Pseudo Force: Definition, Formula, Examples, Effects - JEE PDF Download

Difference between inertial and non inertial reference frame

Pseudo force

The weight displayed on a weighing machine placed inside a moving elevator differs from the actual weight. When the elevator moves upward, the displayed weight is higher than the true weight, while during downward motion, the displayed weight is lower. Does this situation violate Newton's laws? In reality, Newton's laws hold true only when the observer is in an inertial frame of reference. When Newton's laws are not applicable, the frame is considered non-inertial. To analyze motion within a non-inertial frame, a pseudo force is introduced. A pseudo force is a hypothetical force introduced in non-inertial frames of reference to ensure compliance with Newton's laws of motion. It acts as an imaginary force applied to objects within the non-inertial frame.

Negative direction implies that the direction of force is opposite to acceleration of the system.

• It acts opposite to the direction of acceleration of the reference frame.
• It is applied to the non-inertial frame of reference.
• It is a fictitious force so there is no action-reaction pair.
• Pseudo force is also known as Inertial Force, although their need arises because of the use of non-inertial frames.

Apparent Weight In Lift

Consider a block of mass m kept on a weighing machine. Consider the system is kept in an elevator. Let observer O₁ observe this system from the ground frame. Let another observer O₂ be present inside the lift.

Case1: Lift moving upward with acceleration
With respect to O₁
This observer finds the system is moving upwards with acceleration a. Other forces acting on
the block is its weight and normal reaction.

Along the vertical direction,
N - mg = mα
N = mg + mα
N = m(g + α)

As the weighing machine measures the normal reaction acting on it, the weighing machine will show a reading corresponding to ..

n = m(g + α)

With respect to O₂
When the observer is in the elevator, the block is at rest with respect to him. So for this case,
The forces acting on the block are weight (mg) and normal reaction (N).

Along the vertical direction,
n = mg

This is completely wrong as we must get the same reading on the weighing machine. We got the wrong answer because O₂ is in a non-inertial frame so Newton’s laws are not valid. To get the correct answer let’s apply pseudo force. In this case direction of pseudo force will be downward and magnitude is mα . So here is the correct FBD by applying pseudo force:

Along the vertical direction,
N - mg - mα = 0
N = mg + mα
N = m(g + α)
So m(g + α) is the apparent weight of a block of mass m in the lift accelerating upward so geff = (g + α)ms^-2
Case 2: Lift moving downward with acceleration ‘
With respect to O₁
This observer finds the system is moving downwards with acceleration a. Other forces acting on
the block is its weight and normal reaction.

Along the vertical direction,
mg - N = mα
N - mg - mα
N = m(g - α)
As the weighing machine measures the normal reaction acting on it, the weighing machine will show a reading corresponding to .

N = m(g - α)

With respect to O₂
When the observer is in the elevator, the block is at rest with respect to him. So for this case,
The forces acting on the block are weight (mg) and normal reaction (N).

Along the vertical direction,

n = mg
This is completely wrong as we must get the same reading on the weighing machine. We got the wrong answer because O₂ is in a non-inertial frame so Newton’s laws are not valid. To get the correct answer let’s apply pseudo force. In this case direction of pseudo force will be upward and magnitude is mα. So here is the correct FBD by applying pseudo force:

Along the vertical direction,
mg - N = mα = 0
N - mg - mα
N = m(g - α)
So m(g - α)  is the apparent weight of a block of mass m in the lift accelerating downward so. geff = (g - α) ms^-2

Solved Example

Ex.1. A man weighs 80kg. He stands on a weighing machine in a lift that is moving upwards with a uniform acceleration of 5ms^-2. What would be the reading on the machine? (g = 10 ms^-2)

Solution: Here, the observer will feel that he is at rest with respect to the lift as he has the same acceleration as that of the lift. Observer is in a non-inertial frame. Hence, pseudo force should be considered in this case.
N = mg + mα
N = 80(10 +5) = 1200N

Ex.2. Figure shows a pendulum suspended from the roof of a truck that has a constant acceleration 'α'  towards the right relative to the ground. Find the deflection of the pendulum from the vertical.

Solution: If we solve the problem in the truck frame, Let the observer be in the truck.
We have to apply pseudo force in the left direction on the bob as shown in FBD,

Balancing forces in a horizontal direction,

Tsinθ = mα……..(1)
Balancing forces in vertical direction,

Tcosθ = mg ……..(2)
Divide equation (1) and (2),
tanθ = α/g