Hello. Last time, we studied how the Newton-Euler equations of motion describes the motion of the rigid body. Once we learn how to obtain the equations of motion not only F equals ma, but also, moment equals I Alpha, now, we are going to apply this for the general motion of the rigid body. Generally, the rigid body motion is combined by the translation and rotation. First, we are going to study the translation of motion, and rotation for the next session, and then later, in general plane motion. Once we are done, we are going to do the integral over the displacement to obtain the work-energy relationship, and time integral to obtain the impulse and momentum relationship. Let's study how the translational motion of the rigid body could be solved through the Newton-Euler equation. Well, this is a pickup truck of mass m is accelerated from rest to a speed of v over the distance d, and then, effective coefficient is Mu. Then you are supposed to find the normal force under the front and rear wheels. If this is the case where we are handling in Chapter 3, usually, we're assuming the total normal force and the friction force of the two wheels contacting the grounds to be summed up. But here, it seems like first, it's been accelerated from zero to speed v, so that means external force is applied. Also, a normal force of front and rear wheel mean that may be different. Before you go further, you might ask yourself, "Why in this situation has a different normal force front and back? Isn't that supposed to be the same as mg divided by 2, front and back? What causes the difference between them?" That's the good question that you should think of. This is kinetics problem, so I'm going to draw the free body diagram and obtain the equations of motion. First, how many contacts does this one have? Two, front wheel and the rear wheel. There are two forces; friction and normal force and the gravity, and that all generate their linear acceleration and the rotational acceleration. This pictorial relationship between force and motion could be transformed into mathematical form, which is equations of motion. This one is x directional equations of motion, and this is y directional equations of motion. You are supposed to find this one and this one separate, and seems like too many unknowns here. You should apply one more relationship which is moment equals I Alpha, so additional equation will help. Since the truck is moving through the incline, which means there is no rotation, so the Alpha is going to be zero for translational motion. So the torque exerted by the normal force and the friction forces turn out to be zero, the balanced, right? Then if you transfer the friction force as Mu N, then you will have two equations to be solve by series to obtain the N_f and N_r separately. Before we move again, ask yourself, it seems like you may have a different N_f and N_r, maybe you can guess or you would have an interpretation why N_f or N_r is greater than the other? What causes the differences between the two forces? Try to interpret your answer or try to provide some physical explanation what have changed the normal force between the front and back. Also, this is the free body diagram and obtain the equations of motion in x direction. But it seems like there's only friction force, which is a reaction force, and mg in the minus direction, and it seems like can't be accelerated forward. So something looks like missing because there's no force source for the forward acceleration. What I have missed? So think about that. So this is another translational motion problem, which is the three-bar linkage, so one bar in the vertical and two bars in the horizontal, and this pink bar is restless. It seems like first, maybe there is a rotational motion through the pendular motion, right? However, the distance between the bar is linked up and down. Actually, while it moves is something like a translational motion. There is no rotation for the bar. Note that. Then torque is applied on the lower link and you are supposed to find what's going to be the reaction force X at this joints? There are four hinge joint here, so there are four reaction forces pair, okay? Then you are supposed to find what are the reaction force between the pink upper link and the vertical bar, and the acceleration of the link. I'm going to go through to find the coordinate probability diagram and obtain the equations of motion. Since you have a three parts, even though the pink one is mass-less, you need actually three coordinates to describe each of them, so for those hinged pendulum, I would set the coordinate at here, and then I would set the coordinate for the blue one maybe on the ground or in the center of mass. Let's start with the MST-Bar. How many contact does this one have? Two, right? [inaudible] hinge? There are two forces, so RX, RY on the joint hinge D and RX, RY at the hinge B, and then mg. Okay? Those are generating linear and rotational motion. But since it is a translational motion, so there is no Alpha. These pictorial relationship between force and motion can be transformed into equations of motion in x directions and y directions. Okay? Also, the moment equation with respect to the center of mass due to the torque by this horizontal forces will generate soupy balanced out because there is no angular acceleration, okay? It seems like there are too many unknowns. You're supposed to find this one. This is what you're supposed to find, but you don't know this, you don't know that. There are too many unknowns, so