Chapter 4, Problem 84A

(a)

To determine

The tension in the rope which connects two blocks, one of mass $5.0\text{ kg}$ and other of mass $3.0\text{ kg}$ .

Answer to Problem 84A

  $T=37\text{ N}$

Explanation of Solution

Given:

Mass of block, $m_1=5.0\text{ kg}$

Mass of block, $m_2=3.0\text{ kg}$

The rope which connects the two blocks is strung over a massless, resistance-free pulley.

Formula used:

Newton’s second law of motion:

  ${\displaystyle \sum F=ma}$ $\left(1\right)$

Where, $F$ is the force,

  $m$ is mass of the block

  $a$ is acceleration of the block

Figure 1 shows the free body diagrams of masses $m_1$ and $m_2$ . T is tension in the rope which is acting in upward direction, $m_{1}g$ and $m_{2}g$ are weights which are acting in the downward direction. Where, $g$ is acceleration due to gravity, $g=9.8{\text{ m/s}}^2$ .

  

Figure1:Free body diagrams of masses $m_1$ and $m_2$ .

Applying Equation $\left(1\right)$ to the block of mass $m_1$ in the y-direction (which is accelerating in the downward direction),

  ${\displaystyle \sum F=ma}$

Substituting for force, ${\displaystyle \sum F=T-m_1}g$ , above equation becomes,

  $T-m_{1}g=m_1\left(-a\right)$ $\left(2\right)$

(Since the block $m_1$ is accelerating in the downward direction acceleration is taken as negative)

Equation $\left(2\right)$ can be written as

  $m_{1}g-T=m_{1}a$ $\left(3\right)$

Applying Equation $\left(1\right)$ to the block of mass $m_2$ in the y -direction

  ${\displaystyle \sum F=ma}$

Substituting for force, ${\displaystyle \sum F=T-m_2}g$ , above equation becomes

  $T-m_{2}g=m_{2}a$ $\left(4\right)$

From equation $\left(3\right)$ acceleration can be written as

  $a=\frac{m_{1}g-T}{m_1}$ $\left(5\right)$

From equation $\left(4\right)$ acceleration can be written as

  $a=\frac{T-m_{2}g}{m_2}$ $\left(6\right)$

Equating the equations $\left(5\right)$ and $\left(6\right)$

  $\frac{m_{1}g-T}{m_1}=\frac{T-m_{2}g}{m_2}$

  $\Rightarrow m_2\left(m_{1}g-T\right)=m_1\left(T-m_{2}g\right)$

  $\Rightarrow m_1{m}_{2}g-m_{2}T=m_{1}T-m_1{m}_{2}g$

  $\Rightarrow 2m_1{m}_{2}g=m_{1}T+m_{2}T$

  $\Rightarrow 2m_1{m}_{2}g=T\left(m_1+m_2\right)$

  $\Rightarrow T=\frac{2m_1{m}_{2}g}{\left(m_1+m_2\right)}$ $\left(7\right)$

Calculation:

Substituting the numerical values in equation $\left(7\right)$ , tension in the rope is:

  $\begin{array}{l}T=\frac{2\left(5\text{ kg}\right)\left(3\text{ kg}\right)\left(9.8{\text{ m/s}}^2\right)}{\left(5\text{ kg}+3\text{ kg}\right)}\\ T=\frac{294{\text{ kg}}^2{\text{m/s}}^2}{8\text{ kg}}\end{array}$

  $T=36.7{\text{ kgm/s}}^2=37\text{ N}$

Conclusion:

The tension in the rope is $37\text{ N}$ .

(b)

To determine

The acceleration of the blocks.

Answer to Problem 84A

  $a=2.45{\text{ m/s}}^2$ .

Explanation of Solution

Given:

Mass of block, $m_1=5.0\text{ kg}$

Mass of block, $m_2=3.0\text{ kg}$

The rope which connects the two blocks is strung over a massless, resistance-free pulley.

Formula used:

From equation $\left(3\right)$ in the part $\left(a\right)$ ,

  $T=m_{1}g-m_{1}a$ $\left(8\right)$

From equation $\left(4\right)$ in the part $\left(a\right)$ ,

  $T=m_{2}g+m_{2}a$ $\left(9\right)$

Equating the equations $\left(8\right)$ and $\left(9\right)$ ,

  $m_{1}g-m_{1}a=m_{2}g+m_{2}a$

  $\Rightarrow m_{1}g-m_{2}g=m_{2}a+m_{1}a$

  $\Rightarrow \left(m_1-m_2\right)g=\left(m_1+m_2\right)a$

  $\Rightarrow a=\frac{\left(m_1-m_2\right)g}{\left(m_1+m_2\right)}$ $\left(10\right)$

Calculation:

Substituting the numerical values in equation $\left(10\right)$ ,

  $\begin{array}{c}a=\frac{\left(5\text{ kg-}3\text{ kg}\right)\left(9.8{\text{ m/s}}^2\right)}{\left(5\text{ kg}+3\text{ kg}\right)}\\ =\frac{19.6{\text{ kgm/s}}^2}{8\text{ kg}}\\ =2.45{\text{ m/s}}^2\end{array}$

Conclusion:

The acceleration of the blocks is $2.45{\text{ m/s}}^2$ .