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8 - 1 Kinetics of Particles So far, we have only studied the kinematics of particle motion, meaning we have studied the relationships between position,

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Presentation on theme: "8 - 1 Kinetics of Particles So far, we have only studied the kinematics of particle motion, meaning we have studied the relationships between position,"— Presentation transcript:

1 8 - 1 Kinetics of Particles So far, we have only studied the kinematics of particle motion, meaning we have studied the relationships between position, velocity, and acceleration, with only cursory reference to the forces which produced the motion. In some sense, the kinematics is the study of dynamics from a purely experimental standpoint. That is, if we are able to map the positions, velocities, and accelerations of multiple particles, then we can ascertain the force which produced the motion. Note: By the word “particle”, it is implied that the object is contained entirely within a point. In the next part of this course, we will be interested in the forward model. That is, if we know the forces on the particle, what should the resulting motion be? The reason I call this a forward model is because the classification of different forces is purely theoretical and can only be backed up by experimental measurements.

2 8 - 2 NEWTON’S SECOND LAW Denoting by m the mass of a particle, by  F the sum, or resultant, of the forces acting on the particle, and by a the acceleration of the particle relative to a newtonian frame of reference, we write  F = ma Introducing the linear momentum of a particle, L = mv, Newton’s second law can also be written as  F = L. which expresses that the resultant of the forces acting on a particle is equal to the rate of change of the linear momentum of the particle.

3 8 - 3 To solve a problem involving the motion of a particle,  F = ma should be replaced by equations containing scalar quantities. Using rectangular components of F and a, we have  F x = ma x  F y = ma y  F z = ma z x y P anan O atat x y z axax ayay azaz P x P aa O arar Using tangential and normal components,  F t = ma t = m dv dt v2v2 Using radial and transverse components,.....  F r = ma r = m(r - r  2 )  r  F n = ma n = m  F  = ma  = m(r  + 2r  )

4 8 - 4 Central Forces In many forces, such as gravitational, electrical forces, etc., the magnitude of the force only depends on the distance away from the source (i.e. ), and the direction is always towards or away from the source. Using these assumptions, we can derive an expression for the rotation frequency: The result implies that under the action of a central force, the rotation frequency depends only on the radial distance away from the attractor.

5 8 - 5 The above represents a nonlinear differential equation, to which the solution will give us r as a function of time. We will return to this later in the course. One example that we have discussed previously is the flat earth approximation. When is it valid? Using this result, we can arrive at an expression for radial force purely in terms of the radial distance baseball earth ReRe h where when where

6 8 - 6 Example Problem

7 8 - 7 Free Body Diagram for Block B

8 8 - 8 Free Body Diagram for Block A Force on Block A due to substrate

9 8 - 9 What we would like to do is solve for the forces if Block B is on the verge of slipping, but has not yet done so. This allows us to develop a relationship between the accelerations of Blocks A and B Newton’s 1 st law provides us with an expression that the force balance equals zero. (i.e.) These 4 equations have 4 unknowns listed below, and the solution can be solved using linear algebra

10 x y y’ x’ Since we’re only interested in Block B, let’s try a change in coordinates Thus, the critical slipping threshold can be expressed as:

11 Case 1. Assume that the acceleration applied to the block is zero Therefore, the Block will slide under the action of gravity if the angle exceeds θ. Case 2. Neglect Gravity Note: This is the result derived in the book.

12 Case 3. Account for everything Note: We still haven’t solved for the acceleration explicitly. Once the block starts to slip, we have to take into account the relative motion of Blocks A and B. So what is the effect of applying acceleration to the block? You can think of it in two ways. One is that the acceleration delays the action of gravity. Depending on the friction coefficient, the block may slide downwards when the applied force is removed. Alternatively, one can say that the acceleration keeps the block from slipping at higher angles.

13 8 - 13

14 Constraints: Motion of Slider A is confined to the X-axis Motion of Slider B is confined to the Y-Axis Distance between Sliders A and B is constant Free Body Diagrams: Given:

15 Additional Constraints In these two equations, the only thing we don’t know are the accelerations of sliders A and B, however they are interdependent Thus, we really have 2 equations with 2 unknowns…,.

16 Note: The answer shown in the book assumes gravity is zero. What are the correct tensions and accelerations if gravity is taken into account? What also must we do if friction between the sliders and wall are taken into account?

17 P  Problem 1 Block A has a mass of 30 kg and block B has a mass of 15 kg. The coefficients of friction between all plane surfaces of contact are  s = 0.15 and  k = Knowing that  = 30 o and that the magnitude of the force P applied to block A is 250 N, determine (a) the acceleration of block A, (b) the tension in the cord. A B

18 Problem 1 P  Block A has a mass of 30 kg and block B has a mass of 15 kg. The coefficients of friction between all plane surfaces of contact are  s = 0.15 and  k = Knowing that  = 30 o and that the magnitude of the force P applied to block A is 250 N, determine (a) the acceleration of block A, (b) the tension in the cord. A B 1. Kinematics: Examine the acceleration of the particles. 2. Kinetics: Draw a free body diagram showing the applied forces and an equivalent force diagram showing the vector ma or its components.

19 Problem 1 P  Block A has a mass of 30 kg and block B has a mass of 15 kg. The coefficients of friction between all plane surfaces of contact are  s = 0.15 and  k = Knowing that  = 30 o and that the magnitude of the force P applied to block A is 250 N, determine (a) the acceleration of block A, (b) the tension in the cord. A B 3. When a problem involves dry friction: It is necessary first to assume a possible motion and then to check the validity of the assumption. The friction force on a moving surface is F =  k N. The friction force on a surface when motion is impending is F =  s N.

