What is a force

A force is an action that can push or pull an object and cause it to accelerate. It is a vector quantity, and have the following formula:

$\vec {F} = m \vec {a} \quad [N]$

Where $\vec {F}$ represents the force applied onto the object with Newtons as its unit: $N = kg \cdot \frac {m} {s^2}$. Also, $m$ is the mass of the object, and $\vec {a}$ is the acceleration applied onto the object.

Net force

All the forces on an object can be added up to what is known as the net force, the sum of the forces applied onto that object.

It is denoted by either $\vec {F}_{\text{net}}$ or $\Sigma \vec {F}$:

$\Sigma \vec {F} = \vec {F}_1 + \vec {F}_2 + ... + \vec {F}_n$

Where $\Sigma \vec {F}$ is the net force.

Example

Calculate the net force on an object when a force of $3N$ is applied to the right, and a force of $2N$ is applied to the left.

Answer

It would help to visualize this interaction:

As you can see, we should expect the box to be forced to the right, since the force on the right has a greater magnitude.

A simple calculation shows:

$\vec {F}_{ \text{net} } = \vec {F}_1 + \vec {F}_2 = (3N) + (-2N) = 1N \ [ \text{right} ]$

$\therefore$ The net force on the object is $2N$ to the right.

Example

Find the net force needed to accelerate an object with a mass of $3.0kg$ to $10 \frac {m} {s^2}$

Answer

We need to use the formula of the force to determine the answer:

$\vec {F}_{ \text{net} } = m \vec {a} = (3.0kg)(10 \frac {m} {s^2}) = 30 kg \frac {m} {s^2} = 30N$

$\therefore$ A force of $30N$ must be applied to accelerate the object.

Center of mass

As you have already seen, there is always a dot/point in the illustrations where the forces are applied. These points are the center of masses of the objects involved in the interactions. It is the average position of the object weighed by mass.

We use the center of mass of the objects to approximate the effects of the force. Objects that are more symmetrical have a more centered center of mass.

Newton's laws of motion

Newton's laws of motion bring Kinematics and Dynamics together to help describe where and why an object moves.

Newton's first law of motion

An object will remain in a state of uniform motion unless acted upon by an external force.

In other words, an object will retain the same velocity unless acted upon by another force. So if a ball is rolling at $3 \frac {m} {s}$, it will continue to move at that speed given that there are no other forces such as friction, or air resistance acting on it.

Newton's second law of motion

The change in motion of an object is directly proportional to the net force applied onto it.

$\vec {a} \propto \Sigma \vec {F} \quad \text {or} \quad m \vec {a} = \Sigma \vec {F} \newline \vec {a} \propto \frac {1} {m} \quad \text {or} \quad \vec {a} = \frac { \Sigma \vec {F} } {m}$

The symbol $\propto$ means proportional, which indicates that when one value increases, then the other increases two if they are proportional.

Newton's third law of motion

For every action force on another object, there is an opposite but equal reaction force.

For example, if you punch a punching bag with $9N$ of force, it will also apply of force of $9N$ to your arm.

Example

A $2kg$ ball is rolling at a velocity of $7.25 \frac {m} {s}$. The force of friction on the ball is $4N$ opposite to the balls motions. How far does the ball go before stopping?

Answer

The frictional force $\vec {F}_s$ is opposing the ball's motion, meaning it is decelerating the ball. So that means we need to find the acceleration going against the balls motion:

$\vec {F}_s = m \vec {a} \ \Rightarrow \ \vec {a} = \frac { \vec {F}_s } {m} = \frac { 4N } { 2kg } = 2 \frac {m} {s^2}$

Now that we have the deceleration, we can use the equation $\vec {v}_f^2 = \vec {v}_i^2 + 2 \vec {a} \vec {d}$. We know that $\vec {v}_i = 7.25 \frac {m} {s}$, and since the ball comes to a stop, $\vec {v}_f = 0 \frac {m} {s}$.

So, plugging the values in, we get:

$\quad \quad \ \ \ \, \vec {v}_f^2 = \vec {v}_i^2 + 2 \vec {a} \vec {d}$

$(0.00 \frac {m} {s})^2 = (7.25 \frac {m} {s})^2 + 2 (-2 \frac {m} {s^2}) \vec {d}$

$\quad \quad \quad \, \, \vec {d} = 12.1 m$

Forces on an inclined plane

Solving problems involving an inclined may seem difficult, but they are actually quite easy.

Gravity components

The angle of the incline with the horizontal $\theta$ is the same angle that $\vec {F}_g$ makes with the vertical. This makes it easy to determine its parallel and perpendicular components.

You can obtain the values for $\vec {F}_{ g_{\text{parallel}} }$ and $\vec {F}_{ g_{\text{perpendicular}} }$ from the following formulas:

$\vec {F}_{ g_{\text{parallel}} } = \vec {F}_g \sin \theta \newline \vec {F}_{ g_{\text{perpendicular}} } = \vec {F}_g \cos \theta$

Normal force

There is also a force known as the normal force $\vec {F}_N$. We will discuss this in greater detail in the next chapter, however, just know that the normal force balances out the forces perpendicular to the surface at which the object is on.

The formula for the normal force on an inclined plane is:

$\vec {F}_N = - \vec {F}_{ g_{\text{perpendicular}} } = - \vec {F}_g \cos \theta$