Problem - Consider the crankshaft shown in Figure 1. The...

Problem - Consider the crankshaft shown in Figure 1. The system is positioned on a vertical plane (g = 9.81 m s²) and is formed by two massless rods of length L and 2L, respectively. Three pivots, about which the two rods can rotate, are placed at points O, 1 and 2 in figure. The two rods are connected by the pivot corresponding to point 1. Two point-masses, M and 2M, are placed at the lower ends of the two rods (i.e. at points 1 and 2, respectively). Point O is fixed. Point / has free in-plane movement. Point 2 is constrained to move in a vertical rail without friction, passing by the origin. The system has only one degree of freedom. 1 M L1 lo / y Law of cosines Y² = W² + V² - 2WV cos? W² = Y² + V² - 2YV cosp 2L 2 g 2M q = {y} Figure 1 W ? ? Y Figure 2 a) Using q = {y} as the generalized coordinate for the system, derive a system's Lagrangian as a function of y and y which only contains any of L, M, and g (1.5 pt.). L=T(y, y; L, M)-U(y;L,M,g) =

Transcript

00:01 Hi there, so for this problem, we have this situation that is shown in here.

00:04 The coefficient of kinetic friction on the old surfaces is given, and that is 0 .35.

00:14 Now, the first question is to draw all the free body diagrams.

00:19 Okay, so let's just start with the object a, okay? so we're gonna put the axis at its center.

00:32 So then the vertical axis is in this direction, and the x -axis is in this direction, okay? the forces acting on this are the normal force that is always perpendicular to the surface of contact, so this will be the normal force a.

01:01 Then we will have, now, i'm assuming that this is moving to the left, sorry, to the right, so then that will be that a is falling.

01:28 Okay, so well, the sign from the acceleration will give us if it is moving in that direction or in the opposite direction, okay? now, we're gonna assume that then this one is the one that is moving to the right, so then we know that, well, we have the tension.

01:50 This tension is t because it's going to be the same tension for both, and then we have the weight of this, which is done work, okay? this is the weight, and we're gonna label weight a, and since it is going to be moving to the right, then there is some friction that opposes the motion of this, so this is the friction a to the left, okay? it's to the opposite direction, and that's it.

02:26 Those are all of the forces acting on the object a.

02:30 Now, for the object b, we will have, again, the same tension, okay? now, let's put the axis in here, about here, okay? so then let's put the axis, the y -axis in this direction, and the x -axis is going to be positive to this side, okay? now, in this case, since the object a is falling, then the object b is going to be moving up the incline, so the frictional force is in the opposite direction, so the frictional force should be about here, so this is the frictional force b, the normal force that is always perpendicular to the surface, the normal force b, and finally, the weight of this, okay, that is downward.

03:44 Once we have all of this, so that's a solution for the first question of this, for part a of this problem.

03:50 Now, for part b, what we are asked to determine is the acceleration of each block, so let's start with the equations for the object a, okay? so let's assume the force is acting on the x direction, so we know that to the right is positive and to the left is negative, so we will have that the tension is negative, the frictional force a is also negative, and then we will have the x component of the weight.

04:19 Now, remember that this angle is then, will be given by the vertical, okay? so that x component is then given by its magnitude times the sine of that angle, that in this case is the sine of 30 degrees, and then it's moving in that direction.

04:45 We assume that this is moving down, so then, well, it's moving down but in the positive x direction, so this should be positive, and there will be this mass ma times the acceleration, and this acceleration is the same for both because they are moving with the same acceleration because they are connected through the rope.

05:07 Now let's assume the force is acting on the y direction, and here we will obtain just simply that the normal force of a is equal to the weight because the object is not moving in that direction, so there will be then the vertical component of the weight, which is the magnitude of the weight times the cosine of the angle.

05:29 We are gonna use this because remember that the frictional force is defined as the normal force a times the coefficient of friction, okay, so then the equation that we will have from before is that this minus the tension minus the normal force, which is the weight times the cosine of 30 degrees, this times the coefficient of friction plus the weight times the sine of 30 degrees, and then this equals to the mass ma times the acceleration.

06:00 Once we have this, now from, we are gonna do the same but for the object b.

06:09 So for the object b, we apply newton's second law to the forces acting at the x direction.

06:14 Now, in this case, the tension is positive because it's to the right, so we will have the tension minus the frictional force opposites this direction, and then we have minus the x component of the weight, which is the weight times the sine of the angle that in this case is 40 degrees.

06:36 And then in this case, we assume that it's moving also in the x direction, so that will be also positive, so that will be the mass mb times the acceleration.

06:48 Now from newton's second law, from the forces acting on the y direction, we will have that the normal force of this is then the weight times the cosine of 40 degrees, right...

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