# What is Newton’s Second Law of Motion

## Newton’s Second Law of Motion Definition

Newton’s Second Law of Motion states that when a resultant force acts on a body, the body’s acceleration due to the resultant force is directly proportional to the force.

As an equation, we write,

$\sum\vec{F}=m\vec{a}$

The summation sign, $\Sigma$, indicates that one needs to add all the forces using vector addition and find the resultant (or the net) force. According to Newton’s second law of motion, the resultant force is proportional to acceleration. This means that if the resultant force acting on a body is doubled, then the body’s acceleration would also double. If the resultant force is halved, the acceleration will also be halved and so on.

An alternative way to express Newton’s second law of motion is to use momentum. In this definition, the resultant force experienced by a body is equal to the rate of change of momentum of the body

$\sum\vec{F}=\frac{\mathrm{d}\vec{p}}{\mathrm{d}t}$

If we take the case of a body whose mass stays constant, since $\vec{p}=m\vec{v}$, this expression becomes:

$\sum\vec{F}=m\frac{\mathrm{d}\vec{v}}{\mathrm{d}t}=m\vec{a}$

Now, let’s look at a simple example of Newton’s second law of motion.

## Newton’s Second Law of Motion Example

Two pirates tug at a treasure chest, which has a mass of 55 kg. One pirate pulls it towards the Sea with a force of 18 N while the other pulls it away in the opposite direction with a force of 30 N. Find the acceleration of the treasure chest.

The two forces given by the two pirates are in the opposite directions, so the resultant force is (30-18) = 12 N away from the Sea. Now, using Newton’s second law, we have $a=\frac{m}{\sum F}=\frac{55}{18}=\mathrm{3.05 \:N}$ away from the Sea.

## How to Solve Newton’s Second Law Problems

### Problems Involving Lifts (Elevators)

To conclude this article, we will look at a classic physics problem involving the reaction force on a person in a lift. Suppose a person with mass  $m$ is standing inside a lift. Forces acting on the person are the weight $mg$ acting downwards and the reaction force $R$ from the floor of the lift acting upwards.

First, let’s take the case when the lift is still. The forces on the person are balanced. i.e. $R=mg$.

Now, suppose the lift is accelerating downwards. In this case, there is a resultant force acting downwards on the person. The resultant force gives an acceleration $a$. Then, taking downwards direction to be positive, we have

$mg-R=ma\Rightarrow R=m(g-a)$.

Suppose the lift now travels upwards, with an acceleration of the same magnitude. In this case,

$mg-R=-ma\Rightarrow R=m(g+a)$.

So, the person experiences a larger reaction force when the lift is accelerating upwards. This makes intuitive sense: as the floor of the lift is rushing up to meet the person, they should feel a larger force than when the floor is trying to “fall away” from them. The lower reaction force experienced as the lift accelerates downwards is what often makes you feel lighter when you take a lift.

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