it seems like you need more information. Since this is a rigid body and there's no rotation, so that means all the points of your rigid body will share the same translational acceleration since there is no rotational motion. You are supposed to find the acceleration at the center of mass, but if you can get information about acceleration at point D or point B, that actually [inaudible] as acceleration of the center of mass. Let's look at the upper link part. As mentioned before, upper link part is moved with the angular velocity Omega. If this bar is moving with the Omega, you know how to obtain the acceleration at the end with the distance D, which is centripetal acceleration in x-direction and tangential acceleration in the y-direction. They're actually same as the acceleration here and here. You can just use those information here and there. These are what you're supposed to find, the alpha CD and get the reaction force here, and those are unknowns. Still, even though you actually solved out the what's going to be the acceleration term, still there are unknowns, so you need more information. Of course, you can eliminate R_xP from the rotation and motion relationship so that you can find what's going to be the R_xD here because O_mD Omega are given. Still, there are two unknowns to figure out what's going to be the Alpha. Let's work on the three-body diagram of the other two links as well. Even though it doesn't have a mass, there is a force exist. Since there are two to counterpoints, two forces per each so there are four forces applied. Then from that, we can obtain the equations of motion. But note that, since it doesn't have a mass, the right-hand side is going to be zero because mass is zero, same for this one. Mass is zero, that means moment of inertia is zero, so all the forces and the moments should be balanced out. So let's write down the balance equation for the upper link with respect to point C here. That means, only torque by this one should be zero. So which means, our reaction force in the y-direction at D is zero and same for this link. If you write the moment equation with respect to this hinder joint, the moment actively applied should be balanced out by the reaction force there. So what we can get is this one. So this R_yB has been specified. So those information specified from the equations of motion for the other links should be plugged into the original equations of motion for the next bar, then we can find the Alpha. Now, let's solve another problem. This one is also a translational motion. So all of a sudden, you actually move the card on the part along the linear guide, and those bar actually tilted back. It's like a handrail or the train or the bus. When this is happening, what is attention of each cable? Cable in the front and cable in the back. So a similar way, is asking your attention for front and back, you might ask yourself why there's a difference between this tension? Is that because of this different distance of the tension attached to the bar or what? If those are different, which R will be greater? Is attention to the front going to be greater or the back? You should think about that and when you solve the problem, try to interpret your answer. I'm going to go through, first define the coordinate, draw the free-body diagram, and obtain the equations of motion. Where should I put the coordinate? Here, whatever moving on the cards, I would put the coordinate on the card so that I can describe the motion in a relative coordinate frame, that is because that'll be easy. If I set the coordinate and the absolute coordinate frame, the bar just keep moving toward the right-hand side. With respect to the center of mass, I'm going to set the reference frame on the card. But note that this card is moving with the acceleration. This is not inertia frame, this is a non-inertial frame. This is moving with the acceleration, therefore, you should consider the inertia force. Fictitious force that is applied to the center of mass, opposite direction with a magnitude of MAP; P is the acceleration at this point here. How many [inaudible] of this [inaudible] bar have? This one and this one, so there are two cable forces; tension and the gravity, and this is it. This is how you're moving with the bar. As I said, this one doesn't rotate. It's a linear translation, so those R_Phi turns out to be zero. Those fictitious relationship between the force and the motion can be transferred as equations of motion in x-direction, and y-direction, and rotational motion as well. As I said, since this is a translational motion, there is no Alpha; turns out to be zero, should be balanced. Though those torques by this tension front and back should be balanced. You're supposed to find the tension front and back, and the angle, tilt angle, so you have to solve the three equations. Now, since there are three equations and three unknowns, you should have more information. It seems like, am I supposed to know this information? Yes. In the problem, it says a steady state. Steady-state means there is no acceleration, so no time change. If you plug that in, you will have a simplified version of equations of motion, and if you just solved by a series, you can obtain what's going on [inaudible] Theta, and what's going to [inaudible] , front, and back. This ends the Chapter 6.2, about the translation example for applying the Newton Euler equation for the rigid body. Next time, we are going to study how we can apply this for the rotation and motion. Thank you for listening.