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Seventh Edition CHAPTER 12 VECTOR MECHANICS FOR ENGINEERS: DYNAMICS Ferdinand P. Beer E. Russell Johnston, Jr. Kinetics of Particles: Newton’s Second Law Lecture Notes: J. Walt Oler Texas Tech University © 2003 The McGraw-Hill Companies, Inc. All rights reserved. Seventh Edition Vector Mechanics for Engineers: Dynamics Introduction • Newton’s first and third laws are sufficient for the study of bodies at rest (statics) or bodies in motion with no acceleration. • When a body accelerates (changes in velocity magnitude or direction), Newton’s second law is required to relate the motion of the body to the forces acting on it. • Newton’s second law: - A particle will have an acceleration proportional to the magnitude of the resultant force acting on it and in the direction of the resultant force. - The resultant of the forces acting on a particle is equal to the rate of change of linear momentum of the particle. - The sum of the moments about O of the forces acting on a particle is equal to the rate of change of angular momentum of the particle about O. © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 12 - 2 1 Seventh Edition Vector Mechanics for Engineers: Dynamics Newton’s Second Law of Motion • Newton’s second law (acceleration ): • If the resultant force acting on a particle is not zero, the particle will have an acceleration proportional to the magnitude of resultant and in the direction of the resultant. r • When a particle of mass m is acted upon by a force F , the acceleration of the particle must satisfy r r F = ma • Acceleration must be evaluated with respect to a Newtonian frame of reference, i.e., one that is not accelerating or rotating. • If force acting on particle is zero, particle will not accelerate, i.e., it will remain stationary or continue on a straight line at constant velocity. © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 12 - 3 Seventh Edition Vector Mechanics for Engineers: Dynamics Dynamic Equilibrium • Alternate expression of Newton’s second law, r r ∑ F − ma = 0 r − ma ≡ inertial vector • With the inclusion of the inertial vector, the system of forces acting on the particle is equivalent to zero. The particle is in dynamic equilibrium. • Methods developed for particles in static equilibrium may be applied, e.g., coplanar forces may be represented with a closed vector polygon. • Inertia vectors are often called inertial forces as they measure the resistance that particles offer to changes in motion, i.e., changes in speed or direction. • Inertial forces may be conceptually useful but are not like the contact and gravitational forces found in statics. © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 12 - 4 2 Seventh Edition Vector Mechanics for Engineers: Dynamics Linear Momentum of a Particle • Newton’s second law (linear momentum ): - The resultant of the forces acting on a particle is equal to the rate of change of linear momentum of the particle. • Replacing the acceleration by the derivative of the velocity yields r r dv ∑F = m dt r d r dL = (m v ) = dt dt r L = linear momentum of the particle • Linear Momentum Conservation Principle: If the resultant force on a particle is zero, the linear momentum of the particle remains constant in both magnitude and direction. © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 12 - 5 Seventh Edition Vector Mechanics for Engineers: Dynamics Equations of Motion • Newton’s second law provides r r ∑ F = ma • Solution for particle motion is facilitated by resolving vector equation into scalar component equations, e.g., for rectangular components, r r r r r r ∑ (Fx i + Fy j + Fz k ) = m(a x i + a y j + a z k ) ∑ Fx = ma x ∑ Fy = ma y ∑ Fz = ma z ∑ Fx = m&x& ∑ Fy = m&y& ∑ Fz = m&z& • For tangential and normal components, ∑ F t = mat ∑Ft = m © 2003 The McGraw-Hill Companies, Inc. All rights reserved. dv dt ∑ F n = man ∑Fn = m v2 ρ 12 - 6 