Question

Find the magnitude of the vector sum A→+B→+C→ . Each grid square is 2.00 N on a side. If the vector sum is to the west, enter a negative value. If the vector sum is to the east, enter a positive value.

Answer 1

Given data:

* The magnitude of the vector C is,

`\begin{gathered} C=2*2 \\ C=4\text{ N} \end{gathered}`

* The magnitude of the vector B is,

`\begin{gathered} B=4*2 \\ B=8\text{ N} \end{gathered}`

Solution:

From the given diagram, the magnitude of the vector A is,

`\begin{gathered} A=\sqrt[]{Base^2+Perpendicular^2} \\ A=\sqrt[]{(3*2)^2+(4*2)^2} \\ A=\sqrt[]{6^2+8^2} \\ A=\sqrt[]{36+64} \\ A=\sqrt[]{100} \\ A=10\text{ N} \end{gathered}`

The count of square grid in the hypotenuse is,

`\begin{gathered} n=(A)/(2) \\ n=(10)/(2) \\ n=5 \end{gathered}`

The angle of the vector A with the x-axis is,

`\begin{gathered} cos(\theta)=\frac{Base}{\text{Hypotenuse}} \\ \cos (\theta)=(3)/(5) \\ \theta=53.13^(\circ) \end{gathered}`

Thus, the value of vector A is,

`\begin{gathered} \vec{A}=A\cos (53.13^(\circ))+A\sin (53.13^(\circ)) \\ \vec{A}=10*\cos (53.13^(\circ))i+10*\sin (53.13^(\circ))j \\ \vec{A}=6\text{ i + 8 j} \end{gathered}`

The value of vector B is,

`\vec{B}=-8\text{ j}`

The value of vector C is,

`\vec{C}=-4\text{ i}`

Thus, the sum of the vectors is,

`\begin{gathered} \vec{A}+\vec{B}+\vec{C}=6\text{ i+8 j-8 j-4 i} \\ \vec{A}+\vec{B}+\vec{C}=2\text{ i} \\ |\vec{A}+\vec{B}+\vec{C}|=\sqrt[]{2^2} \\ |\vec{A}+\vec{B}+\vec{C}|=\text{ 2 N} \end{gathered}`

Thus, the magnitude of the sum of three given vectors is 2 N towards the east (positive of the x-axis).