Force, Mass, and Acceleration

Transcription

1 Introduction Force, Mass, and Acceleration At this point you append you knowledge of the geometry of motion (kinematics) to cover the forces and moments associated with any motion (kinetics). The relations governing such force systems are Newton s Laws of Motion, Work Energy relationships, conservation laws, etc. You begin with bodies which can be modeled by the path followed by their center of mass, and then go on to more general rigid body motions that must consider the rotation of the body. Usually these studies are restricted to planar motions since general three dimensional rigid body motions are covered in a course in advanced dynamics. The dynamic motion of non rigid (real) bodies is usually covered in a course in finite element analysis and/or vibrations of a continuous media. In this chapter motion along a straight line or curvilinear path are solved by Newton s Laws for particle motion, namely that the resultant force on a body equals its mass times the acceleration of the center of mass:. Example 14.2 This is an extension of the prior Chapter 9 study of friction to include possible accelerations. The block on the inclined plane is connected, by an inextensible cable, to a vertically hanging weight. The cable constrains the two objects to have the same acceleration value, in the direction of their connection to the cable. Knowing the two masses, and the static and kinetic coefficients of friction, you are to determine the acceleration (if any). The system and its FBDs are in Figure 1. Figure 1 A coupled pair of blocks From Chapter 9 you know that the blocks may remain stationary or may slide and accelerate. The kinetics rules are written assuming that any acceleration occurs in the downward direction for each block. That implies a consistent assumption that any friction force on A is up the plane. For some downward angles the motion would be different, and the accelerations will change sign. To get a feel for whether this will be a statics problem or a dynamics problem, consider at which angle would the system remain at rest with no friction ( 0, 0). The general rules in Figure 2 allow for that case. Making those assumptions the results in Figure 3 show that would occur at an angle of about 49 degrees. In that position, the static tension from the hanging weight B is just sufficient to balance the portion of the weight of A tangent to the plane. Larger downward angles would require a friction force on block A to the right. If that required force exceeds the maximum friction force then A would accelerate to the left and downward. For positive angles, block A could sit at rest Copyright J.E. Akin. All rights reserved. Page 1 of 9

2 Figure 2 Rules for static or dynamic response of the system on the plane, or accelerate to the right and downward if the friction limit is exceeded. To check the static case, for the given positive angle of 20 degrees, assume that the acceleration of A (and thus B) is zero. The results in Figure 4 shows that assumption requires a friction force of about 429 N, but that is far above the maximum static and kinetic friction forces that could be developed. So the TK logical variable status shows a result of slides. Therefore, a new solution must be found assuming that an acceleration of A exists. In that case the cable tension is also unknown. Thus, it is necessary to start the iterative TK solution by assigning Guess values to ax_b and T. The results for that case, agreeing with the text values, are found in Figure 5. To close this example, a large negative plane angle of 60 degrees was investigated in Figure 6. As expected, the sign of the acceleration and the friction force on block A both change. Note that the magnitude of the friction force is at its maximum kinetic limit. Copyright J.E. Akin. All rights reserved. Page 2 of 9

3 Figure 3 Results for the assumption of no slip with no friction Figure 4 Results assuming a positive angle static solution are not possible Copyright J.E. Akin. All rights reserved. Page 3 of 9

4 Figure 5 Text dynamic result for acceleration down the plane Figure 6 Acceleration results for A to the left for a large negative plane angle Copyright J.E. Akin. All rights reserved. Page 4 of 9

5 As noted in the text, the above example did not consider the mass of the pulley nor the cable. Both can be included, but adding the effect of the pulley is more common. That change will be presented next. Assuming that the cable does not slip on the pulley the (tangential) acceleration of either block causes an angular acceleration of the pulley. That angular acceleration times the pulley mass moment of inertia equals the net moment on the pulley. The presence of a moment at the pulley center means that the tension in the cable changes magnitude as it passes around the pulley. The new kinetic and kinematic rules for that refinement are in Figure 7. Figure 7 Kinetics and kinematics when including the pulley Copyright J.E. Akin. All rights reserved. Page 5 of 9

6 The relative importance of including or ignoring a pulley depends on its mass and radius. For the first example output, in Figure 8, the pulley mass is a small percentage of the smaller block mass. Its radius is also taken to be small. There, the tensions change by less than 1%, compared to the neglected pulley in Figure 5. Figure 9 shows that letting the pulley mass be about one third of the smallest block mass causes a larger effect. The tension ratios change by about 20% and the linear accelerations reduce. Figure 8 Results for a relatively small pulley show small tension and acceleration changes Returning to the original text example in Figures 2 and 5, note that they were for an instant in time and that the resulting acceleration is constant because the forces are constant. For constant acceleration, you have seen that it is easy to compute the later position and velocity. Figure 10 shows additional rules, appended to Figure 2, for that purpose. The associated output for the time and velocity at a later position, to supplement Figure 5, is also given. In Chapter 15, a simpler process, to obtain the same velocity (2.07 m/sec), solves this problem again. Copyright J.E. Akin. All rights reserved. Page 6 of 9

7 Figure 9 A large inertia pulley substantially changes the tensions and accelerations Figure 10 Extending Figure 2 to obtain time, position and velocity Copyright J.E. Akin. All rights reserved. Page 7 of 9

8 In Example 14.2, the kinematic constraint from the cable occurs in its simplest form. Generally, the effect of an inextensible (constant length) cable is slightly more involved and introduces a linear constrain on the velocities and accelerations. In practice, the constant length of a cable is made up of an effective length, dependent on the location of moving masses, and other constant length sections such a partial arcs around a pulley. Referring to Figure 11, the cable system on the left has an effective length of 2. Taking time derivatives of the cable length yields linear constraints on the velocity and acceleration of 0 2 and 0 2, where the positive direction is downward. Likewise, for the cable on the right side the constraints are 2 2., , and , where again positive values are downward. Figure 11 Constant length cables form kinematic constraints Example 14.5 Train on inclined circular horizontal track The train (Figure 12) is moving with constant speed, so there is no tangential acceleration. The engine thrust just balances the air resistance and rolling friction (neither is given). Your goal is to find the necessary track forces acting on the car in the front view, for the specified angle of incline. The rules are in Figure 13, while the outputs are in Figure 14. Note that the sideways force on the railcar, and its passengers, is about 17% of the car weight. That would be an undesirable feeling for the passengers, so the problem was re arranged to specify no sideways force on the car. The new result in Figure 15 shows that the desired angle of elevation (for the given speed) is about 31 degrees. Figure 12 Top and front views of a train on an inclined horizontal circular track Copyright J.E. Akin. All rights reserved. Page 8 of 9

9 Figure 13 Train kinetics and kinematics for constant velocity circular motion Figure 14 Force results for specified incline angle Figure 15 Incline angle for no sideways force on a passenger Copyright J.E. Akin. All rights reserved. Page 9 of 9