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 `vec F_"net" = m vec a`.

  3. For every action force there is an equal and opposite reaction force. That is, if A applies a force on B, then `vec F_"A–B" = -vec F_"B–A"`.

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:

23ºrailcarFa
Railcar being pulled at a 23º angle

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.

FfFa[bwd][fwd]
Free body diagram of the railcar

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 `vec F_"a"`. The sum of this force and `vec F_"a"` is simply the parallel component of `vec F_"a"`, which is its scalar projection onto the tracks, given by `|vec F_"a"|cos theta`.

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

`vec F_"net" = |vec F_"a"|cos theta + vec F_"f" = mvec a`.

Rearranging to solve for acceleration gives us

`vec a = (|vec F_"a"|cos theta + vec F_"f")/m`,

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

`vec a = ((2.75xx10^4\ "N")cos23º + (-1.0xx10^2\ "N"))/(1.4xx10^4\ "kg") = 1.8009\ "m/s"^2`,

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