https://youtu.be/Vg0_tm6QzOo
PHYS 1101: Lecture Twenty-Two, Part Seven
We’re back to another equilibrium problem. Our Alpha is going to be 0; our torques have to balance. Here’s what we’ve got. A 30-kilogram beam that’s suspended as shown. We’ve got to figure out what the tension is in the cable. So to start the problem off, I’m going to slow down and first realize I’ve got to focus on an object. I’m going to take this beam, the 30-kilogram beam, as my object in red. So I need to think of a rotation axis for this object and forces on this object and therefore torques on this object. Well, what’s my rotation axis? Well where it’s attached here is a natural choice. I’m going to make that choice and now I’m going to draw forces on this object to be sure I consider all of them properly when I calculate my torques.
My force of gravity is going to be in the middle of this beam. It’s a nice, symmetric beam. My tension, of course ropes always pull. They pull in the direction of the rope so there’s my tension. And then the other place where I have contact is here, and contact, this is a surface. I’m going to just consider that it pushes.
Even if I wanted to consider friction here, which would possibly be in a different direction to this, perhaps in addition to this normal force, it’s not going to have any torque because still, where these forces are applied, is right here at this rotation axis, and so my r value will be 0 for all of those forces. Any force that’s applied right at the hinge, right at the rotation axis, won’t provide any torque. It’s not at all helping to rotate it. The r is 0.
Let me label these forces. Here’s t, the tension. Force due to gravity, and I’m going to label this Fn. Okay, I need to have a clear picture of the r values for each of these forces so I have to go from the rotation axis out to where the force is applied. Let’s call that r1. For my tension, I’m going to need what I’ll call r2. I’m going to need the length all the way out to the end of that beam. Okay, with this picture as I’ve drawn, consider these next quiz questions here.
With this hinge as being the rotation point, rotation axis, what’s the sine of the torque caused by this force? What’s the sign of the torque caused by the weight, force of gravity? What’s the torque caused by the force at this hinge location that’s on the beam? Question 22: is the beam in equilibrium? Okay, here’s my picture drawn again. I’ve got my values, my knowns, written again here at the top. Let’s solve this problem now.
The variable that I want is T, the tension in the line. It’s the hypotenuse. What is the fundamental equation I’m going to start with to solve this problem? What I need to set out and define is, and insist is that the equation Alpha equals the sum of the torques over I is true for this object. Okay, what’s the left side? What is Alpha? Is the thing in equilibrium? Well, it’s not spinning faster or slowing down. It’s just sitting there. Alpha’s 0, so I’m in equilibrium. I don’t have to worry about this denominator. This equation simplifies to 0 equals the sum of the torques. All of my torques have to balance.
Well, what is the right side of the equation? Using this as my rotation axis, I’ve got to put the torques in to each of these forces. Let’s just write them all out explicitly, and then we’ll cross off the zeroes and start plugging in. I potentially have a torque due to the normal force. I have a torque due to the gravity that would cause this thing to rotate in the clockwise direction about this point. That’s a negative torque. The only other force I have then is the tension. It’s only this component of the tension that has any contribution to the rotation. This is the tangential part. This would cause rotation in the counterclockwise direction, which would be positive.
What are these values? For each one, I have to put in that a torque equals the distance to the force application point times the part of the force that’s tangent. The normal force. What’s the r for the normal force? Well, that force is applied at the rotation axis. There’s zero distance between the rotation axis and that force. So this torque is 0. Zero minus.
Force of gravity. What’s the r and the f for this torque that I have to multiply together? Well ,the r is r1, and what’s the part of this force that’s nice and perpendicular to r? Well, the whole force is. It’s at right angles, so that’s mg. The magnitude of the force due to gravity plus what’s the r for this force?
Well, I now need the distance from the rotation axis out all the way to the end of the beam. That’s r2 times the part of the force that’s tangent. This component. Given the geometry that this is 60 degrees, if this is 60 degrees then the part of the force I care about is the hypotenuse times the cosine of 60 degrees; t cosine of 60 degrees. This angle information I left off of the original problem, and it would be given to you for this problem. That was a typo. You may be told for example this is 30 degrees, in which case this would be 60. You have to have this geometric information to work out the component that you care about.
So here is the fT, the part of the force I care about for this torque and here’s the hypotenuse. That’s the variable I’m trying to solve for. Rearranging this equation, if I add this to the other side I end up with r1mg has to be equal to r2T cos60. Solving for T, you end up with T is equal r1 over r2 times mg divided by the cosine of 60 degrees.
When you plug that in, r1 is one-half of a meter. r2 is 1 meter. These were givens in the problem, so this ratio is one-half. My mass is 30 kilograms. g is 9.8, cosine of 60. I end up with… let me just put those numbers down for you. I end up with about 170 Newtons.
So my hypotenuse, the total tension in the cable, is 170 Newtons, but notice that in the fundamental equation I used to work it out it had to do with torques balancing, and I had to represent just the component of that force in this direction to represent the torque. Nonetheless, that component contains the hypotenuse information, so has the variable that I’m able to solve for.
Let’s round out this lecture with just a few more quiz questions. Question 23 has you thinking about the torque due to the force of gravity. Sometimes that’s called the gravitational force. So this would be the torque due to gravity would be r times mg, the force of gravity. In order to figure out that torque, though, you have to think carefully about the location of the force of gravity and the r, the distance from rotation axis and that force.
So here’s the geometry that you have. It’s an I-shaped bar, and remember the force of gravity acts at the center of a symmetric object. Whoa. I got carried away there. Didn’t look very symmetric there. Okay, so it’s symmetric up-down, so the center of gravity has to be somewhere here in the center up-down, and also right-left it’s symmetric. So my gravity will act. I can draw that gravitational force starting right there. But notice what you’re looking at here is a rotation axis that’s the dashed line in each of these comparisons, and this is the top view.
So you’re holding this I-shaped piece so it’s horizontal and it’s flat, and I’m considering a rotation axis about a line through the center. The force of gravity from the top view is acting right at the center here and straight down into the page, into your computer screen. So the r has to be the distance, this r, from the rotation axis to the place where the gravitational force is acting, the center of the object. How do the torques rank for these combinations? Rank them from largest to smallest.
Question 24: you have a 5-meter steel beam of uniform cross section and composition that weighs 100 Newtons. Notice it’s already given to you as m times g; it’s already in Newtons. What’s the minimum force required to lift one end of the beam? By minimum force, consider that the beam is in equilibrium, so the minimum force would be applying just enough torque that it balances the gravitational force on this object.
Here’s my steel beam. I’m going to draw it partially lifted up into the air as though I’m able to overcome the torque of gravity. I’ve drawn that too extreme here. Let me draw it just barely above the ground here. So I’ve just started to rotate this. The force that I’m applying balances the torque due to gravity. So what does this force have to be in order to do that? That brings us to the end of lecture 22.