3 Seventh Edition Vector Mechanics for Engineers: Dynamics Equations of Motion • For radial and transverse components, ∑ Fr = mar = m(&r& − rθ& 2 ) ∑ Fθ = maθ = m(rθ&& + 2r&θ& ) © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 12 - 7 Seventh Edition Vector Mechanics for Engineers: Dynamics Sample Problem 12.1 SOLUTION: • Resolve the equation of motion for the block into two rectangular component equations. A 200-lb block rests on a horizontal plane. Find the magnitude of the force P required to give the block an acceleration or 10 ft/s2 to the right. The coefficient of kinetic friction between the block and plane is µk = 0.25. • Unknowns consist of the applied force P and the normal reaction N from the plane. The two equations may be solved for these unknowns. © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 12 - 8 4 Seventh Edition Vector Mechanics for Engineers: Dynamics Sample Problem 12.1 SOLUTION: • Resolve the equation of motion for the block into two rectangular component equations. ∑ Fx = ma : ( O )( P cos 30° − 0.25 N = 6.21 lb ⋅ s 2 ft 10 ft s 2 y ) = 62.1lb ∑ Fy = 0 : x m= W 200 lb = g 32.2 ft s 2 lb ⋅ s 2 ft F = µk N = 6.21 = 0.25 N N − P sin 30° − 200 lb = 0 • Unknowns consist of the applied force P and the normal reaction N from the plane. The two equations may be solved for these unknowns. N = P sin 30° + 200 lb P cos 30° − 0.25( P sin 30° + 200 lb ) = 62.1 lb P = 151lb © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 12 - 9 Seventh Edition Vector Mechanics for Engineers: Dynamics Sample Problem 12.3 SOLUTION: • Write the kinematic relationships for the dependent motions and accelerations of the blocks. • Write the equations of motion for the blocks and pulley. • Combine the kinematic relationships with the equations of motion to solve for the accelerations and cord tension. The two blocks shown start from rest. The horizontal plane and the pulley are frictionless, and the pulley is assumed to be of negligible mass. Determine the acceleration of each block and the tension in the cord. © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 12 - 10 5 Seventh Edition Vector Mechanics for Engineers: Dynamics Sample Problem 12.3 O x y SOLUTION: • Write the kinematic relationships for the dependent motions and accelerations of the blocks. y B = 12 x A a B = 12 a A • Write equations of motion for blocks and pulley. ∑ Fx = m A a A : T1 = (100 kg )a A ∑ Fy = m B a B : m B g − T2 = m B a B (300 kg )(9.81 m s 2 )− T2 = (300 kg )a B T2 = 2940 N - (300 kg )a B ∑ Fy = mC aC = 0 : T2 − 2T1 = 0 © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 12 - 11 Seventh Edition Vector Mechanics for Engineers: Dynamics Sample Problem 12.3 • Combine kinematic relationships with equations of motion to solve for accelerations and cord tension. O x y y B = 12 x A a B = 12 a A T1 = (100 kg )a A T2 = 2940 N - (300 kg )a B ( = 2940 N - (300 kg ) 12 a A ) T2 − 2T1 = 0 2940 N − (150 kg )a A − 2(100 kg )a A = 0 a A = 8.40 m s 2 a B = 12 a A = 4.20 m s 2 T1 = (100 kg )a A = 840 N T2 = 2T1 = 1680 N © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 12 - 12 6 Seventh Edition Vector Mechanics for Engineers: Dynamics Sample Problem 12.4 SOLUTION: • The block is constrained to slide down the wedge. Therefore, their motions are dependent. Express the acceleration of block as the acceleration of wedge plus the acceleration of the block relative to the wedge. The 12-lb block B starts from rest and slides on the 30-lb wedge A, which is supported by a horizontal surface. • Write the equations of motion for the wedge and block. • Solve for the accelerations. Neglecting friction, determine (a) the acceleration of the wedge, and (b) the acceleration of the block relative to the wedge. © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 12 - 13 Seventh Edition Vector Mechanics for Engineers: Dynamics Sample Problem 12.4 SOLUTION: • The block is constrained to slide down the wedge. Therefore, their motions are dependent. r r r aB = a A + aB A • Write equations of motion for wedge and block. ∑ Fx = m A a A : N1 sin 30° = m A a A y 0.5 N1 = (W A g )a A x ∑ Fx = mB a x = mB (a A cos 30° − a B A ) : − WB sin 30° = (WB g )(a A cos 30° − a B aB A A ) = a A cos 30° + g sin 30° ∑ Fy = mB a y = mB (− a A sin 30°) : N1 − WB cos 30° = −(WB g )a A sin 30° © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 12 - 14 7 Seventh Edition Vector Mechanics for Engineers: Dynamics Sample Problem 12.4 • Solve for the accelerations. 