System force

The force servo system. Figure 10. Sinusoidal response curve of the force servo system (1 Hz). Figure 10. Sinusoidal response curve of the force servo system (1 Hz). Figure 11. Sinusoidal response curve of the force servo system (5 Hz). Figure 11. Sinusoidal response curve of the force servo system (5 Hz). Figure 12. Sinusoidal response curve of the force servo system (10 Hz). Figure 12. Sinusoidal response curve of the force servo system (10 Hz). Figure 13. System model block diagram with the feedforward correction element added. Figure 13. System model block diagram with the feedforward correction element added. Figure 14. Simulation model of the force servo system. Figure 14. Simulation model of the force servo system. Figure 15. Response curve of the force servo system. Figure 15. Response curve of the force servo system. Figure 16. Step response curve of the force servo system. Figure 16. Step response curve of the force servo system. Figure 17. Sinusoidal response curve of the force servo system (5 Hz). Figure 17. Sinusoidal response curve of the force servo system (5 Hz). Figure 18. Sinusoidal response curve of the force servo system (10 Hz). Figure 18. Sinusoidal response curve of the force servo system (10 Hz). Figure 19. Open-loop model block diagram of the position servo system. Figure 19. Open-loop model block diagram of the position servo system. Figure 20. Simulation model of the position servo system. Figure 20. Simulation model of the position servo system. Figure 21. Response curve of the position servo system. Figure 21. Response curve of the position servo system. Figure 22. Dual-channel EMA performance testing system. Figure 22. Dual-channel EMA performance testing system. Figure 23. Static anti-disturbance performance. (a) Static anti-disturbance performance of the DEMA. (b) Static anti-disturbance performance of the GEMA. Figure 23. Static anti-disturbance performance. (a) Static anti-disturbance performance of the DEMA. (b) Static anti-disturbance performance of the GEMA. Figure 24. Dynamic anti-disturbance performance. (a) Dynamic anti-disturbance performance of the DEMA. (b) Dynamic anti-disturbance performance of the GEMA. Figure 24. Dynamic anti-disturbance performance. (a) Dynamic anti-disturbance performance of the DEMA. (b) Dynamic anti-disturbance performance of the GEMA. Disclaimer/Publisher’s Note: The statements, opinions and data contained in all publications are solely those of the individual author(s) and contributor(s) and not of MDPI and/or the editor(s). MDPI and/or the editor(s) disclaim responsibility for any injury to people or property resulting from any ideas, methods, instructions or products

Value of 166 N; this is the upward force upon the bottom box. Determine the acceleration of the elevator (and boxes) and determine the forces acting between the boxes.Both approaches will be used to solve this problem. The first approach involves the dual combination of a system analysis and an individual object analysis. For the system analysis, the two boxes are considered to be a single system with a mass of 14.0 kg. There are two forces acting upon this system - the force of gravity and the normal force. The free-body diagram is shown at the right. The force of gravity is calculated in the usual manner using 14.0 kg as the mass.Fgrav = m•g = 14.0 kg • 9.8 N/kg = 137.2 NSince there is a vertical acceleration, the vertical forces will not be balanced; the Fgrav is not equal to the Fnorm value. The normal force is provided in the problem statement. This 166-N normal force is the upward force exerted upon the bottom box; it serves as the force on the system since the bottom box is part of the system. The net force is the vector sum of these two forces. SoFnet = 166 N, up + 137.2 N, down = 28.8 N, upThe acceleration can be calculated using Newton's second law:a = Fnet /m = 28.8 N/14.0 kg = 2.0571 m/s2 = ~2.1 m/s2Now that the system analysis has been used to determine the acceleration, an individual object analysis can be performed on either box in order to determine the force acting between them. As in the previous problems, it does not matter which box is chosen; the result will be the same in either case. The top box is used in this analysis since it encounters one less force. The free-body diagram is shown at the right. The force of gravity on the top box is m•g where m = 6.0 kg. The force of gravity is 58.8 N. The upward force is not known but can be calculated if the Fnet = m•a equation is applied to the free-body diagram. Since the acceleration is upward, the Fnet side of the equation would be equal to the force in the direction of the acceleration (Fcontact) minus the force that opposes it (Fgrav). SoFcontact - 58.8 N = (6.0 kg)•(2.0571 m/s2)(Notice that the unrounded value of acceleration is used here; rounding will occur when the final answer is determined.) Solving for Fcontact yields 71.14 N. This figure can be rounded to two significant digits - 71 N. So the dual combination of the system analysis and the individual body analysis leads to an acceleration of 2.1 m/s2 and a contact force of 71 N.Now

