https://youtu.be/8hY4kmc3SDc
PHYS 1101: Lecture Twenty-Two, Part Four
There are a lot of great biology examples here in the context of torques and how the body moves.
Here’s a great one where a person is in this configuration here, so the rotation axis would be the elbow, and this forearm is attached to this fixture, and this person is pulling on this fixture such that they feel a 190-Newton force, is what they’re pulling against up here at the wrist. That pulling is happening because a muscle is tugging on this forearm at this distance above the elbow.
We’re asked that for this configuration, what is the horizontal force, M, that the muscle exerts in order for this to be the case, for this scale here to read 190 Newtons?
Okay, Step One, what’s the object we want to focus on? There’s nothing rotating here, but ask yourself, “What would rotate if, for example, this broke, this force went to zero?” Or the person relaxed their muscle. It’s the forearm that would rotate.
So I’m going to grab a thick marker here and highlight for you in red the forearm here that I want us to consider as the object that’s going to rotate. Rotate about where? Here is the axis.
Okay, now decide, “Is this object in equilibrium or not?” Let me put that up here, “Equilibrium: is the angular acceleration zero.” Yeah, well for sure it is, because I want in fact this to be stationary. I don’t want it to rotate at all, so the answer is yes. That means that my basic equation that the sum of the torques divided by I is equal to zero. That reduces to zero equals the sum of the torques. In other words, all the torques on this object have to balance to explain this behavior.
Well, let me right that basic equation here, because I’m now going to break down the right side. I have to be sure I’ve represented all forces on this object and calculate the proper torque for each one. What forces do I have?
I have the muscle. It is applying this force at a distance of .054 meters, that’s the r, let’s call this rM for muscle. Then this 190 Newtons is definitely in contact with this forearm and it’s applying a force that’s to the right. If you think about this forearm, from the perspective of the forearm it feels this band around the wrist, pulling to the right.
So I’m going to go in with blue and I’m going to put a force here that says–I’m just going to call it F that’s equal to 190 Newtons. The radius for that force is this distance. Let’s call this just plain r for this plain F, but then for this force M, it’s radius, it’s r value is rM.
Now I also have the force of gravity acting on this forearm, on this object. Gravity–you’ll see in a minute–I’ll talk more specifically about effectively the force of gravity acts–but at this point I’m sure you’d appreciate that the force of gravity on this forearm is definitely straight down. If it’s straight down, it’s not contributing to rotating this forearm either counterclockwise or clockwise. It’s like pushing straight along into a door; you’re not going to get the door to rotate. So the only forces that I have to consider on the right are those that have a torque. The F perpendicular, the F tangent if you will, is zero for the force due to gravity. I only need to include these two forces.
So zero equals, I’m going to break this up into the torque due to the muscle plus the torque due to force F. Let’s plug in what they are now. For each of these, it’s r times the part of the force that’s tangent. Let me be sure the sign is correct. This force is applying a torque that would cause counterclockwise rotation, so this should be positive, which I should perhaps make a little clearer here. Let’s write it out explicitly. This force is applying a torque that would cause clockwise rotation, therefore it’s going to be negative. I have numbers for all three of these variables. This is the variable that I want.
In fact, I just realized that lets label it–I’ve mixed my variables here. I should call this they do here in the problem–M–You know what? I don’t like that choice. Your textbook, I don’t know about it sometimes. Let’s still call it force muscle. Let’s do this, let’s call this force muscle. I think that’s a better choice, more consistent with what we’ve done all along.
This is what I want to solve for, and here is my basic equation that defines what this value has to be. It has to make this statement true. It has to be a value that causes the torques to balance. We do a little algebra rearrangement there and you’ll see that F, the force of the muscle, has to be equal to the r, the distance out to this applied force times that value, divided by r of the muscle, the r, the distance up to where that muscle force is applied.
When you put in the values for that, you end up with the big r is .34 meters. I have to divide that by the small r, .054 meters, and I’ve got to multiple that by the magnitude of this force F, 190 Newtons.
Maybe I scripted over this explanation. Notice for each of these torques that I just put the product in of r and the force. Each torque, remember, is r times the F tangent, but a lot of times, as is the case here, all of the force is nice and perpendicular or in the tangent direction for this rotation, or what would be rotation if other torques weren’t there. Nice and perpendicular to r. Both of them are, the muscle and this wristband force. That’s why it ended up being just the product for each of these.