0.5 N1 = (W A g )a A N1 − WB cos 30° = −(WB g )a A sin 30° 2(W A g )a A − WB cos 30° = −(WB g )a A sin 30° aA = aA = gWB cos 30° 2W A + WB sin 30° (32.2 ft s 2 )(12 lb)cos 30° 2(30 lb ) + (12 lb ) sin 30° a A = 5.07 ft s 2 aB A = a A cos 30° + g sin 30° aB A = 5.07 ft s 2 cos 30° + 32.2 ft s 2 sin 30° ( ) ( aB © 2003 The McGraw-Hill Companies, Inc. All rights reserved. ) A = 20.5 ft s 2 12 - 15 Seventh Edition Vector Mechanics for Engineers: Dynamics Sample Problem 12.5 SOLUTION: • Resolve the equation of motion for the bob into tangential and normal components. • Solve the component equations for the normal and tangential accelerations. The bob of a 2-m pendulum describes an arc of a circle in a vertical plane. If the tension in the cord is 2.5 times the weight of the bob for the position shown, find the velocity and acceleration of the bob in that position. • Solve for the velocity in terms of the normal acceleration. © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 12 - 16 8 Seventh Edition Vector Mechanics for Engineers: Dynamics Sample Problem 12.5 SOLUTION: • Resolve the equation of motion for the bob into tangential and normal components. • Solve the component equations for the normal and tangential accelerations. mg sin 30° = mat ∑ Ft = mat : at = g sin 30° ∑ Fn = man : at = 4.9 m s 2 2.5mg − mg cos 30° = man an = g (2.5 − cos 30°) an = 16.03 m s 2 • Solve for velocity in terms of normal acceleration. an = v2 ρ v = ρan = (2 m )(16.03 m s 2 ) v = ±5.66 m s © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 12 - 17 Seventh Edition Vector Mechanics for Engineers: Dynamics Sample Problem 12.6 SOLUTION: • The car travels in a horizontal circular path with a normal component of acceleration directed toward the center of the path.The forces acting on the car are its weight and a normal reaction from the road surface. Determine the rated speed of a highway curve of radius ρ = 400 ft banked through an angle θ = 18o. The rated speed of a banked highway curve is the speed at which a car should travel if no lateral friction force is to be exerted at its wheels. © 2003 The McGraw-Hill Companies, Inc. All rights reserved. • Resolve the equation of motion for the car into vertical and normal components. • Solve for the vehicle speed. 12 - 18 9 Seventh Edition Vector Mechanics for Engineers: Dynamics Sample Problem 12.6 • Resolve the equation of motion for the car into vertical and normal components. R cosθ − W = 0 ∑ Fy = 0 : W R= cosθ W ∑ Fn = man : R sin θ = an g SOLUTION: • The car travels in a horizontal circular path with a normal component of acceleration directed toward the center of the path.The forces acting on the car are its weight and a normal reaction from the road surface. W W v2 sin θ = g ρ cosθ • Solve for the vehicle speed. v 2 = gρ tan θ ( ) = 32.2 ft s 2 (400 ft ) tan 18° v = 64.7 ft s = 44.1 mi h © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 12 - 19 Seventh Edition Vector Mechanics for Engineers: Dynamics Angular Momentum of a Particle r r r • H O = r × mV = moment of momentum or the angular momentum of the particle about O. r r r • H O is perpendicular to plane containing r and mV r r r H O = rmV sin φ i j k r = rm vθ HO = x y z 2& mv x mv y mv z = mr θ • Derivative of angular momentum with respect to time, r r r r r r r r r H& O = r& × mV + r × mV& = V × mV + r × ma r = r×∑F r = ∑ MO © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 12 - 20 10 Seventh Edition Vector Mechanics for Engineers: Dynamics Angular Momentum of a Particle • Newton’s second law (angular momentum ): r& r r HO = r × ∑ F = ∑ M O • It follows from Newton’s second law that the sum of the moments about O of the forces acting on the particle is equal to the rate of change of the angular momentum of the particle about O. © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 12 - 21 Seventh Edition Vector Mechanics for Engineers: Dynamics Conservation of Angular Momentum • When only force acting on particle is directed toward or away from a fixed point O, the particle is said to be moving under a central force. • Since the line ofr actionr of the central force passes through O, ∑ M O = H& O = 0 and r r r r × mV = H O = constant • Position vector and motion r of particle are in a plane perpendicular to H O . • Magnitude of angular momentum, H O = rm V sin φ = constant = r0 m V0 sin φ 0 or H O = mr 2θ& = constant HO angular momentum = r 2θ& = h = unit mass m © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 12 - 22 11 Seventh Edition Vector Mechanics for Engineers: Dynamics Conservation of Angular Momentum • Radius vector OP sweeps infinitesimal area dA = 12 r 2 dθ • Define dA 1 2 dθ 1 2 & = r = r θ = areal velocity dt 2 dt 2 • Recall, for a body moving under a central force, h = r 2θ& = constant • When a particle moves under a central force, its areal velocity is constant. © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 12 - 23 Seventh Edition Vector Mechanics for Engineers: Dynamics Eqs of Motion in Radial & Transverse Components • Consider particle at r and θ, in polar coordinates, ∑ Fr = mar = m(&r& − rθ& 2 ) ∑ Fθ = maθ = m(rθ&& + 2r&θ& ) • This result may also be derived from conservation of angular momentum, H O = mr 2θ& ( ( ) d mr 2θ& dt = m r 2θ&& + 2rr&θ& r ∑ Fθ = ∑ Fθ = m(rθ&& + 2r&θ& ) © 2003 The McGraw-Hill Companies, Inc. All rights reserved. ) 12 - 24 12 Seventh Edition Vector Mechanics for Engineers: Dynamics Sample Problem 12.7 SOLUTION: • Write the radial and transverse equations of motion for the block. • Integrate the radial equation to find an expression for the radial velocity. • Substitute known information into the transverse equation to find an expression for the force on the block. A block B of mass m can slide freely on a frictionless arm OA which rotates in a horizontal plane at a constant rate θ&0 . Knowing that B is released at a distance r0 from O, express as a function of r a) the component vr of the velocity of B along OA, and b) the magnitude of the horizontal force exerted on B by the arm OA. © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 12 - 25 Seventh Edition Vector Mechanics for Engineers: Dynamics Sample Problem 12.7 SOLUTION: • Write the radial and transverse equations of motion for the block. ∑ Fr = m ar : 0 = m(&r& − rθ ) ∑ F θ = m aθ : F = m(rθ&& + 2r&θ& ) • Integrate the radial equation to find an expression for the radial velocity. dv dv dr dv = vr r &r& = v&r = r = r dr dt dr dt dv dvr dvr dr = = vr r &r& = v&r = dr dt dr dt 2 2 & & vr dvr = rθ dr = rθ 0 dr vr r 0 r0 2 ∫ vr dvr = θ&0 ∫ r dr &2 ( vr2 = θ 02 r 2 − r02 ) • Substitute known information into the transverse equation to find an expression for the force on the block. ( ) 12 F = 2mθ 02 r 2 − r02 © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 12 - 26 13 Seventh Edition Vector Mechanics for Engineers: Dynamics Sample Problem 12.8 SOLUTION: • Since the satellite is moving under a central force, its angular momentum is constant. Equate the angular momentum at A and B and solve for the velocity at B. A satellite is launched in a direction parallel to the surface of the earth with a velocity of 18820 mi/h from an altitude of 240 mi. Determine the velocity of the satellite as it reaches it maximum altitude of 2340 mi. The radius of the earth is 3960 mi. © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 