Question

Carry through the details of getting $\mathbf{(}\mathbf{7}\mathbf{.4}\mathbf{)}$from$\mathbf{(}\mathbf{7}\mathbf{.2}\mathbf{)}$ and $\mathbf{(}\mathbf{7}\mathbf{.3}\mathbf{)}$ . Hint: You need the dot product of $\mathit{e}{\mathbf{\text{'}}}_{\mathbf{\beta }}$and $e_j$ . This is the cosine of an angle between two axes since each eis a unit vector. Identify the result from matrixAin$\mathbf{(}\mathbf{2}\mathbf{.10}\mathbf{)}$ .

Short answer

The statement has been proven.

Step-by-step solution

Given Information

Force per unit area $P_{ij}$ is perpendicular to the $i^{th}$direction and acting in the $j^{th}$direction

Definition of a force per unit area.

The force exerted on a unit area of the surface is known as force per unit area.

Verify the statement.

Letthe force is exerted by the material on$dS{\text{'}}_{\alpha }=dS$.

Component${\beta }^{th}$of a force in K system per unit area is $P{\text{'}}_{\alpha \beta }$.

The total force exerted is $P{\text{'}}_{\alpha \beta }e{\text{'}}_{\beta }dS$.

The force acts on surfaces perpendicular to the axes is given below.

$P{\text{'}}_{\alpha \beta }e{\text{'}}_{\beta }dS=\sum _i{F}_i$per area $.d{S}_i$

$\begin{array}{l}P{\text{'}}_{\alpha \beta }e{\text{'}}_{\beta }dS=P{\text{'}}_{\alpha \beta }e{\text{'}}_{\beta }d{S}_i\\ P{\text{'}}_{\alpha \beta }e{\text{'}}_{\beta }dS=P_{ij}{e}_j{a}_{\alpha i}dS\end{array}$

Dot both sides of the above equation by $e{\text{'}}_{\beta }$

The equation becomes as follows.

$\begin{array}{c}P{\text{'}}_{\alpha \beta }e{\text{'}}_{\beta }.e{\text{'}}_{\beta }=P_{ij}(e_j.e{\text{'}}_{\beta })a_{\alpha i}\\ P{\text{'}}_{\alpha \beta }=a_{\alpha i}{a}_{\beta j}{P}_{ij}\end{array}$

Hence, the statement has been proven.