Chapter 4, Problem 87A

To determine

The acceleration of the larger mass in the arrangement, where two blocks one of mass $4.0\text{ kg}$ and other of mass $1.0\text{ kg}$ are connected by a rope which is strung over a massless, resistance-free pulley.

Answer to Problem 87A

  $a=5.8{\text{ m/s}}^2$ in the downward direction.

Explanation of Solution

Given:

Mass of block ${\text{m}}_1=4.0\text{ kg}$

Mass of block ${\text{m}}_2=1.0\text{ kg}$

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 ${\text{m}}_1$ and ${\text{m}}_2$ . In the figure $\text{2}$ , $\text{T}$ is tension in the rope which is acting in upward direction, ${\text{m}}_1\text{g}$ and ${\text{m}}_2\text{g}$ are weights which are acting in the downward direction. $g$ is acceleration due to gravity, $g=9.8{\text{ m/s}}^2$ .

  

Figure 1: Free body diagrams of masses ${\text{m}}_1$ and ${\text{m}}_2$ .

Applying Equation $\left(1\right)$ to the block of mass ${\text{m}}_1$ 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 ${\text{m}}_1$ 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 ${\text{m}}_2$ 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

  $\begin{array}{c}T=\frac{2\left(\text{4 kg}\right)\left(\text{1 kg}\right)\left(9.8{\text{ m/s}}^2\right)}{\left(\text{4 kg}+1\text{ kg}\right)}\\ =\frac{78.4{\text{ kg}}^2{\text{m/s}}^2}{\text{5 kg}}\end{array}$

  $\begin{array}{c}T=15.68{\text{ kgm/s}}^2\\ =\text{ 16 N}\end{array}$

Substituting the value of T in equation $\left(5\right)$ ,

  $a=\frac{\left(4\text{ kg}\right)\left(9.8{\text{ m/s}}^2\right)-\left(\text{16 N}\right)}{\left(4\text{ kg}\right)}$

  $=\frac{39.2{\text{ kgm/s}}^2{\text{-16 kgm/s}}^2}{\left(4\text{ kg}\right)}$

  $=\frac{23.2{\text{ kgm/s}}^2}{4\text{ kg}}$

  $=5.8{\text{ m/s}}^2$ in the downward direction.

Conclusion:

The acceleration of the larger mass is $5.8{\text{ m/s}}^2$ in the downward direction.