Chapter 8, Problem 107A

(a)

To determine

The tension (T) in the rope.

Answer to Problem 107A

  $23.52\text{ N}$

Explanation of Solution

Given that, Masses of the two blocks are $m_1=2$ kg and $m_2=3$ kg

The required diagram is shown above.

Where, g is gravitational acceleration, $g=9.8{\text{ m/s}}^{\text{2}}$ .

Since $m_2$ is greater than $m_1$ so $m_2$ will move downward with a acceleration a.

And $m_1$ will move upward with the same acceleration as shown in figure and a tension T will be produced on each side of the rope

So, from the figure, for 2 kg mass,

  $m_{1}a=T-m_{1}g$

And, for $m_2$ ,

  $m_{2}a=m_{2}g-T$

Now, by adding these equations

  $a=\frac{(m_2-m_1)g}{m_1+m_2}$

This is the expression of acceleration for both of the block, because they are moving with same acceleration

And, by subtracting the second equation from first one we get the expression of tension T,

  $T=\frac{2m_1{m}_{2}g}{m_1+m_2}$

Substitute the values and solve:

  $T=23.52\text{ N}$

Conclusion is, The tension of the rope will be $23.52\text{ N}$ .

(b)

To determine

The acceleration of the blocks.

Answer to Problem 107A

  $1.96{\text{ m/s}}^{\text{2}}$

Explanation of Solution

Given:

Masses of the two blocks are $m_1=2$ kg and $m_2=3$ kg

Calculation:

From the figure, determine two equations for two blocks of mass $m_1\text{and }{m}_2$ :

  $m_{1}a=T-m_{1}g$

And, $m_{2}a=m_{2}g-T$

Acceleration a is:

  $a=\frac{(m_2-m_1)g}{m_1+m_2}$

Substitute the values and solve:

  $a=1.96{\text{ m/s}}^{\text{2}}$

Conclusion:

So, the acceleration of the blocks is $a=1.96{\text{ m/s}}^{\text{2}}$ .