Chapter 8.5, Problem 66E

To determine

Use Cramer’s rule to find $t_1$ , $t_2$ and $a$ .

Answer to Problem 66E

  $\boxed{t_1=\frac{1157}{7},t_2=\frac{576}{7},a=\frac{-64}{7}}$

Explanation of Solution

Given information:

A system of pulleys that is assumed frictionless and without mass is loaded with $192$ pound and $64$ pound weights (see figure). The tensions $t_1$ and $t_2$ in the ropes and the accelaration $a$ of $64$ pound weight are found by solving the system

  $\begin{array}{l}{t}_1-2t_2=0\\ t_1-3a=192\\ t_2+2a=64\end{array}$

Where $t_1$ and $t_2$ are measured in pounds and $a$ is in feet per second square. Use Cramer’s rule to find $t_1$ , $t_2$ and $a$ .

Calculation:

Consider the system of equations.

  $\begin{array}{l}{t}_1-2t_2=0\\ t_1-3a=192\\ t_2+2a=64\end{array}$

This will be solved by Cramer’s rule.

Cosider the coefficient matrix

  $\begin{array}{ccc}1& -2& 0\\ 1& 0& -3\\ 0& 1& 2\end{array}$

Determinant of the matrix is $D=\begin{array}{ccc}1& -2& 0\\ 1& 0& -3\\ 0& 1& 2\end{array}$

  $D=1\begin{array}{cc}0& -3\\ 1& 2\end{array}+2\begin{array}{cc}1& -3\\ 0& 2\end{array}+0\begin{array}{cc}1& 0\\ 0& 1\end{array}$

  $=1\left(0+3\right)+2\left(2-0\right)+0$

  $=7\ne 0$

So, Cramer’s rule can be applied.

For Cramer’s rule $t_1=\frac{D_{t_1}}{D},t_2=\frac{D_{t_2}}{D},a=\frac{D_a}{D}$

  $t_1=\frac{D_{t_1}}{D}$

  $=\frac{1}{7}\begin{array}{ccc}0& -2& 0\\ 192& 0& -3\\ 64& 1& 2\end{array}$

  $=\frac{1}{7}\begin{array}{cc}0& -3\\ 1& 2\end{array}+2\begin{array}{cc}192& -3\\ 64& 2\end{array}+0\begin{array}{cc}192& 0\\ 64& 1\end{array}$

  $=$$\frac{1152}{7}$

  $t_2=\frac{D_{t_2}}{D}$

  $=\frac{1}{7}\begin{array}{ccc}1& 0& 0\\ 1& 192& -3\\ 0& 64& 2\end{array}$

  $=\frac{1}{7}\begin{array}{cc}192& -3\\ 64& 2\end{array}$

  $=\frac{576}{7}$

  $a=\frac{D_a}{D}$

  $=\frac{1}{7}\begin{array}{ccc}1& -2& 0\\ 1& 0& 192\\ 0& 1& 64\end{array}$

  $=\frac{1}{7}\left[\left(0-192\right)+2\left(64-0\right)+0\right]$

  $=\frac{-64}{7}$

Hence, the solution is $\boxed{t_1=\frac{1157}{7},t_2=\frac{576}{7},a=\frac{-64}{7}}$