# Unbalanced Force – Newton’s Second Law of Motion

Show/Hide Sub-topics (Forces & Turning Effect Of Forces | O Level Physics)
Show/Hide Sub-topics (Forces And Dynamics | A Level Physics)

When two or more external forces acting on a body produce a net resultant force, i.e, the vector sum of forces is not zero, the forces are unbalanced.

## Newton’s Second Law Of Motion

Newton’s Second Law of Motion says that the resultant force acting on a body produces a net acceleration and causes the body to accelerate in the direction of the resultant force.

• If the net resultant force is in the direction of the motion, the body will accelerate (positive acceleration).
• If the net resultant force is against the direction of motion, the body will decelerate (negative acceleration).

For a body of constant mass, Newton’s Second Law of Motion can be expressed as:

$$F = ma\tag{1}$$

, where

F = resultant force (N)
m = mass (kg)
a = acceleration ($\text{m s}^{-2}$)

Since mass (m) is scalar quantity, the direction of acceleration (a) is thus in the same direction as the applied force (F).

To calculate the acceleration of a body, one should use Equation 1, with $F$ being the net resultant force exerted on the body. The equation also tells us that:

• Doubling the resultant force $F$ on an object doubles its acceleration $a$
• With the same resultant force $F$, doubling the mass $m$ halves the acceleration $a$
• One newton is the force that produces an acceleration of $1 \text{ m s}^{-2}$ on a mass of 1 kg

### [A Level] Newton’s Second Law In 3-Dimensional Form

The Newton’s Second Law of Motion can be written as:

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

Newton’s Second Law is vectorial in nature and can be applied independently along each of the three perpendicular axes (x, y, and z) in a Cartesian coordinate system. Specifically, $\sum \vec{F}_{x} = m \vec{a}_{x}$ denotes the sum of all forces acting on the object in the x-direction is equal to the mass of the object times its acceleration in that direction. Similarly, $\sum \vec{F}_{y} = m \vec{a}_{y}$ and $\sum \vec{F}_{z} = m \vec{a}_{z}$ represent the same relationship for the y and z directions, respectively. This decomposition allows for the analysis of complex motions by examining the forces and accelerations along each axis independently, facilitating the solution of dynamics problems in three-dimensional space.

Hence, the law can also be written as:

\begin{aligned} \sum \vec{F}_{x} &= m \vec{a}_{x} \\ \sum \vec{F}_{y} &= m \vec{a}_{y} \\ \sum \vec{F}_{z} &= m \vec{a}_{z} \end{aligned}

## More Advanced Definition Of Newton’s Second Law (A Level)

### More Complete Form Of Newton’s Second Law

Newton’s second law states that the rate of change of momentum of a body is proportional to the resultant force acting on it and the change takes place in the direction of the force.

$$F = ma + v \frac{dm}{dt}$$

1. $F = ma$: This is the traditional form of Newton’s second law, where force $F$ is equal to the product of mass $m$ and acceleration $a$. This component describes the force required to accelerate a mass.
2. $v \frac{dm}{dt}$: This additional term involves the velocity $v$ of the object multiplied by the rate of change of mass $\frac{dm}{dt}$. This term represents the force required to account for changes in mass over time.
• The second term, $v \frac{dm}{dt}$, suggests that the force needed to accelerate an object is influenced not only by its mass but also by how that mass changes with respect to time. This kind of scenario might arise in certain contexts, such as a rocket expelling mass during propulsion. In such cases, the force required for acceleration involves not only the mass of the rocket but also the rate at which mass is being expelled.

## Worked Examples

### Example 1

A box of mass 20 kg is pushed with a force of 50 N. What is the acceleration of the box? Neglect the effects of friction.

From Newton’s Second Law,

\begin{aligned} F &= ma \ a &= \frac{F}{m} \\ &= \frac{50}{20} \\ &= 2.5 \text{ m s}^{-2} \end{aligned}

### Example 2

A car of mass 1000 kg accelerates from rest to $20 \text{ m s}^{-1}$ in 5 s. Calculate the driving force of the engine. (Neglect the effects of friction)

Let’s calculate the acceleration first.

\begin{aligned} a &= \frac{v-u}{t} \\ &= \frac{20-0}{5} \\ &= 4 \text{ m s}^{-2} \end{aligned}

From Newton’s Second Law,

\begin{aligned} F &= ma \\ &= 1000 \times 4 \\ &= 4000 \text{ N} \end{aligned}

### Example 3

A particle which moves from rest is acted upon by two forces: a constant forward force and a retarding force which is directly proportional to its velocity. Which one of the following statements about the subsequent motion of the particle is true?

1. Its velocity increases from zero to a maximum.
2. Its acceleration increases from zero to a maximum.
3. Its velocity increases from zero to a maximum and then decreases.
4. Its acceleration increases from zero to a maximum and then decreases.

The net force on the particle decreases as the velocity increases. When the retarding force is equal to the constant forward force, the net force becomes zero. The velocity of the particle is at the maximum.

### Example 4

A mass of 0.20 kg is suspended from the roof of a lift by a light thread. The thread is cut while the lift is ascending upwards at 3.0 m s-2. What is the magnitude of the acceleration of the 0.20 kg mass at the instant the thread is cut?

1. 0 m s-2
2. 3.0 m s-2
3. 9.8 m s-2
4. 12.8 m s-2

When the string is cut, the only acceleration acting on the mass is due to gravity which is equal to 9.81 m s-2.

### Example 5

A large box is pushed across a horizontal ground at constant speed. If the push is removed, the box will

1. stop at once.
2. start to slow down to a stop.
3. continue to travel at constant speed.
4. continue to travel at constant speed before slowing down to a stop

Friction must be present initially to balance the pushing force, so that the box moves with constant speed. Once the pushing force is removed, friction will cause the box to slow down to a stop.