8.0 Pseudo force

Before studying pseudo force let us first understand the frame of reference.

Systems of coordinate axes, which defines the position of a particle or an event in two or three-dimensional space is called a frame of reference.

Frame of reference is of two types,

1. Inertial frame of reference: A non-accelerating frame of reference is called an inertial frame of reference.

A frame of reference at rest or moving with a constant velocity is an inertial frame of reference.

2. Non-inertial frame of reference: An accelerating frame of reference is called a non-inertial frame of reference.

A rotating frame of reference is also a non-inertial frame of reference because it has a centripetal acceleration.

Note: In non-inertial frame of reference Newton laws of motion is not valid.

Let us understand inertial frame of reference with the help of an illustration.

If an observer observes from an Inertial frame of reference (ground) as shown in the figure, the equation of motion of the block is,

$$\begin{equation} \begin{aligned} N - mg = m{a_n} \\ N = m\left( {g + {a_n}} \right)\quad ...(i) \\\end{aligned} \end{equation} $$

where ${\overrightarrow a _n}$ is the acceleration of non-inertial frame of reference.

Now, if an observer observes from a non-inertial frame of reference (elevator) as shown in the figure, the equation of motion of the block is,

$$N = mg\quad ...(ii)$$

The above equation of motion is incorrect as the acceleration of the non-inertial frame is not considered.

So, the Newton laws of motion are not valid in non-inertial frame of reference.

To overcome this problem, a concept of pseudo force was introduced.

Pseudo force: is actually a fictions force, which acts on a mass whose motion is when observed from a non-inertial frame of reference.

Pseudo force is an extra force applied on a mass, which converts the non-inertial to an inertial frame of reference, so as to make Newton laws of motion valid.

Mathematically, pseudo force is given as, $${\overrightarrow F _P} = m\left( { - {{\overrightarrow a }_n}} \right)$$

where, ${\overrightarrow a _n}$ is the acceleration of non-inertial frame of reference

Note: Pseudo force is applied in a direction opposite to the acceleration of non-inertial frame of reference.

Therefore, after applying Pseudo force as shown in the figure, the equation finally becomes,

$$N = m\left( {g + {a_n}} \right)\quad ...(iii)$$

Hence, equation $(i)$ and $(iii)$ are both identical.

Note: Here we applied Pseudo force in downward direction because the acceleration of non-inertial frame of reference was in upward direction.

Question 13. Find the acceleration and normal reaction on the block of mass $m$ as shown in the figure.

Solution: The elevator is going upwards with an acceleration ${\overrightarrow a _n}$, therefore the pseudo force will act in the downward direction.

From FBD we can write, $$\begin{equation} \begin{aligned} m\left( {g + {a_n}} \right)\sin \theta = ma \\ a = \left( {g + {a_n}} \right)\sin \theta \\ {N_1} = m\left( {g + {a_n}} \right)\cos \theta \\\end{aligned} \end{equation} $$

Question 14. A bob of mass $m$ is suspended from the ceiling of a wagon moving with an acceleration $a$, as shown in the figure. Find angle $\theta $ and tension $T$ in equilibrium position.

Solution: The wagon moves with forward acceleration $a$, therefore the pseudo force acts in the backward direction.

From FBD we can write, $$\begin{equation} \begin{aligned} T\sin \theta = ma\quad ...(i) \\ T\cos \theta = mg\quad ...(ii) \\\end{aligned} \end{equation} $$

From equation $(i)$ and $(ii)$ we get, $$T = \sqrt {{{\left( {ma} \right)}^2} + {{\left( {mg} \right)}^2}} $$

Dividing equation $(i)$ and $(ii)$ we get, $$\begin{equation} \begin{aligned} \tan \theta = \frac{a}{g} \\ \theta = {\tan ^{ - 1}}\left( {\frac{a}{g}} \right) \\\end{aligned} \end{equation} $$

Question 15. Find the mass $m_3$ of the hanging block in figure, which will prevent the smaller block from slipping over the wedge. All surface are frictionless and the string and pulley are light.

Solution: Due to the hanging block there will be a net acceleration of the system.

However the wedge and the hanging block will have different accelerations.

Constrained equation, $$\begin{equation} \begin{aligned} \overrightarrow T .\;{\overrightarrow a _1} + 2\overrightarrow T .\;{\overrightarrow a _2} = 0 \\ T{a_1} - 2T{a_2} = 0 \\ {a_1} = 2{a_2}\quad ...(i) \\\end{aligned} \end{equation} $$

From FBD we can write, $${N_1} = {m_1}\left( {g + {a_1}} \right)\cos \theta \quad ...(ii)$$

Condition for no slipping along the wedge, $$\begin{equation} \begin{aligned} {m_1}g\sin \theta = {m_1}{a_1}\cos \theta \quad ...(iii) \\ {m_3}g - 2T = {m_3}{a_2}\quad ...(iv) \\\end{aligned} \end{equation} $$

If the block does not slip on wedge, then we can consider wedge and block as single mass, $$T = \left( {{m_1} + {m_2}} \right){a_1}\quad ...(v)$$

From equation $(i)$, $(iii)$, $(iv)$ and $(v)$ we get,