Chapter 7.3, Problem 50E

To determine

To find: the values of tensions $t_1,t_2$ and the acceleration a of the 32-pound weight by solving the given system.

Answer to Problem 50E

The value of $t_1$ is 96 pounds, the value of $t_2$ is 48 pounds and the value of acceleration is -16 feet per second

Explanation of Solution

Given information:

A system of pulleys is loaded with 128-pound and 32-pound weights. The tensions $t_1$ and $t_2$ in the ropes and the acceleration a of the 32-poundweight are found by solving the system

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

  

Calculation:

Using Kirchhoff’s rule on the circuit the system of equation formed will be

  $\begin{array}{l}{t}_1-2t_2=0\cdots \left(i\right)\\ t_1-2a=128\cdots \left(ii\right)\\ t_2+a=32\cdots \left(iii\right)\end{array}$

From equation $\left(i\right)$

We get

  $t_1=2t_2$

Put $t_1=2t_2$ in $\left(ii\right)$

We get

  $2t_2-2a=128\cdots \left(v\right)$

Now add twice of equation $\left(iii\right)$ to $\left(v\right)$

  $\begin{array}{l}2t_2+2t_2+2a-2a=2\times 32+128\\ 4t_2=64+128\\ 4t_2=192\\ t_2=\frac{192}{4}\\ t_2=48\end{array}$

Put $t_2=48$ in equation $\left(v\right)$

  $\begin{array}{l}2\left(48\right)-2a=128\\ 96-2a=128\\ 2a=96-128\\ 2a=-32\\ a=-16\end{array}$

Put $t_2=48$ in $t_1=2t_2$

We get,

  $\begin{array}{l}{t}_1=2\left(48\right)\\ t_1=96\end{array}$

Hence the value of $t_1$ is 96 pounds, the value of $t_2$ is 48 pounds and the value of acceleration is -16 feet per second