Okay, when you plug in numbers for that you end up with that the muscle force has to be 1,200 Newtons, roughly. These are realistic numbers. Let’s think this through again, what does this say? This always surprises me. I think it’s neat to realize the only way that your forearm can rotate and can move is because your muscle is attached, not right at the elbow, but there has to be a moment arm, this is called. There has to be a distance from the rotation axis that this force is applied. This muscle has to put a torque, but because that muscle is so close to the rotation axis it’s like pushing on the door right near the axis. Boy, you have to push really hard to have an effect. This muscle has to apply a force of 1,200 apples, 1,200 Newtons in order to balance a 190 Newton pull out at the end of the wrist.
This force is leveraged and magnified by having such a large distance to the rotation axis. That leverage means that for the same torque I can have a much smaller force than the muscle has to provide. That’s why some people are intrinsically stronger than others. A lot of it is just physiological of where their muscle attaches, the distance, compared to their elbow, the rotation point. That’s why monkeys are so much stronger too.
Okay, so to do these problems for the sum of the torques we have to include the torque caused by every force on that object. We didn’t have to worry about the force of gravity for this forearm that was straight up and down, but in general we do. Gravity will put a significant torque on an object. Where does that force belong? Where should we imagine that force being located on this object?
Here’s the recipe. I’ll have you do a little experiment here to convince yourself of it. We need to put this force of gravity at what we’re going to call the center of Gravity. This is also sometimes called the center of mass. It’s the point where is you collapsed all of the mass of this object, all of the molecules, if you collapsed them and condensed them into a single point, that’s where gravity would effectively be acting is where the center of mass of this object.
For a symmetric object, that center of mass, that center of gravity, is going to be the point where the object balances. Here’s what I mean by that. Take you pen that you’re probably holding and put your figure out and try to balance you pen across your finger. Obviously, if the pen is sitting too far off to one side it’s going to tip; it’s going to rotate and fall off your finger. If it’s too far the other way it’s going to rotate the other way. Because this has–unless you’ve got a really odd pen with maybe a big fluffy or a big duck on the end, or something–it’s nice and uniform and the mass is going to be evenly distributed, and the balancing point, the center of mass, the center of gravity, is going to be about near the center. Okay? It’s on that spot that we need to draw our big blue arrow that represents the force of gravity pointing straight down. It’s at the balance point. If it’s a symmetric object like a nice, uniform pen it’s going to be dead in the middle.
Okay, that’s quiz question six for you. There should be an easy–You know what? Let’s make this worth five points. Yeah, that’s good.
So let’s consider that in the context of a beam. Let’s think about the torques on a beam. The beam is the object that we’re focused on. I need to consider all the forces acting on the beam in order to properly think about the torques. Well, what’s it in contact with? Let’s do those first.
I have contact point here. It’s a surface. It’s an edge here. Surfaces always push up. I’m going to call this FN1. What else is in contact with the beam? Oops, it feels a surface here due to this purple box; that’s going to put a force down. That’s going to be the weight of the box. This is going to be the magnitude of this contact force there. And I’m going to keep going around here. I have another contact at this end, and I’m going to continue around, no more contact, but then I know the force of gravity I have to include. Well, this beam, it’s supposed to be drawn nice and symmetrically, so I would put my force of gravity dead in the middle.
Okay, those are all the forces on the object I’m interested in. Now let’s think through and just get a visual check on the lengths of these forces, the magnitudes that I’ve sketched. I’m happy with the location that I put them and the direction. Let’s think about the magnitude. It’s certainly true what we’ve learned before that if the thing is in equilibrium, which it is, it’s not translating, speeding up to the right, left, up or down. It’s not spinning up, or slowing down; it’s in equilibrium, meaning a is zero, and Alpha is zero. I know all the up, down, right, left forces have to balance for this to be true and in know all my torques have to balance for this to be true.
Let’s do a visual check and see it this looks to be the case. Do all the up-down forces balance? Well, I’ve got a big down there, another down here, and two ups. It looks like I’d have to make these a little bit bigger to properly add up to this plus this is the equivalent length to that and that.
But, I may not yet have enough to confirm whether this is true, whether the torques balance. From considering just up-down forces balancing, I don’t know how the size of this normal force compares to the size of this one, and where this box is located will definitely change the relative size of these. I suspect your intuition tells you that. If you were holding up and providing the support at 1 and 2 here, if this block or this weight were sitting entirely over one of your hands you would feel like this, your right hand, was supporting all the load, and this wasn’t supporting much at all. B, if you put the block dead in the middle you would feel an equal support on both sides.
So quiz question seven here gets you thinking about the size of these forces in a configuration something like this, and that plays an important role in thinking through whether the torques balance or not, or if that’s a proper representation. So let me add here just a question mark beside each of these forces in terms of the size and how they compare.