Professor Anderson

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>> Hello Class, Professor Anderson here. Let's take a look at an equilibrium problem, where we're dealing with forces and torques. And let's try the following. Let's say we have a beam and we are going to suspend that beam from a wire. And we're going to put a mass on the end of the beam, and we're going to suspend this end of the beam from another wire attached to the floor. Okay, so we have the roof up here, we have the floor down here, and now this is some sort of mobile. And we have some tensions in these wires, okay. So, this one is acting down, it's trying to hold the beam down in position. This one is acting up, it's carrying the weight. And of course, this is acting down with a force m g. So, let's label a few of these things. Let's say this is tension one, and this is tension two. And let's give them some positions, okay. So, this position right here, we will call x two. The position of this one is all the way that the left of the beam, we'll call that x equals zero. And, the position of the mass we will say is x sub m. Alright, so that's what our picture looks like. A little bit messy, but I think you get the idea. And now let's analyze the forces. Okay, there's no forces in the x direction. There are forces in the y direction. We have t two going up, we have t one going down, and we have m g going down. Now, we're assuming that this beam is massless. If you have a mass on your beam, you have to add that, and that would be acting at the center of mass of the beam. So, forces have to add up to zero if it's in static equilibrium. And now let's analyze the torques. So, the torque we need to pick an axis, and let's pick the left end of the beam for our axis of rotation. So, we'll say it's right there. And now let's analyze the torques. t one is a force, but the lever arm for that is zero. t two is going to rotate it counterclockwise, we give it a positive value and its lever arm is x two. And mg is acting to rotate it in a clockwise direction, and therefore we give it a minus sign, and its lever arm is what we call x sub m. And all of that has to add up to zero. Okay, we have two equations now, and we have two unknowns. And so, we can solve or these. So, if we solve the second equation for t two, what do we get? We get t two equals, first term goes away, last term I move over to the other side. It becomes positive m g x sub m. And then I have to divide by x sub two. So, that is the tension in cable two. And now, if I want to solve for the tension in cable one, I go back to the first equation, and the first equation says t one is equal to t two minus mg. And so, this becomes mg times x m over x two minus one. Okay, we just put all those terms together, you can double check my math. And now you can look at the limits. So, if x m was equal to x two, then it says that t one would be zero. And so, that's a condition where we moved our mass right to where the cable t two is attached. And if that is indeed the case, then t two is holding the whole thing up, it's perfectly balanced, and you need zero tension in t one. Okay, so that limit makes sense. You can try this with your various numbers for different parameters here, and make sure it makes sense to you, but hopefully that's clear now. So, if you have any offit [phonetic] any questions, come see me in my office. Cheers.

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>> Hello Class, Professor Anderson here. Let's take a look at an equilibrium problem, where we're dealing with forces and torques. And let's try the following. Let's say we have a beam and we are going to suspend that beam from a wire. And we're going to put a mass on the end of the beam, and we're going to suspend this end of the beam from another wire attached to the floor. Okay, so we have the roof up here, we have the floor down here, and now this is some sort of mobile. And we have some tensions in these wires, okay. So, this one is acting down, it's trying to hold the beam down in position. This one is acting up, it's carrying the weight. And of course, this is acting down with a force m g. So, let's label a few of these things. Let's say this is tension one, and this is tension two. And let's give them some positions, okay. So, this position right here, we will call x two. The position of this one is all the way that the left of the beam, we'll call that x equals zero. And, the position of the mass we will say is x sub m. Alright, so that's what our picture looks like. A little bit messy, but I think you get the idea. And now let's analyze the forces. Okay, there's no forces in the x direction. There are forces in the y direction. We have t two going up, we have t one going down, and we have m g going down. Now, we're assuming that this beam is massless. If you have a mass on your beam, you have to add that, and that would be acting at the center of mass of the beam. So, forces have to add up to zero if it's in static equilibrium. And now let's analyze the torques. So, the torque we need to pick an axis, and let's pick the left end of the beam for our axis of rotation. So, we'll say it's right there. And now let's analyze the torques. t one is a force, but the lever arm for that is zero. t two is going to rotate it counterclockwise, we give it a positive value and its lever arm is x two. And mg is acting to rotate it in a clockwise direction, and therefore we give it a minus sign, and its lever arm is what we call x sub m. And all of that has to add up to zero. Okay, we have two equations now, and we have two unknowns. And so, we can solve or these. So, if we solve the second equation for t two, what do we get? We get t two equals, first term goes away, last term I move over to the other side. It becomes positive m g x sub m. And then I have to divide by x sub two. So, that is the tension in cable two. And now, if I want to solve for the tension in cable one, I go back to the first equation, and the first equation says t one is equal to t two minus mg. And so, this becomes mg times x m over x two minus one. Okay, we just put all those terms together, you can double check my math. And now you can look at the limits. So, if x m was equal to x two, then it says that t one would be zero. And so, that's a condition where we moved our mass right to where the cable t two is attached. And if that is indeed the case, then t two is holding the whole thing up, it's perfectly balanced, and you need zero tension in t one. Okay, so that limit makes sense. You can try this with your various numbers for different parameters here, and make sure it makes sense to you, but hopefully that's clear now. So, if you have any offit [phonetic] any questions, come see me in my office. Cheers.