https://youtu.be/blm6UUUGNO8
PHYS 1101: Lecture Twenty-Two, Part Three
Let’s walk through this example to see how this might work. My problem here is I’m installing a spark plug in the car, and the manual specifies that it has to be tightened to a torque with a magnitude of 45 Newton meters. So, I need to use the information in this drawing to determine the magnitude of the force that you have to exert on the wrench.
Okay. Let me emphasize a few things. My rotation axis. You got to hone in on that as the first step. That’s going to be the center of my bolt. The r that I need to think about is the length of the line that goes from my center of my rotation axis or my rotation axis out to where the force is applied. That is the length r. Now, to this line r that I need to sketch the perpendicular direction. Perpendicular to that r, I need to find the component of f in this direction. Think of shining a light toward this axis. What’s the shadow that this total force… What’s the shadow that it casts on this axis? That means I need to dash a line straight down toward this axis. This is the part of the force that I need. This is the tangential component. Okay.
So this is finding components again. This is hypotenuse, and this is a component. You’d have to use trigonometry on this force, on this vector, to find this component. The first quiz question here is: what variable would you have to focus on or solve for? What’s it asking? It wants us to determine the magnitude f of the force that you must exert on the wrench. It’s asking for the hypotenuse, but the equation that we’re going to use to determine how large the hypotenuse has to be is the equation that involves just the component, f tangential, the component. So this, I’m going to use my torque equation. For this, I have to substitute in the component of the total force in the tangent direction.
Okay. Question four asks: At the end of the day, what is the variable you need a value for? Okay. Question five: Look carefully at this drawing. You’re told that this is 50 degrees. So, let me just do some similar triangles here and point out for you that then this is 50 degrees. What’s the correct trigonometric equation that represents this component and the hypotenuse and our angle of 50 degrees? So, here I’ve blown up that geometry for you and sketched in r again, from the rotation axis out to where the force is applied. Here’s my hypotenuse. This is the magnitude, my resultant force. I dash in my directions here, one perpendicular to r, and then this one that kind of continues in the direction of r.
What we need is f tangential, which is this component, this side of the right triangle. So, my total f is the hypotenuse. I need a right triangle where this tangent direction is one of my sides of this right triangle. So, let me go in with a really thick blue pen here and write. This force, this component is what you’re after. fT. That’s what has to go into my torque equation. The torque is the radius times fT. What’s the right trigonometric relationship for fT, given this geometry? f is the hypotenuse. That’s the variable you need to solve for.
Okay. So, it’s this torque that causes something to change its rotation, to start spinning faster or slowing down. If you want to see the derivation, I invite you to look through the text. What it’ll show you is that it’s just a rearrangement of our fundamental Newton’s Second Law. Tangential forces cause tangential acceleration. It’s just a connection of these variables to their rotational analogs that leads us to our fundamental equation that we need to highlight here.
Okay. The sum of the torques here has to be all torques, all forces with their location information. The torque implications of all of them have to be considered. Be careful with the sign. You have to represent positive and negative torques here to properly capture competing torques. The denominator is going to be the moment of inertia. I’ll talk to you more about that in just a minute.
The left side of the equation. It’s either going to be zero or some number of real radians per second2. I just took a minute here to rewrite this top statement, because I realized I was a little bit misleading. We want to be really clear. A net torque causes the rotation of an object to change. A net torque causes an angular acceleration, not a velocity, but angular acceleration. This means Alpha… Either the net torque is zero, in which case Alpha is zero. The thing’s in equilibrium.
Or, I can have a net torque, in which case it’s not in equilibrium, and the thing is actually spinning faster or starting to slow down. Remember, if something’s in equilibrium, it could either be sitting there stationary, or don’t forget, it could be sitting there rotating around at a nice constant rate. That also corresponds to Alphas equal to zero, of the thing being in equilibrium. Remember, that was a real subtle point that was difficult to appreciate about Newton’s Second Law. The analog was something could be moving along with a constant velocity and have no net force.
Here are my problem-solving suggestions or hints, kind of sequence of events to try and follow, that will help you get off to a good start when you’re solving these problems. As always, you have to be sure you’ve identified the right object to focus on. In this context, it’s the object that is rotating or might rotate. Isolate that object, and then you’re thinking about torques applied to that object.
What’s the moment of inertia, the variable I, for that object? Which we’ll discuss in a minute here. What’s the angular acceleration for that object? All of these rotation variables are defined by having a very specific axis in the problem. In these problems, you’ll see that it’s either clear what the rotation axis is, it’s given to you, or for equilibrium problems, you’ll see it’s up to you to pick the rotation axis. You’ll see more about that in a minute.
The third step, I would say to slow down and be sure that you’ve in your mind decided if this object is in equilibrium or not, that’ll help you appreciate this distinction. Are you looking for a case if it’s in equilibrium where the torques have to balance? Or is there a real value for the angular acceleration because there is a net torque divided by I? So, our basic equation, maybe I could highlight here for you, turns into really two choices. If it’s in equilibrium, the equation just reduces to the torques adding up to zero. If it’s not in equilibrium, then I have to use this full equation, which I just highlighted again for you.