\left | 2\hat{i}+3\hat{j}-6\hat{k} \right |}=\frac{2}{7}\hat{i}+\frac{3}{7}\hat{j}-\frac{6}{7}\hat{k}$

(3) Force F_B in cartesian vector will be;

   $\small \overrightarrow{F_{B}}=\left | F_{B} \right |(\hat{AB})=840\left (\frac{3}{7}\hat{i}-\frac{2}{7}\hat{j}-\frac{6}{7}\hat{k} \right )=360\hat{i}-240\hat{j}-720\hat{k}$

and the force F_C in cartesian vector will be;

$\small \overrightarrow{F_{C}}=\left | F_{C} \right |(\hat{AB})=420\left (\frac{2}{7}\hat{i}+\frac{3}{7}\hat{j}-\frac{6}{7}\hat{k} \right )=120\hat{i}+180\hat{j}-360\hat{k}$

(4) The resultant force at A is;

$\small \overrightarrow{F}=\overrightarrow{F_{B}}+\overrightarrow{F_{C}}$

   $\small =(360+120)\hat{i}+(-240+180)\hat{j}+(-720-360)\hat{k}=480\hat{i}-60\hat{j}-1080\hat{k}$

Therefore the magnitude is;

$\small \left | F \right |=\sqrt{(480)^{2}+(-60)^{2}+(-1080)^{2}}=1183.385N$