Newton’s laws of motion

1. An object either remains at rest or moves in a straight line at constant velocity unless it is acted on by an external, unbalanced force. This behaviour is a property of matter known as inertia.

2. The vector sum of all forces acting on an object is related to the object’s mass and acceleration by ${\overrightarrow{F}}_{\text{net}}=m\overrightarrow{a}\text{.}$

3. For every action force there is an equal and opposite reaction force. That is, if A applies a force on B, then ${\overrightarrow{F}}_{\text{A}-\text{B}}=-{\overrightarrow{F}}_{\text{B}-\text{A}}\text{.}$

Example

A tractor pulls a railcar with mass 1.4 × 10⁴ kg at a 23º angle to the tracks with 2.75 × 10⁴ N of force. If there is a frictional force of 1.0 × 10² N, what is the railcar’s acceleration?

Let’s draw a sketch of the situation:

Next, we need to draw a free body diagram. This is a simple diagram that represents all the forces acting on a body, and you should get into the habit of using them for all dynamics problems. If we weren’t looking at this problem from above, we would also include the gravitational force and the normal force, even though they cancel each other out.

This diagram isn’t completely accurate—the railcar does not leave the tracks. There is another force, exerted by the tracks themselves, that cancels out the perpendicular component of ${\overrightarrow{F}}_{\text{a}}\text{.}$ The sum of this force and ${\overrightarrow{F}}_{\text{a}}$ is simply the parallel component of ${\overrightarrow{F}}_{\text{a}}\text{,}$ which is its scalar projection onto the tracks, given by $\left|{\overrightarrow{F}}_{\text{a}}\right|\cos \theta \text{.}$

We can now use Newton’s second law of motion to find acceleration:

${\overrightarrow{F}}_{\text{net}}=\left|{\overrightarrow{F}}_{\text{a}}\right|\cos \theta +{\overrightarrow{F}}_{\text{f}}=m\overrightarrow{a}\text{.}$

Rearranging to solve for acceleration gives us

$\overrightarrow{a}=\frac{\left|{\overrightarrow{F}}_{\text{a}}\right|\cos \theta +{\overrightarrow{F}}_{\text{f}}}{m}\text{,}$

and if we let [fwd] be positive, then we can substitute and get

$\overrightarrow{a}=\frac{(2.75\times {10}^4\text{N})\cos 23\text{º}+(-1.0\times {10}^2\text{N})}{1.4\times {10}^4\text{kg}}=1.8009{\text{m/s}}^2\text{,}$

therefore the railcar’s acceleration is 1.8 m/s² [fwd].