# Newtons Laws EX 19

(a) ${\displaystyle a=4m/s^{2}}$

There are three forces acting on mass ${\displaystyle m_{2}}$: ${\displaystyle F_{G}}$, ${\displaystyle F_{N}}$ (contact with the horizontal surface) and ${\displaystyle F_{T}}$ (contact with the massless rope). The forces acting on the mass ${\displaystyle m_{1}}$ are: ${\displaystyle F_{G}}$ and ${\displaystyle F_{T}}$ (contact with the massless rope). We note there is no other object in contact with ${\displaystyle m_{1}}$ and therefore no other force. We also note that the tension in the rope is the same at both ends of the rope since the pulley and the rope are massless. The diagram below shows the forces and the acceleration of the system:

We choose the coordinate system for each of the blocks so that the x-axis points to the right (the acceleration is positive) for ${\displaystyle m_{2}}$ and the x-axis pointing down (the acceleration is positive as well) for ${\displaystyle m_{1}}$.

Now we will compute the components of all forces for each block:

${\displaystyle m_{1}}$ ${\displaystyle m_{2}}$ x-axis x-axis ${\displaystyle F_{G}}$ 20 N 0 ${\displaystyle -m_{2}g}$ ${\displaystyle F_{T}}$ ${\displaystyle -F_{T}}$ ${\displaystyle F_{T}}$ 0 ${\displaystyle F_{N}}$ 0 ${\displaystyle F_{N}}$

Now we can write the equations for Newton's Second Law for both masses:

```${\displaystyle 20-F_{T}=m_{1}a=(2kg)(4m/s^{2})=8N}$
${\displaystyle F_{T}=m_{2}a=4m_{2}}$
${\displaystyle F_{N}-m_{2}g=0}$
```

Our task is to determine ${\displaystyle m_{2}}$ so we will solve for ${\displaystyle F_{T}}$ from the first equation finding ${\displaystyle F_{T}=12}$ N. We will then replace this value in the second equation and find ${\displaystyle m_{2}=3}$ kg.

(b) Now we will use the same system of axis and determine for what value of ${\displaystyle m_{2}}$ is the tension equal to 8 N. The components will now have these values:

${\displaystyle m_{1}}$ ${\displaystyle m_{2}}$ x-axis x-axis ${\displaystyle F_{G}}$ 20 N 0 ${\displaystyle -m_{2}g}$ ${\displaystyle F_{T}}$ -8 N 8 N 0 ${\displaystyle F_{N}}$ 0 ${\displaystyle F_{N}}$

Now we can write the equations for Newton's Second Law for both masses:

```${\displaystyle 20-8=m_{1}a=2a}$
${\displaystyle 8=m_{2}a}$
${\displaystyle F_{N}-m_{2}g=0}$
```

Our task is to determine ${\displaystyle m_{2}}$ so we will solve for ${\displaystyle a}$ from the first equation finding ${\displaystyle a=6m/s^{2}}$. We will then replace this value in the second equation and find ${\displaystyle m_{2}=1.33}$ kg.