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Q10.

Expert-verified
Found in: Page 68

### Physics Principles with Applications

Book edition 7th
Author(s) Douglas C. Giancoli
Pages 978 pages
ISBN 978-0321625922

# Given the vectors $\stackrel{\to }{A}$ and $\stackrel{\to }{B}$ in figure below, determine $\stackrel{\to }{B}-\stackrel{\to }{A}$. (b) Determine $\stackrel{\to }{A}-\stackrel{\to }{B}$ without using your answer in (a). Then compare your results your results and see if they are opposite.

(a) The X and Y components of vector $\stackrel{\to }{B}-\stackrel{\to }{A}$ are -54.57 and 2.68, respectively.

(b) The X and Y components of vector $\stackrel{\to }{A}-\stackrel{\to }{B}$ are 54.57 and -2.68, respectively.

See the step by step solution

## Step 1. Addition of vectors

Vector addition can be done by resolving components of individual vectors along the axes and adding corresponding components.

## Step 2. Given data and assumptions

The magnitude of vector $\stackrel{\to }{A}$, A = 44.0

The magnitude of vector $\stackrel{\to }{B}$, B = 26.5

## Step 3. Resolving vector components of A→ and B→ along the axes

Resolving components of vector $\stackrel{\to }{A}$

X-component of vector $\stackrel{\to }{A}$:

$\begin{array}{c}{A}_{\mathrm{x}}=A\mathrm{cos}26°\\ =44.0×\mathrm{cos}26°\\ =39.55\end{array}$

Y-component of vector $\stackrel{\to }{A}$:

$\begin{array}{c}{A}_{\mathrm{y}}=A\mathrm{cos}64°\\ =44.0×\mathrm{cos}64°\\ =19.29\end{array}$

Resolving components of vector $\stackrel{\to }{B}$

X-component of vector $\stackrel{\to }{B}$:

$\begin{array}{c}{B}_{\mathrm{x}}=-B\mathrm{cos}26°\\ =-26.5×\mathrm{cos}56°\\ =-14.82\end{array}$

Y-component of vector $\stackrel{\to }{B}$:

$\begin{array}{c}{B}_{\mathrm{y}}=B\mathrm{cos}34°\\ =26.5×\mathrm{cos}34°\\ =21.97\end{array}$

## Step 4. Calculating the value of B→-A→

The component of $\stackrel{\to }{B}-\stackrel{\to }{A}$ along the x-direction is given as:

$\begin{array}{c}{\left(\stackrel{\to }{B}-\stackrel{\to }{A}\right)}_{\mathrm{x}}={B}_{\mathrm{x}}-{A}_{\mathrm{x}}\\ =-14.82-39.55\\ =-54.57\end{array}$

The component of $\stackrel{\to }{B}-\stackrel{\to }{A}$ along the y-direction is given as:

$\begin{array}{c}{\left(\stackrel{\to }{B}-\stackrel{\to }{A}\right)}_{\mathrm{y}}={B}_{\mathrm{y}}-{A}_{\mathrm{y}}\\ =21.97-19.29\\ =2.68\end{array}$

$\stackrel{\to }{B}-\stackrel{\to }{A}$ vector has components -54.57 and 2.68 in the x and y-direction, respectively.

## Step 5. Calculating the value of A→-B→

The component of $\stackrel{\to }{A}-\stackrel{\to }{B}$ along the x-direction is given as:

$\begin{array}{c}{\left(\stackrel{\to }{A}-\stackrel{\to }{B}\right)}_{\mathrm{x}}={A}_{\mathrm{x}}-{B}_{\mathrm{x}}\\ =39.55+14.82\\ =54.57\end{array}$

The component of $\stackrel{\to }{A}-\stackrel{\to }{B}$ along the y-direction is given as:

$\begin{array}{c}{\left(\stackrel{\to }{A}-\stackrel{\to }{B}\right)}_{\mathrm{y}}={A}_{\mathrm{y}}-{B}_{\mathrm{y}}\\ =19.29-21.97\\ =-2.68\end{array}$

$\stackrel{\to }{A}-\stackrel{\to }{B}$ vector has components 54.57 and -2.68 in the x and y-direction, respectively.

## Step 6. Checking whether vectors B→-A→ and A→-B→ are equal and opposite or not

As the values of ${\left(\stackrel{\to }{A}-\stackrel{\to }{B}\right)}_{\mathrm{x}}=-{\left(\stackrel{\to }{B}-\stackrel{\to }{A}\right)}_{\mathrm{x}}$ and ${\left(\stackrel{\to }{A}-\stackrel{\to }{B}\right)}_{\mathrm{x}}=-{\left(\stackrel{\to }{B}-\stackrel{\to }{A}\right)}_{\mathrm{x}}$; thus, we can say that vectors $\stackrel{\to }{A}-\stackrel{\to }{B}$ and $\stackrel{\to }{B}-\stackrel{\to }{A}$ are equal in magnitude but opposite in direction.