12 - 27 Seventh Edition Vector Mechanics for Engineers: Dynamics Sample Problem 12.8 SOLUTION: • Since the satellite is moving under a central force, its angular momentum is constant. Equate the angular momentum at A and B and solve for the velocity at B. rm v sin φ = H O = constant rA m v A = rB m v B r vB = v A A rB = (18820 mi h ) (3960 + 240)mi (3960 + 2340 )mi v B = 12550 mi h © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 12 - 28 14 Seventh Edition Vector Mechanics for Engineers: Dynamics %Quiz 1 clear all % clear all variables and breakpoints t = [0:0.01:1.2]; theta = t.^3+4*t; r = t.^3+2*t.^2; x = r.*cos(theta); y = r.*sin(theta); v_r = 3*t.^2+4*t; v_theta = (t.^3+2.*t.^2).*(3.*t.^2+4); v = sqrt(v_r.^2+v_theta.^2); a_r = 6*t+4-(t.^3+2.*t.^2).*(3.*t.^2+4).^2; a_theta = (t.^3+2.*t.^2).*(6*t)+2.*(3.*t.^2+4).*(3*t.^2+4*t) ; a = sqrt(a_r.^2+a_theta.^2); © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 12 - 29 Seventh Edition Vector Mechanics for Engineers: Dynamics figure(1) plot(x,y); xlabel('x (in.)') ylabel('y (in.)') legend('Trajectory',2) grid on figure(2) plot(t,v_r,t,v_theta,t,v) xlabel('t (sec)') ylabel('v_x, v_t, v (in./sec)') legend('v_r','v_t','v',2) grid on figure(3) plot(t,a_r,t,a_theta,t,a) xlabel('t (sec)') ylabel('a_r, a_t, a (in./sec^2)') legend('a_r','a_t','a',2) grid on © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 12 - 30 15 Seventh Edition Vector Mechanics for Engineers: Dynamics © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 12 - 31 Seventh Edition Vector Mechanics for Engineers: Dynamics © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 12 - 32 16 Seventh Edition Vector Mechanics for Engineers: Dynamics © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 12 - 33 Seventh Edition Vector Mechanics for Engineers: Dynamics © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 12 - 34 17 Seventh Edition Vector Mechanics for Engineers: Dynamics % Problem 11.C5 % syms t theta = pi*t; r = 1.75/(1-0.75*cos(pi*t)); x = r.*cos(theta); y = r.*sin(theta); r_dot = diff(r,t); r_ddot = diff(r_dot,t); t_dot = diff(theta,t); t_ddot = diff(t_dot,t); v_r = r_dot; v_theta = r*t_dot; v = sqrt(v_r.^2+v_theta.^2); a_r = r_ddot-r*(t_dot)^2; a_theta = r*t_ddot+2*r_dot*t_dot; a = sqrt(a_r.^2+a_theta.^2); © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 12 - 35 Seventh Edition Vector Mechanics for Engineers: Dynamics t = [0:.01:2]; v_r = subs(v_r,t); v_theta = subs(v_theta,t); v = subs(v,t); a_r = subs(a_r,t); a_theta = subs(a_theta,t); a = subs(a,t); x = subs(x,t); y = subs(y,t); © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 12 - 36 18 Seventh Edition Vector Mechanics for Engineers: Dynamics figure(1) plot(x,y); xlabel('x (in.)') ylabel('y (in.)') legend('Trajectory',2) grid on %axis([0 10000 -3e5 0]) figure(2) plot(t,v_r,t,v_theta,t,v) xlabel('t (sec)') ylabel('v_x, v_t, v (in./sec)') legend('v_r','v_t','v',4) grid on figure(3) plot(t,a_r,t,a_theta,t,a) xlabel('t (sec)') ylabel('a_r, a_t, a (in./sec^2)') legend('a_r','a_t','a',4) grid on © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 12 - 37 Seventh Edition Vector Mechanics for Engineers: Dynamics © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 12 - 38 19 Seventh Edition Vector Mechanics for Engineers: Dynamics % function problemc2 % fprintf('Mu Theta\n') for mk = 0:0.05:0.4 y0 = [3;0]; g = 9.81; ro = 1.5; tspan = [0 0.275]; % [t,y] = ode45(@f,tspan,y0,[],mk,ro,g); % eq1 = g*cos(y(:,2)) - y(:,1).^2/ro; [x,k] = min(abs(eq1)); theta = y(k,2)*180/pi; fprintf('%-4.2f %4.2f\n',mk,theta); end % -------------------------------------------------------------------------function dydt = f(t,y,mk,ro,g) dydt = [ g*(sin(y(2))-mk*cos(y(2)))+mk*y(1)^2/ro y(1)/ro ]; © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 12 - 39 Seventh Edition Vector Mechanics for Engineers: Dynamics © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 12 - 40 20