Question

1. determine the vector that must be added to the sum of 𝐴 + 𝐵 in figure 1 to give a resultant displacement of (a) 0 and (b) 4.0 km w.

Explanation:

Step1: Resolve vectors $\vec{A}$ and $\vec{B}$ into components

Let the positive x - axis be towards the East and positive y - axis be towards the North.
For $\vec{A}$ with $|\vec{A}| = 5.1$ km, $A_x=0$ km, $A_y = 5.1$ km.
For $\vec{B}$ with $|\vec{B}|=6.8$ km, $B_x=-6.8\sin(10^{\circ})\approx - 1.18$ km, $B_y=-6.8\cos(10^{\circ})\approx - 6.69$ km.
The sum of the x - components of $\vec{A}+\vec{B}$ is $R_x=A_x + B_x=- 1.18$ km and the sum of the y - components is $R_y=A_y + B_y=5.1-6.69=-1.59$ km. So, $\vec{R}=\vec{A}+\vec{B}=(-1.18\hat{i}-1.59\hat{j})$ km.

Step2: Find the vector for resultant 0

Let the vector to be added be $\vec{C}=(C_x\hat{i}+C_y\hat{j})$. If $\vec{R}+\vec{C}=\vec{0}$, then $C_x=-R_x = 1.18$ km and $C_y=-R_y = 1.59$ km. So, $\vec{C}=(1.18\hat{i}+1.59\hat{j})$ km. In magnitude - direction form, $|\vec{C}|=\sqrt{C_x^{2}+C_y^{2}}=\sqrt{(1.18)^{2}+(1.59)^{2}}\approx1.98$ km. $\theta=\tan^{-1}(\frac{C_y}{C_x})=\tan^{-1}(\frac{1.59}{1.18})\approx53.4^{\circ}$ N of E.

Step3: Find the vector for resultant 4.0 km [W]

The desired resultant $\vec{R_d}=-4.0\hat{i}$ km. Let the vector to be added be $\vec{D}=(D_x\hat{i}+D_y\hat{j})$. Then $\vec{R}+\vec{D}=\vec{R_d}$. So, $D_x=R_d_x - R_x=-4.0-(-1.18)=-2.82$ km and $D_y=R_d_y - R_y=0 - (-1.59)=1.59$ km. $\vec{D}=(-2.82\hat{i}+1.59\hat{j})$ km. In magnitude - direction form, $|\vec{D}|=\sqrt{D_x^{2}+D_y^{2}}=\sqrt{(-2.82)^{2}+(1.59)^{2}}\approx3.24$ km. $\theta=\tan^{-1}(\frac{D_y}{|D_x|})=\tan^{-1}(\frac{1.59}{2.82})\approx29.4^{\circ}$ N of W.

Answer:

(a) A vector of approximately $1.98$ km at an angle of $53.4^{\circ}$ N of E.
(b) A vector of approximately $3.24$ km at an angle of $29.4^{\circ}$ N of W.