Free Body Diagram Videos
Intermediate
Mike: Hey
Orbax: Hi, I’m the Great Orbax and this is Dr. Massa. And today we’re here to discuss free body diagrams. Mike?
Mike: So, we know that forces can act on an object to cause an object’s motion.
Orbax: It’s true.
Mike: One way to sort of make sure that we’re accounting for all the relevant forces in a problem is to draw what we call a free body diagram.
Orbax: Or an FBD.
Mike: Alright, so let’s take an example of a guy or a girl sitting on a sled, on a slope, and look at the forces acting on them, and relate that to the motion.
Orbax: Alright, well one of the ways that we look at free body diagrams is to first take our mass and reduce it to a point mass.
Mike: Perfect.
Orbax: There.
Mike: Okay. So, what forces that are acting on the slope.
Orbax: [Very quickly] Force of gravity acting down. [Slower] So, the force of gravity acting down.
Mike: Okay, the sled’s resting against an incline. So, that contact means that there’s going to be a normal force. And the normal force always is perpendicular to the surface of contact. So let me draw a normal force here.
Orbax: And it hasn’t been stated whether this is frictionless or not. Assuming that there is some sort of friction and that the sledder is actually going down the hill, then the friction always acts opposite the direction of motion, meaning that if they’re sledding down, that friction is acting back up against.
Mike: And if it’s motion involved, then we’re going to use kinetic friction. But if it was that the sledder was there and just stuck in place, there would be static friction holding.
Orbax: Either way, it’s still acting in the same direction.
Mike: Right. So, there’s a free body diagram. There’s all the forces acting on it. And we can use that to relate to the motion of the sledder.
Orbax: Now we have these rules here for free body diagrams. And how do those relate? First one is we only draw the forces that are acting on our object.
Mike: So, we have to make sure we include everything.
Orbax: Okay, so then, what’s the difference between the first and the second rule? This one says no force is caused by the object.
Mike: Okay, well, so the sledder might be exerting forces on their surroundings. And we can’t include those because it’s only the forces acting on the sledder that dictates the sledder’s motion.
Orbax: Okay. What’s an internal force?
Mike: The internal force so we’re treating the sled and the sledder as a single object. And if you were sitting on a sled, there might be friction between you and the seat of the sled. But as that’s all inside the object, the force of friction here would be ignored. That’s internal to the –
Orbax: And motion.
Mike: Okay, well, alright, we talked about whether the sled’s moving downwards or not. We can describe that with, say, a velocity or maybe an acceleration vector. We just need to make sure that, that vector doesn’t make it in here, because we’re going to be using this diagram to look at forces that cancel out or add together. So, an extraneous or another vector there might just mess up the math.
Difficult
[Orbax is swinging a tennis ball on a string overhead, Mike, is ducking out of the way.]
Orbax: Hello! I’m the Great Orbax. I’m here today with Dr. Mike Massa to discuss circular motion.
[Text FBD: Conical Pendulum appears at the bottom of the screen.]
Mike: Okay. So, we’ve been talking about the idea that an object can travel at a constant speed, but because its direction changes, there’s going to be an acceleration, and that’s the centripetal acceleration.
[Orbax keeps swinging the tennis ball, points to his hand at the ‘centre’.]
Mike: It always points to the centre of the circular path. Now, because there’s an acceleration, there’s a net force. And we might call the tension in the string here, for example, pulling towards the centre of the circle as the centripetal force.
Let’s take a look at an example.
[Orbax slows the pendulum to a gentle spin.]
Mike: But instead of something this aggressive, let’s look at the ball spinning. It’s still traveling in a circular path, but it’s hanging from the rope now. There’s still tension, but the tension now is not directed to the centre of the path. So, to figure out what the centripetal force is here, why don’t we draw a free body diagram.
Orbax: Let’s do it!
Mike: Okay. Alright, let’s take the incident where the ball was coming around here, it’s heading away from the camera.
[Mike draws a large dot in the centre of the screen and indicates an arc motion.]
Mike: The rope is going to be –
[Mike draws an arrow up to the left of the screen from the large dot.]
Mike: -- make an angle this way with some angle theta [Draws a dotted line straight up from the large dot.] relative to the vertical, and that’s just some tension force T [labels the original arrow with a T].
Orbax: And of course, the ball itself actually has a weight associated with it, meaning that you’re going to have the force of gravity acting directly down. [Draws an arrow straight down from the original dot.]
Mike: With any free body diagram, the goal is to list all the forces acting on the ball. There’s just these two. Now, centripetal acceleration, v squared over R, is the response, is the motion and that’s traveling always at the centre of the circle. We could say the centripetal acceleration is pointing in this direction at that instant in time. [Indicates the direction of the motion on the diagram.] But, now we’re done, I mean –
Orbax: And that’s the thing right? We’re looking at literally doing a free body diagram. All we want to discuss is the sum of forces, and I this case, we just looked at the sort of forces in the x direction. It’s not equal to 0, it’s equal to a mass times an acceleration term where that acceleration term is that centre-seeking acceleration or that inwards pulls [Writes the equation on the screen.]
Mike: Right. So, in this case I mean, we would say that T sine Theta is acting in the x direction, [Orbax adds this information to the equation] and that’s causing ma, the mv squared over r, centripetal motion.
Orbax: And that’s it. You literally have an acceleration towards the centre that’s defined by the fact you have a component of the tension acting in that direction.
[Final equation reads: \(\begin {align} \sum F_x = ma_c \\= T \sin \theta \end{align}\) on screen.]
Mike: There you go.
[Orbax picks up string and starts swinging the ball around.]
Mike: Okay. Just -- [Ducks]