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.

S F = ma . S F = L NEWTON’S SECOND LAW Denoting by m the mass of a particle, by S 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 S F = ma Introducing the linear momentum of a particle, L = mv, Newton’s second law can also be written as . S 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.

S Fx = max S Fy = may S Fz = maz To solve a problem involving the motion of a particle, S F = ma should be replaced by equations containing scalar quantities. Using rectangular components of F and a, we have P ax az S Fx = max S Fy = may S Fz = maz x z Using tangential and normal components, y an dv dt S Ft = mat = m at P v2 r S Fn = man = m O x Using radial and transverse components, aq . .. ar S Fr = mar= m(r - rq2) r P .. . q . S Fq = maq = m(rq + 2rq) x O

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.

Using this result, we can arrive at an expression for radial force purely in terms of the radial distance 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? baseball where h Re when earth where

Example Problem

Free Body Diagram for Block B

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

Newton’s 1st law provides us with an expression that the force balance equals zero. (i.e.) 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 These 4 equations have 4 unknowns listed below, and the solution can be solved using linear algebra

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

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.

Case 3. Account for everything 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. 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.

Given: 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:

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… , .

Note: The answer shown in the book assumes gravity is zero 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?

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 ms = 0.15 and mk = 0.10. Knowing that q = 30o 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 q A B

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 ms = 0.15 and mk = 0.10. Knowing that q = 30o 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 q 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.

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 ms = 0.15 and mk = 0.10. Knowing that q = 30o 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 q 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 = mk N. The friction force on a surface when motion is impending is F = ms N.

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 ms = 0.15 and mk = 0.10. Knowing that q = 30o 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 q A B 4. Apply Newton’s second law: The relationship between the forces acting on the particle, its mass and acceleration is given by S 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.

aA = aB Problem 1 Solution Kinematics. 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. P q A B q A B aA aB aA = aB

= = WA = mA g WA = (30 kg)(9.81 m/s2) WA = 294.3 N WB = mB g Problem 1 Solution P q A B Kinetics; draw a free body diagram. Block A : mA a = 30 a N T 250 N WA= 294.3 N Fk = mk N = WA = mA g WA = (30 kg)(9.81 m/s2) WA = 294.3 N WB = mB g WB = (15 kg)(9.81 m/s2) WB = 147.15 N Block B : WB= 147.15 N T N N’ Fk = mk N F’k = mk N’ mB a = 15 a =

Problem 1 Solution Block A : WA= 294.3 N T mA a = 30 a 30o = 250 N N Fk = mk N Apply Newton’s second law. + SFy = 0: N - (294.3) cos 30o = 0 N = 254.87 N Fk = mk N = 0.10 (254.9) = 25.49 N + SFx = ma: 250 + (294.3) sin 30o - 25.49 - T = 30 a 371.66 - T = 30 a (1)

= + SFy = 0: N’ - N - (147.15) cos 30o = 0 N’ = 382.31 N Problem 1 Solution Block B : WB= 147.15 N T N N’ Fk = mk N F’k = mk N’ = 30o mB a = 15 a + SFy = 0: N’ - N - (147.15) cos 30o = 0 N’ = 382.31 N F’k = mk N’ = 0.10 (382.31) = 38.23 N + SFx = ma: T - Fk - F’k - (147.15) sin 30o = 15 a T - 137.29 = 15 a (2) Solving equations (1) and (2) gives: T = 215 N a = 5.21 m/s2

Problem 1 Solution P q A B Verify assumption of motion. Check: We should verify that blocks actually move by determining the value of the force P for which motion is impending Find P for impending motion. For impending motion both blocks are in equilibrium: Block A Block B N T P WA= 294.3 N Fm = ms N 30o WB= 147.15 N N’ F’m = ms N’

N = 254.87 N and N’ = 382.31 N, F’m = ms N’ = 0.15 (382.31) = 57.35 N Problem 1 Solution N T P WA= 294.3 N Fm = ms N 30o WB= 147.15 N N’ F’m = ms N’ A B From + SFy = 0 find again N = 254.87 N and N’ = 382.31 N, and thus Fm = ms N = 0.15 (254.87) = 38.23 N F’m = ms N’ = 0.15 (382.31) = 57.35 N

+ SFx = 0: P + (294.3) sin 30o - 38.23 - T = 0 (3) Problem 1 Solution N T P WA= 294.3 N Fm = ms N 30o WB= 147.15 N N’ F’m = ms N’ A B For block A: + SFx = 0: P + (294.3) sin 30o - 38.23 - T = 0 (3) For block B: + SFx = 0: T - 38.23 - 57.35 - (147.15) sin 30o = 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.

Problem 2 12 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. B 30 lb A 30o

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. 30o 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.

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. 30o 3. Apply Newton’s second law: The relationship between the forces acting on the particle, its mass and acceleration is given by S 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.

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

= = Problem 2 Solution 12 lb Kinetics; draw a free body diagram. B Block B : B 12 lb N1 mB aB/A = aB /A 12 32.2 mA aA = aA = A 30 lb N1 N2 mA aA = aA 30 32.2 Block A : =

= + SFx = ma: (N1 + 30) sin 30o = aA (1) Problem 2 Solution Block A : Apply Newton’s second law. 30o N1 A A = x N2 x 30 32.2 30 lb mA aA = aA + SFx = ma: (N1 + 30) sin 30o = aA (1) 30 32.2

= + SFy = may : 12 - N1 = aA sin 30o (2) Block B : Problem 2 Solution 12 32.2 mB aB/A = aB /A 12 lb = B 30o B 12 32.2 mA aA = aA N1 + SFy = may : 12 - N1 = aA sin 30o (2) + SFx = max: 0 = aB /A - aA cos 30o (3) 12 32.2 Solving equations (1), (2), and (3) gives: aA = 20.5 ft/s2 30o aB /A = 17.75 ft/s2