Laws Of Motion

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Common Misconceptions

4 min read

- Let's bust some common misconceptions

Are Newton's Laws of motion universally applicable?
Newtonian mechanics can be used to explain the dynamics of a particle in most of the real life cases, however in some cases where the reference frame in consideration is accelerating with some finite acceleration, the Laws of Motion cannot be directly applied.
Ex: Consider the case of a pendulum bob suspended inside a car accelerating with acceleration a.

For a stationary observer the bob is seen to accelerate with the same acceleration as a.

In the inertial frame, the net forces acting on the bob are:

Weight, $F_{g} = m g$
Tension, $T$

Applying Newton's Second Law of motion,

\sum F_{g} + T = 0

In component form it can be written as,

\sum F_{x} = T sin θ = m a

\sum F_{y} = T cos θ - m g = 0

Hence for a stationary observer the bob seems to be accelerating along with the car and Newton's Laws are valid. However in the reference frame of the car, the bob is stationary and there should be no net force acting on it.

\sum F = 0

, (Car's reference frame)

This can only happen if we assume an extra inertial force acting on the bob.

\sum F_{g} + T + F_{I} = 0

\sum m a + F_{I} = 0

F_{I} = - m a

The free body diagram of the bob in the frame of reference of the car is

Thus, we have seen that the newton's second law isn't valid in the non inertial frame, unless we consider the inertial force on the object in consideration. This force is also called "pseudo force" or fictitious force as it has no existence in the inertial frame.

Is centrifugal force real?
Now that you understand the meaning of pseudo force you are in a better position to argue whether it is real or not. You must have observed while taking a right turn in a car you experience a force towards the left. The faster you take a turn, the stronger the force is.

The centripetal force required to turn a car on a circular track of radius r with a velocity v is given by

\frac{m v ^{2}}{r}

. However in the reference frame of the car, which is a non-inertial frame, the driver will experience an equal and opposite force which we call as the centrifugal force. Thus centrifugal force is the pseudo force which is experienced only in rotating frame of reference (non- inertial frame of reference).

Does acceleration of a point depend on its history?
To answer this question let us imagine a scenario, where a train is moving forward with a horizontal acceleration, a. Hence all the objects within the train, including the a person standing inside will have the same acceleration

Now what will happen if the person was to drop a ball from the accelerating train. The moment the ball is dropped from the train, it loses its horizontal acceleration and only gravity( vertical acceleration) acts on it. It simply has no memory of the acceleration it had when it was inside the train at the person's hands.

Newton's Second Law( $F = m a$ ) is a local relation, which means the acceleration possessed by an object at an instant depends only on the forces acting on it at that instant. Acceleration here and now is determined by the force here and now, not by any history of the motion of the particle.

But can we say the same thing about velocity?

Following the expression for Newton's Second Law we can say,

a = 0

then,

F = 0

Hence The First Law can be arrived at from the Second Law itself. If the force acting on a particle is zero, then the particle will have zero acceleration, or in other words, the velocity of the particle will stay at whatever value it was initially. Hence, the particle velocity depends on the history of the motion of the particle.

The Horse Cart Paradox

Imagine a horse-cart. The cart moves forward because of the force exerted by the horse. Now, the Newton's Third Law says that, for every action there is an equal and opposite reaction. Say the horse pulls the cart with a forward force

F_{h}

then the cart must also pull it back with a force,

F_{c}

which is equal and opposite to

F_{h}

. If both forces are equal and opposite then the net force on the horse-cart system should be zero. Then how does the horse cart move forward? Where is this paradox arising from?
The confusion arises from common misconceptions in applying the Newton's Third Law. The forces

F_{c}

and

F_{h}

do not balance each other out despite being equal and opposite because they are acting on two different objects. Equal and opposite forces, both acting on the same object would cancel out. Let us try to understand in detail.

The horse and cart are two separate systems, attached with a harness. Let us understand the forces acting on the horse and the cart separately and draw there free body diagrams.

Forces acting on the horse are

1. Weight of the horse, downward force, (

m_{h} g

)

2. Reaction force exerted by the ground on the horse, (

F_{r}

)

The horse exerts an inclined force on the ground. It can be shown by Newton's Third Law that the ground exerts an equal and opposite force on it, which we are taking into account here.

3. Backward force exerted by the cart, (

F_{c}

)

Let us incorporate all these into the free body diagram of the horse as a separate system

The force exerted by the road

F_{r}

can be resolved into parallel and perpendicular components. The perpendicular component balances out the weight. If the parallel component

F_{r}^{∥}

is greater than the backward force exerted by the cart, the horse can move forward.

Similarly lets draw the free body diagram for the cart. The forces acting on the cart are,

Weight of the cart, downward force, ( $m_{c} g$ )
Reaction force exerted by the ground on the cart, ( $F_{r}^{'}$ )
Forward force exerted by the horse, ( $F_{h}$ )

The reaction force can be resolved to parallel and perpendicular components as before. The perpendicular component of

F_{r}^{^{'}}

balances the weight of the cart. The parallel component however opposes the forward force exerted by horse. If forward force (

F_{h}

) is higher than this parallel component , then there is an overall unbalanced force on the cart and it moves forward.

The same paradox can be resolved by taking horse and the cart as a single system. The forces acting on the system as a whole are:

The weight of the horse cart, ( $(m_{h} + m_{c}) g$ )
The reaction force exerted by the ground, $F_{r}$

The reaction force can be resolved into parallel and perpendicular components as you can see in the free body diagram. The parallel component, is the unbalanced force acting on the system, which moves the system forward.