20 Problem 1 Block A has a mass of 30 kg and block B has a mass of 15 kg. The coefficients of friction between all plane surfaces of contact are  s = 0.15 and  k = Knowing that  = 30 o and that the magnitude of the force P applied to block A is 250 N, determine (a) the acceleration of block A, (b) the tension in the cord. P  A B 4. Apply Newton’s second law: The relationship between the forces acting on the particle, its mass and acceleration is given by  F = m a. The vectors F and a can be expressed in terms of either their rectangular components or their tangential and normal components. Absolute acceleration (measured with respect to a newtonian frame of reference) should be used.

21 Problem 1 Solution Kinematics. P  A B Assume motion with block A moving down. If block A moves and accelerates down the slope, block B moves up the slope with the same acceleration.  A B aAaA aBaB a A = a B

22 Problem 1 Solution P  A B Kinetics; draw a free body diagram. Block A : Block B : m A a = 30 a N T 250 N W A = N F k =  k N = W B = N T N N’ F k =  k N F’ k =  k N’ m B a = 15 a = W A = m A g W A = (30 kg)(9.81 m/s 2 ) W A = N W B = m B g W B = (15 kg)(9.81 m/s 2 ) W B = N

23 Problem 1 Solution Apply Newton’s second law. m A a = 30 a = +  F y = 0: N - (294.3) cos 30 o = 0 N = N F k =  k N = 0.10 (254.9) = N +  F x = ma: (294.3) sin 30 o T = 30 a T = 30 a (1) Block A : N T 250 N W A = N F k =  k N 30 o

24 Problem 1 Solution Block B : m B a = 15 a +  F y = 0: N’ - N - (147.15) cos 30 o = 0 N’ = N F’ k =  k N’ = 0.10 (382.31) = N +  F x = ma: T - F k - F’ k - (147.15) sin 30 o = 15 a T = 15 a (2) Solving equations (1) and (2) gives: T = 215 N a = 5.21 m/s 2 W B = N T N N’ F k =  k N F’ k =  k N’ = 30 o

25 Problem 1 Solution P  A B Check: We should verify that blocks actually move by determining the value of the force P for which motion is impending Verify assumption of motion. Find P for impending motion. For impending motion both blocks are in equilibrium: N T P W A = N F m =  s N 30 o W B = N T N N’ F m =  s N F’ m =  s N’ 30 o Block B Block A

26 Problem 1 Solution N T P W A = N F m =  s N 30 o W B = N T N N’ F m =  s N F’ m =  s N’ 30 o A B From +  F y = 0 find again N = N and N’ = N, and thus F m =  s N = 0.15 (254.87) = N F’ m =  s N’ = 0.15 (382.31) = N

27 Problem 1 Solution N T P W A = N F m =  s N 30 o W B = N T N N’ F m =  s N F’ m =  s N’ 30 o A B For block A: +  F x = 0: P + (294.3) sin 30 o T = 0 (3) For block B: +  F x = 0: T (147.15) sin 30 o = 0 (4) Solving equations (3) and (4) gives P = 60.2 N. Since the actual value of P (250 N) is larger than the value for impending motion (60.2 N), motion takes place as assumed.

28 Problem 2 A B 12 lb 30 lb A 12-lb block B rests as shown on the upper surface of a 30-lb wedge A. Neglecting friction, determine immediately after the system is released from rest (a) the acceleration of B relative to A, (b) the acceleration of B relative to A. 30 o

29 Problem 2 A B 12 lb 30 lb 1. Kinematics: Examine the acceleration of the particles. 2. Kinetics: Draw a free body diagram showing the applied forces and an equivalent force diagram showing the vector ma or its components. 30 o A 12-lb block B rests as shown on the upper surface of a 30-lb wedge A. Neglecting friction, determine immediately after the system is released from rest (a) the acceleration of B relative to A, (b) the acceleration of B relative to A.

30 Problem 2 A B 12 lb 30 lb 3. Apply Newton’s second law: The relationship between the forces acting on the particle, its mass and acceleration is given by  F = m a. The vectors F and a can be expressed in terms of either their rectangular components or their tangential and normal components. Absolute acceleration (measured with respect to a newtonian frame of reference) should be used. 30 o A 12-lb block B rests as shown on the upper surface of a 30-lb wedge A. Neglecting friction, determine immediately after the system is released from rest (a) the acceleration of B relative to A, (b) the acceleration of B relative to A.

31 Problem 2 Solution Kinematics. A B 12 lb 30 lb 30 o A aAaA B aAaA a B/A Since the wedge is constrained to move on the inclined surface, its acceleration a A is in the direction of the slope. The acceleration a B of the block can be expressed as the sum of the acceleration of A and the acceleration of B relative to A. a B = a A + a B/A

32 Problem 2 Solution A B 12 lb 30 lb 30 o Kinetics; draw a free body diagram. Block B : A 30 lb A N1N1 N2N2 m A a A = a A Block A : B 12 lb B N1N1 m B a B/A = a B /A m A a A = a A = =

33 Apply Newton’s second law. Problem 2 Solution Block A : A 30 lb A N1N1 N2N2 m A a A = a A = 30 o x x +  F x = ma: (N ) sin 30 o = a A (1)

34 Problem 2 Solution B 12 lb N1N1 B m B a B/A = a B /A m A a A = a A = 30 o +  F y = ma y : 12 - N 1 = a A sin 30 o (2) +  F x = ma x : 0 = a B /A - a A cos 30 o (3) Solving equations (1), (2), and (3) gives: a A = 20.5 ft/s 2 30 o a B /A = ft/s 2 Block B :


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