Question

Find the scalar product of the vectors in Figure P7.10.

Short answer

The scalar product of the two vectors is 5.33 J.

Step-by-step solution

Given data

Magnitude of the first vector

A =32.8 N

Magnitude of the second vector

$$\begin{gathered} B=17.3\mathrm{cm} \\ =17.3·\left(1\mathrm{cm}\times \frac{1\mathrm{m}}{100\mathrm{cm}}\right) \\ =0.173\mathrm{m} \end{gathered}$$

Large angle made by the first vector and y axis = ${118}^{\circ }$

Large angle made by the second vector and x axis = ${132}^{\circ }$

Dot product of two vectors

The dot product of two vectors $\overrightarrow{\mathbf{A}}$and $\overrightarrow{\mathbf{B}}$ is

$\overrightarrow{\mathbf{A}}\mathbf{·}\overrightarrow{\mathbf{B}}\mathbf{=}\mathbf{ABcos\theta }\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(I\right)$

Determining the dot product of the given vectors

Large angle between the positive y axis and the positive x axis = ${270}^{°}$

Thus, angle between the two vectors

$$\begin{gathered} \theta ={270}^{\circ }-{118}^{\circ }-{132}^{\circ } \\ ={20}^{\circ } \end{gathered}$$

From equation (I), the dot product of the two vectors is

$$\begin{gathered} \overrightarrow{A}·\overrightarrow{B}=32.8\mathrm{N}\times 0.173\mathrm{m}\times \cos {20}^{\circ } \\ =5.33·\left(1\mathrm{N}·\mathrm{m}\times \frac{1\mathrm{J}}{1\mathrm{N}·\mathrm{m}}\right) \\ =5.33\mathrm{J} \end{gathered}$$

Thus the required dot product is 5.33 J.