AKA Freebody Diagrams, Part Two.
For a definition of/conceptual take on freebody diagrams and the forces within them, make sure you check out Part One, then head on back here to plug in the numbers.
Let’s begin by reviewing our basic freebody diagram from last time- but with one small change.
Last time, our vertical forces cancelled each other out and our horizontal forces cancelled each other out. Now our vertical forces still cancel out- just as you predicted they would if you did the thought experiment at the end of the previous post- but our horizontal forces don’t. Our applied force is stronger than our frictional force (again, you expected this after our thought experiment), so our net force isn’t zero. What exactly is our net force and what does that mean about the motion of the box? Let’s stick in some numbers and see.
- Normal force = 10N up
- Earth’s gravitational force = 10N down
- Applied force = 20N to the right
- Frictional force = 15N to the left
This makes our net force 5N to the right- the box must be accelerating to the right. Since we see that the box is also being pushed to the right, it must be moving in that direction and speeding up.
This point seems self-evident, but it’s totally possible to have a net force that opposes the direction of motion. Imagine you were pushing your standard physics box down a long, smooth hallway. What would happen if you stopped pushing it? You’re no longer applying a force to it, but it doesn’t stop instantly. You know from experience that it will keep going in the same direction; friction will eventually slow it to a stop. Let’s combine your intuition with a freebody diagram of this situation (this diagram assumes you were initially pushing the box to the right).
Our vertical forces cancel each other out. You aren’t pushing the box anymore, so there is no applied force. The frictional force is to the left because it always opposes the direction of motion. With no rightward-force to counteract it, the frictional force is the net force. Since the direction of the net force is opposite the direction of motion, the object is slowing down instead of speeding up.
We’ll wrap up freebody diagrams (for now, at least!) by linking them back to Newton’s Second Law of Motion, which can be summed up by this formula:
- Force = mass * acceleration
In our first example, we saw that the net force on the box was 5N to the right. We knew that it was accelerating, and with this formula we can quantify that acceleration. Well, we can quantify it after I actually tell you the mass of the box. Let’s say it’s 5kg.
- 5N = 5kg * acceleration
- acceleration = 1 m/sec^2
So far we’ve started with the forces on an object and calculated the acceleration of the object from there. But in physics, math, and life in general, we frequently perform operations backwards as well as forwards. What if we were given the acceleration of an object and the frictional force on it and then asked to calculate the applied force? That kind of question would look like this:
A 10kg box is being pushed to the left across a horizontal surface. It is accelerating to the left at a rate of 3 m/sec^2. If the frictional force is 15N, what is the magnitude (fancy word for strength) of the applied force?
First let’s figure out what the net force has to be.
- force = mass * acceleration
- force = 10kg * 3 m/sec^2 = 30N
We know the frictional force is 15N. To get a net force of 30N, our applied force has to be 30N greater than that frictional force, giving us an applied force of 45N.
You may have noticed that we’ve mostly ignored Earth’s gravitational force and the normal force, since they’ve ended up cancelling out in all of our examples. Even though we haven’t focused on them in this discussion, they’re totally calculable, and we’ve actually covered all the individual pieces that we need. In a coming post we’ll combine Newton’s Second Law of Motion and Newton’s Law of Universal Gravitation. This will not only give a shortcut for calculating Earth’s gravitational force on any object, but it will also show us why a feather and a rock fall at the same rate in a vacuum.