Chapter 7.3, Problem 68E

To determine

The system of equations of system pulley.

Answer to Problem 68E

  $t_1=96,t_2=48,$ and $a=-16$ .

Explanation of Solution

Given information:

A system of pulleys is loaded with $128$ -pound and $32$ -pound weights (see figure). Thetensions and in the ropes and the acceleration of the $32$ -pound weight are found by solving the system of equations:

  $\{\begin{array}{l}{t}_1-2t_2\text{ =0}\\ t_1-2a=128\\ t_2+a=32\end{array}$

Where $t_1$ and $t_2$ are in pounds and is in feet per secondsquared. Solve this system.

  

Calculation:

Consider the provided pulley system,

The system of equations formed on the application of Kirchhoff’s rule on the circuit is:

  $\begin{array}{l}{t}_1-2t_2=\mathrm{0...........}(1)\\ t_1-2a=\mathrm{128........}(2)\\ t_2+a=\mathrm{32............}(3)\end{array}$

Simplify the equations first:

  $t_1=2t_2\mathrm{.....}(4)$

Substitute $2t_2$ for $t_1$ in the equation $(2)$ ,

  $2t_2-2a=\mathrm{128........}(5)$

Multiply the equation $(3)$ by $-2$ :

  $\begin{array}{l}-2(t_2+a)=-2(32)\\ -2t_2-2a=-\mathrm{64..............}(6)\end{array}$

Add the equation five and sixth:

  $\begin{array}{l}\underset{\_}{\begin{array}{l}2t_2-2a=128\\ -2t_2-2a=-64\end{array}}\\ -4a=64\\ a=-16\end{array}$

Now solve for $t_2$ , substitute $-16$ for $a$ in the equation $(5)$ :

  $\begin{array}{l}2t_2-2(-16)=128\\ 2t_2+32=128\\ 2t_2=96\\ t_2=48\end{array}$

Plug the value of $t_2$ in equation first to get the value of $t_1$ :

  $\begin{array}{l}{t}_1-2t_2=0\\ t_1-2(48)=0\\ t_1=96\end{array}$

Thus, the value of $t_1=96,t_2=48,$ and $a=-16$ .