Chapter 1, Problem 1.59P

Given $r=4i-6j+2km$ (position vector) $F=20i+40j-30kN$ (force vector) $\lambda =0.8j+0.6k$ (dimensionless unit vector) compute (a) $r\times F\cdot \lambda ;$ and (b) $\lambda \times r\cdot F.$

To determine

(a)

The scalar triple product $r\times F·\lambda .$

Answer to Problem 1.59P

The scalar triple product $r\times F·\lambda$ is given by:

  $r\times F·\lambda =296N.m$

Explanation of Solution

Given that:

Position vector, $r=4i-6j+2k$ m

Force Vector, $F=20i+40j-30k$ N

Dimensionless unit vector, $\lambda =0.8j+0.6k$

Concept Used: - We know that the scalar triple product $A\times B·C$ can be written in the following determinant form:

  $A\times B·C=\left|\begin{array}{ccc}{A}_x& A_y& A_z\\ B_x& B_y& B_z\\ C_x& C_y& C_z\end{array}\right|$

  $A\times B·C=\left|\begin{array}{cc}{B}_y& B_z\\ C_y& C_z\end{array}\right|A_x-\left|\begin{array}{cc}{B}_x& B_z\\ C_x& C_z\end{array}\right|A_y+\left|\begin{array}{cc}{B}_x& B_y\\ C_x& C_y\end{array}\right|A_z$

Calculation: So the scalar triple product $r\times F·\lambda$ is calculated as given by:

  $r\times F·\lambda =\left|\begin{array}{ccc}4& -6& 2\\ 20& 40& -30\\ 0& 0.8& 0.6\end{array}\right|$

  $r\times F·\lambda =\left|\begin{array}{cc}40& -30\\ 0.8& 0.6\end{array}\right|4-\left|\begin{array}{cc}20& -30\\ 0& 0.6\end{array}\right|\left(-6\right)+\left|\begin{array}{cc}20& 40\\ 0& 0.8\end{array}\right|2$

  $\Rightarrow r\times F·\lambda =\left(24+24\right)4+\left(12\right)6+\left(16\right)2$

  $\Rightarrow r\times F·\lambda =\left(48\right)4+\left(12\right)6+\left(16\right)2=296N.m$

To determine

(b)

The scalar triple product $\lambda \times r·F$

Answer to Problem 1.59P

The scalar triple product $\lambda \times r·F$ is given by:

  $\lambda \times r·F=296N.m$

Explanation of Solution

Given that:

Position vector, $r=4i-6j+2k$ m

Force Vector, $F=20i+40j-30k$ N

Dimensionless unit vector, $\lambda =0.8j+0.6k$

Concept Used: - We know that the scalar triple product $A\times B·C$ can be written in the following determinant form:

  $A\times B·C=\left|\begin{array}{ccc}{A}_x& A_y& A_z\\ B_x& B_y& B_z\\ C_x& C_y& C_z\end{array}\right|$

  $A\times B·C=\left|\begin{array}{cc}{B}_y& B_z\\ C_y& C_z\end{array}\right|A_x-\left|\begin{array}{cc}{B}_x& B_z\\ C_x& C_z\end{array}\right|A_y+\left|\begin{array}{cc}{B}_x& B_y\\ C_x& C_y\end{array}\right|A_z$

Calculation: So the scalar triple product $\lambda \times r·F$ is calculated as given by:

  $\lambda \times r·F=\left|\begin{array}{ccc}0& 0.8& 0.6\\ 4& -6& 2\\ 20& 40& -30\end{array}\right|$

  $\lambda \times r·F=\left|\begin{array}{cc}-6& 2\\ 40& -30\end{array}\right|0-\left|\begin{array}{cc}4& 2\\ 20& -30\end{array}\right|\left(0.8\right)+\left|\begin{array}{cc}4& -6\\ 20& 40\end{array}\right|0.6$

  $\Rightarrow \lambda \times r·F=\left(180-80\right)0-\left(-120-40\right)0.8+\left(160+120\right)0.6$

  $\Rightarrow \lambda \times r·F=\left(100\right)0-\left(-160\right)0.8+\left(280\right)0.6=296N.m$