Physics > Laws of Motion > 2.0 Newton's second law of motion
Laws of Motion
1.0 Newton's first law of motion
2.0 Newton's second law of motion
3.0 Newton's third law of motion
4.0 Force
5.0 Normal Force
6.0 Gravitional Force
7.0 Tension Force
8.0 Pseudo force
9.0 Friction
10.0 Centripetal force
2.1 Equations of Newton's second law of motion
2.1.1 When block is stationary
From the FBD we will write the vector equation of force.
Note: Direction of force is very important.
Along $x$ axis,
$${\overrightarrow F _x} = {\overrightarrow N _1} + {\overrightarrow F _2}$$ or $$\begin{equation} \begin{aligned} {\overrightarrow F _x} = {\overrightarrow N _1} + {\overrightarrow F _2} \\ {\overrightarrow F _x} = \left( {{N_1} - {F_2}} \right)\widehat i\quad ...(i) \\\end{aligned} \end{equation} $$ Along $y$ axis, $$\begin{equation} \begin{aligned} {\overrightarrow F _y} = {\overrightarrow N _2} + {\overrightarrow F _1} + m\overrightarrow g \\ {\overrightarrow F _y} = \left( {{N_2} - mg - {F_1}} \right)\widehat j\quad ...(ii) \\\end{aligned} \end{equation} $$ Along $z$ axis, $${\overrightarrow F _z} = 0$$
As the block is stationary so, $\begin{equation} \begin{aligned} {\overrightarrow a _x} = 0 \\ {\overrightarrow a _y} = 0 \\ {\overrightarrow a _z} = 0 \\\end{aligned} \end{equation} $
Therefore, $$\begin{equation} \begin{aligned} {\overrightarrow F _x} = m{\overrightarrow a _x} \\ {\overrightarrow F _x} = 0 \\\end{aligned} \end{equation} $$
Similarly, $${\overrightarrow F _x} = {\overrightarrow F _y} = {\overrightarrow F _z} = 0\quad ...(iii)$$
From equation $(i)$, $(ii)$ and $(iii)$ we get, $$\begin{equation} \begin{aligned} 0 = {N_1} - {F_2}\quad \left( {As,\;{{\overrightarrow F }_x} = 0} \right) \\ {N_1} = {F_2} \\\end{aligned} \end{equation} $$ Also, $$\begin{equation} \begin{aligned} 0 = {N_2} - \left( {mg + {F_1}} \right)\quad \left( {As,\;{{\overrightarrow F }_y} = 0} \right) \\ {N_2} = mg + {F_1} \\\end{aligned} \end{equation} $$
2.1.2 When block is moving with constant speed
This type of problem will be solved in the same way as the previous one.
From the FBD we will write the vector equation of force.
Note: Direction of force is very important.
Along $x$ direction, $$\begin{equation} \begin{aligned} {{\vec F}_x} = {{\vec F}_1} + {{\vec F}_2} \\ {{\vec F}_x} = \left( {{F_1} - {F_2}} \right)\widehat i\quad ...(i) \\\end{aligned} \end{equation} $$ Along $y$ direction, $$\begin{equation} \begin{aligned} {\overrightarrow F _y} = \overrightarrow N + m\overrightarrow g \\ {\overrightarrow F _y} = \left( {N - mg} \right)\widehat j\quad ...(ii) \\\end{aligned} \end{equation} $$
As the system is moving with constant velocity (means zero acceleration) so,
$\begin{equation} \begin{aligned} {\overrightarrow a _x} = 0 \\ {\overrightarrow a _y} = 0 \\ {\overrightarrow a _z} = 0 \\\end{aligned} \end{equation} $
Therefore, $$\begin{equation} \begin{aligned} {\overrightarrow F _x} = m{\overrightarrow a _x} \\ {\overrightarrow F _x} = 0 \\\end{aligned} \end{equation} $$
Similarly, $${\overrightarrow F _x} = {\overrightarrow F _y} = {\overrightarrow F _z} = 0\quad ...(iii)$$
From equation $(i)$, $(ii)$ and $(iii)$ we get,
$$\begin{equation} \begin{aligned} 0 = {F_1} - {F_2}\quad \left( {As,\;{{\overrightarrow F }_x} = 0} \right) \\ {F_1} = {F_2} \\ \\ 0 = N - mg\quad \left( {As,\;{{\overrightarrow F }_y} = 0} \right) \\ N = mg \\\end{aligned} \end{equation} $$
2.1.3 When block is moving with constant acceleration
From the FBD we will write the vector equation of force.
Note: Direction of force is very important.
$$\begin{equation} \begin{aligned} {\overrightarrow F _x} = {\overrightarrow F _1} + \overrightarrow F 2 \\ {\overrightarrow F _x} = \left( {{F_1} - {F_2}} \right)\widehat i\quad ...(i) \\ \\ {\overrightarrow F _y} = \overrightarrow N + m\overrightarrow g \\ {\overrightarrow F _y} = \left( {N - mg} \right)\widehat j\quad ...(ii) \\\end{aligned} \end{equation} $$
As we can see the acceleration is only along $x$ direction. So,
$\begin{equation} \begin{aligned} {\overrightarrow a _x} = a\left( { - \widehat i} \right) \\ {\overrightarrow a _y} = 0 \\ {\overrightarrow a _z} = 0 \\\end{aligned} \end{equation} $
Therefore, $$\begin{equation} \begin{aligned} {\overrightarrow F _x} = m{\overrightarrow a _x} \\ {\overrightarrow F _x} = ma\left( { - \widehat i} \right) \\ {\overrightarrow F _x} = - ma\widehat i\quad ...(iii) \\\end{aligned} \end{equation} $$ And, $${\overrightarrow F _y} = {\overrightarrow F _y} = 0\quad ...(iv)$$
From equation $(i)$, $(ii)$, $(iii)$ and $(iv)$ we get, $$\begin{equation} \begin{aligned} - ma\widehat i = \left( {{F_1} - {F_2}} \right)\widehat i\quad \left( {As,\;{{\overrightarrow F }_x} = - ma\widehat i} \right) \\ {F_1} = {F_2} - ma \\ \\ 0 = \left( {N - mg} \right)\quad \left( {As,\;{{\overrightarrow F }_y} = 0} \right) \\ N = mg \\\end{aligned} \end{equation} $$
