Hi and welcome to module eight of two dimensional dynamics. Today's learning outcome will be to continue to solve problems, solve them for the motion of the mass center of bodies using Newton-Euler motion equations. So if you recall. Here's Euler's first law for a system of particles or bodies. The sum of the external forces equals the mass times the acceleration of the mass center of the body. So I use a graphical tool free body diagram equals kinetic diagram and so lets go ahead and solve another problem. Again this has a real world application. We've got a truck shown. It's travelling at 45 miles an hour, and we want to find the minimum stopping distance, such that the 250 pound crate will not slide. And assume that the crate cannot tip over. So that's the problem statement. What do I do first? Okay, back in statics we always drew a free body diagram we're going to draw a free body diagram equals kinetic diagram in this course. But before we do that even, let's go ahead and put everything in common units, since I'm working in the English system of units. I've got 250 pound crate. Let's change, miles per hour to feet per second. And so I've got 45 miles per hour times 5,280 feet per mile times. One hour for 60 minutes and one minute for 60 seconds. So that converts to 66 feet per second for my initial velocity of the truck. Then let's go ahead and, again to the free body diagram equals kinetic diagram. You should do that on your own. Come back and let's do it together. And the body of interest in this case is the crate. So I'll draw it here from my free body diagram and I'll draw it from my kinetic diagram. And we'll say this is a uniform crate so the mass center is right in the center of the crate. So we'll call that C. And on my free body diagram, I've got my weight, which is 250 pounds. And I've got my friction force. Friction always opposes motion. The impending friction is going to oppose the pending motion. So as this truck goes. Forward, the crate's going to, when it's slowing down going to want to go forward so we're going to have the friction force going back to the right, I'll call it f, and then I also have my normal force. And if you remember back to my modules in applications in engineering mechanics the normal force. We'll migrate as we have impending slipping or tipping, so it's going to migrate in this case at the crate tends forward to the left, and it'll be at a distance of d. So let's draw that distance in here, an unknown distance. D. 'Kay. On my kinetic diagram, I've gotta draw my two orthoganal mass times acceleration, components. And so I've got, mass time, I'm using to the left, positive, since the, crate's going to tend to slip to the left, and the truck is going to the left. I, I, I could do it, to the right, and if I got a negative number. I would now what's in the opposite direction but I'll go ahead and draw it this way. So I have mass times acceleration in the x direction and then I'll have a mass times an acceleration in the y direction. But since the crate is moving. Neither up, nor down, we know that this mass times acceleration y, is equal to zero. And so that's a good free-body diagram equals kinetic diagram. What should we do next? And always what you should say is, let's apply the equations of equilibrium, and so, I've got, let's do it in the y direction to start. So I've got. I'll choose up positive, again this is just a signed convention for assembling the equation. And so I've got sum of the forces in the Y direction on my free body diagram equals sum of the forces in the Y direction on my kinetic diagram. And that gives me. Let's see, N is up, 250 is down, equals zero. So N equals 250 pounds. Kay, let's do the same thing for the X direction. You can do it on your own, come back and check to make sure you've got it correct. I'll choose to the right positive and I have got some of the forces in the X direction on my free body diagram, equals some of the forces in the X on my [UNKNOWN] diagram. So I have the friction force f That's it. And the X direction Y free body diagram, equals minus M, A, X. I know that the fiction force is going to be Mu times N. The co-efficient of friction times N. In this case, that's, that's the max friction force I can develop before the crate would slide. In this case I'm saying that Mu-slipping and Mu-kinetic is the same. It's 0.3. By Coulomb friction law, the friction force is mu times n so let's go ahead and substitute that in. And I've got Mu times N. So, I've now got mu is 0.3. Times N which was 250, which I found right over here, equals minus the mass, well the weight is 250 pounds and to find the mass I've got to divide that by the acceleration due to gravity in the English system of units which I'll use as 32.2. Feet per second squared times AX. And if then you can find that the acceleration must be no more than a sub X, i'll right it up here. A sub X. Is equal to minus 9.66 feet per second squared. So it's in the opposite direction of what's shown. And what that means is the truck is decelerating, which we know it is. It can't decelerate anymore than 9.66 feet per second squared. Or that crate's going to slide forward. And so that's the condition. Now we need to find the minimum stopping distance based on that condition. So, here we go. I've got my acceleration, minus 9.66. The initial velocity we found to be 66 feet per second. And we said that if the deceleration. If the deceleration exceed that, there's not enough friction and the crate will begin to slide. I want to find the minimum stopping distance, I've got acceleration, what I'm going to have to do is start integrating and I'll integrate acceleration first to get velocity. And then, later, integrate velocity to get a, displacement or distance. And, so, when I integrate. I've got V of T is equal to the integral of AXDT, which we said was minus 9.66 T plus the initial velocity. And I'm saying that the minimum stopping distance is when I maintain that constant deceleration of minus 6 6 feet per second squared. So, I integrate that constant. I get the times T plus the initial velocity. I use the IC for the initial velocity. You recall that the IC. I see being the initial condition. We've got at T equals zero. The initial velocity we found to be 66 feet per second. I can substitute that in. And so I get. Therefore, V sub T equals minus 9.66 T plus 66. We also know that when V equals zero, the truck will have stopped. I want to find the stopping distance. So I know that v of zero equals zero. The truck. Has stopped. So I'll put that in my velocity equation and I get zero equals minus 9.66 T plus 66. This will allow me so solve for the amount of time that it takes for the truck to stop. And we find out that t equals 6.83 seconds to stop. Okay. Let's keep on going. So I've got my acceleration. I've got my velocity. I ass, my assumed initial displacement, my datum will be zero feet. And I know it takes, 6.83 seconds to stop, so I'm going to go ahead and integrate the velocity to find that distance. [BLANK_AUDIO] So, the integral. X of t is the integral of minus 9.99 T plus 66. I got two, two plus in there, I only need one. Oh, I'd had a T there, that's what it was. D t, so it's minus 9.66 T plus 66. If I integrate that I get X of T equals the minus 9.66 T squared over 2, plus 66 T. Plus the initial displacement. And again, I can use my initial condition of the initial displacement being zero. So, my initial condition is that at T equals zero, X zero equals zero. Substitute that in. And I have got, therefore, my equation for my distance travelled X at T equals 9 minus 9.66 T squared over 2 plus 66 T. I know it takes 6.83 seconds to. Stop so let's substitute in 6.83 and I get minus 9.66, 6.83 squared over 2 plus 66 times 6.83 or 225. 225 feet to stop. And that's our answer. Okay. Let's have you finally do a worksheet on your own. You've seen me do a couple of these, one last module and problem this time. So now on this one I would like, we have two blocks here. We've got their their mass in kilograms. We've got the coefficients of friction between the two blocks and between the bottom block and the surface. It's on a 25 degree slope. I want you to find the accelration of blocks when released from rest. And please note, you're going to have to check several cases here. The blocks might slide together, B2 may slide, and B1 slides faster. You know, there's a whole bunch of cases here. There are not a whole bunch, but there are cases, different cases that you need to check. Once you get that solution. I have the, you can check it with the solutions that I have in the module handouts and I will see you next time.