Individual object analysis generates an equation with an unknown. The result is a system of two equations with two unknowns. The system of equations is solved in order to determine the unknown values. As a first example of the two approaches to solving two-body problems, consider the following example problem. Example Problem 1:A 5.0-kg and a 10.0-kg box are touching each other. A 45.0-N horizontal force is applied to the 5.0-kg box in order to accelerate both boxes across the floor. Ignore friction forces and determine the acceleration of the boxes and the force acting between the boxes.The first approach to this problem involves the dual combination of a system analysis and an individual object analysis. As mentioned, the system analysis is used to determine the acceleration and the individual object analysis is used to determine the forces acting between the objects. In the system analysis, the two objects are considered to be a single object. The dividing line that separates the objects is ignored. The mass of the system of two objects is 15.0 kg. The free-body diagram for the system is shown at the right. There are three forces acting upon the system - the gravity force (the Earth pulls down on the 15.0 kg of mass), the normal force (the floor pushes up on the system to support its weight), and the applied force (the hand is pushing on the back part of the system). The force acting between the 5.0-kg box and the 10.0-kg box is not considered in the system analysis since it is an internal force. Just as the forces holding atoms together within an object are not included in a free-body diagram, so the forces holding together the parts of a system are ignored. These are considered internal forces; only external forces are considered when drawing free-body diagrams. The magnitude of the force of gravity is m•g or 147 N. The magnitude of the normal force is also 147 N since it must support the weight (147 N) of the system. The applied force is stated to be 45.0 N. Newton's second law (a = Fnet/m) can be used to determine the acceleration. Using 45.0 N for Fnet and 15.0 kg for m, the acceleration is 3.0 m/s2.Now that the acceleration has been determined, an individual object analysis can be performed on either object in order to determine the force acting between them. It does not matter which object is chosen; the result will be the same in either case. Here the individual object analysis is conducted on the 10.0 kg object (only because there is one less force acting on it). The free-body diagram for the 10.0-kg object is shown at the

PID parameters selected for the force loop are 10 rpm/m, 3 rpm/(m·s), and 0 (rpm·s)/m, respectively.A load force of 5 kN was applied to the GEMA system, and under the conditions of a step command of 100 mm, a sinusoidal command of ±50 mm at 1 Hz, and a sinusoidal command of ±25 mm at 2 Hz, a performance simulation analysis was conducted on the GEMA system with an impact force of 5 kN and a random disturbance force within the range of ±1 kN added. The response curve is shown in Figure 21.(1)Under the condition of no disturbance force, the rise time of the 100 mm step response was 0.3 s, and the system exhibited a smooth output curve without steady-state error; for the response to the sinusoidal command of ±50 mm at 1 Hz, the output curve was smooth, with a tracking error of 4.2 mm (8.4%); for the response to the sinusoidal command of ±25 mm at 2 Hz, the output curve was smooth, with a tracking error of 3.8 mm (15.2%).(2)Under the condition of a random disturbance force within the range of ±1 kN, the accuracy and dynamic performance of the step response and sinusoidal response were almost unaffected, and the system exhibited a smooth output curve with almost no jitter.(3)Under the condition of an impact force of 5 kN, the output curves of the step response and sinusoidal response fluctuated slightly, and subsequently, the system immediately recovered to a dynamic steady state.These simulation results indicated that under multiple external disturbance forces, the motion accuracy and dynamic performance of the GEMA position servo system introduced with a PID controller were almost unaffected. Nevertheless, the tracking accuracy of the GEMA system decreased as the frequency of the command signal increased. 6. Building of the Test Bed and Analysis of the Test DataTo further test the dynamic performance of the EMA, verify the anti-disturbance control strategy for the single-channel EMA, and lay the foundation for subsequent research on dual-channel synchronous control strategies, a dual-channel EMA performance testing system was built for this study, which consists of an electronic control unit (ECU) and a testing system.As shown in Figure 22, the testing system comprises a GEMA, a DEMA, a force sensor, an angle sensor, joint bearings, and mounting components. The ECU consists of an upper computer, a controller, a driver, and an electric system. Specifically, the upper computer is

Ir .Xs(i), xs Discuss your answers to section I with a tutorial instructor before continuing. Tutorials in Introductory Physics McDermott, Shaffer, & P.E.G., U. Wash. ©Prentice Hall, Inc. First Edition, 2002 Newton's second and third laws Mech 33 II. Applying Newton's laws to interacting objects: varying speed Suppose the bricks were pushed by the hand with the same force as in section I; however, the coefficient of kinetic friction between the bricks and the table is less than that in section I. A. Describe the motions of systems A and B. How does the motion compare to that in part I? Hand pushes with A same force as in section I ~7777] 77;77i 7. '(_Coefficient of friction less than in section I B. Compare the net force (magnitude and direction) on system A to that on system B. Explain. C. Draw and label separate free-body diagrams for systems A and B. Free-body diagram for system A Free-body diagram for system B D. Consider the following discussion between two students. Student l: "System A and system B are pushed by the eame force ae before, eo they will have the ea me motion ae in eection I." Student 2: "/ disagree. I think that they are speeding up eince friction is Iese. So now system A ie puehing on system B with a greater force than system B ie puehing on syetem A." With which student, if either, do you agree? Explain your reasoning. E. Rank the magnitudes of all the