CEE 271: Applied Mechanics II, Dynamics Lecture 9: Ch.13, Sec.45


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1 1 / 40 CEE 271: Applied Mechanics II, Dynamics Lecture 9: Ch.13, Sec.45 Prof. Albert S. Kim Civil and Environmental Engineering, University of Hawaii at Manoa
2 2 / 40 EQUATIONS OF MOTION:RECTANGULAR COORDINATES Today s objectives: Students will be able to 1 Apply Newton s second law to determine forces and accelerations for particles in rectilinear motion. Inclass activities: Reading Quiz Applications Equations of Motion Using Rectangular (Cartesian) Coordinates Concept Quiz Group Problem Solving Attention Quiz
3 3 / 40 READING QUIZ 1 In dynamics, the friction force acting on a moving object is always (a) in the direction of its motion. (b) a kinetic friction. (c) a static friction. (d) zero. ANS: (b) 2 If a particle is connected to a spring, the elastic spring force is expressed by F = ks. The s in this equation is the (a) spring constant. (b) undeformed length of the spring. (c) difference between deformed and undeformed lengths. (d) deformed length of the spring. ANS: (c)
4 APPLICATIONS If a man is trying to move a 100 lb crate, how large a force F must he exert to start moving the crate? What factors influence how large this force must be to start moving the crate? If the crate starts moving, is there acceleration present? What would you have to know before you could find these answers? 4 / 40
5 5 / 40 APPLICATIONS(continued) Objects that move in air (or other fluid) have a drag force acting on them. This drag force is a function of velocity. If the dragster is traveling with a known velocity and the magnitude of the opposing drag force at any instant is given as a function of velocity, can we determine the time and distance required for dragster to come to a stop if its engine is shut off? How?
6 6 / 40 RECTANGULAR COORDINATES (Section 13.4) The equation of motion, F = ma, is best used when the problem requires finding forces (especially forces perpendicular to the path), accelerations, velocities, or mass. Remember, unbalanced forces cause acceleration! Three scalar equations can be written from this vector equation. The equation of motion, being a vector equation, may be expressed in terms of its three components in the Cartesian (rectangular) coordinate system as ΣF = ma or (ΣF x ) i + (ΣF y ) j + (ΣF z ) k = m(a x i + a y j + a z k) or, as scalar equations, ΣF x = ma x, ΣF y = ma y, ΣF z = ma z
7 7 / 40 PROCEDURE FOR ANALYSIS Free Body Diagram Establish your coordinate system and draw the particle s free body diagram showing only external forces. These external forces usually include the weight, normal forces, friction forces, and applied forces. Show the ma vector (sometimes called the inertial force) on a separate diagram. Make sure any friction forces act opposite to the direction of motion! If the particle is connected to an elastic linear spring, a spring force equal to ks should be included on the FBD.
8 8 / 40 PROCEDURE FOR ANALYSIS (continued) Equation of Motion If the forces can be resolved directly from the freebody diagram (often the case in 2D problems), use the scalar form of the equation of motion. In more complex cases (usually 3D), a Cartesian vector is written for every force and a vector analysis is often best. A Cartesian vector formulation of the second law is ΣF = ma or ΣF x i + ΣF y j + ΣF z k = m(a x i + a y j + a z k) Three scalar equations can be written from this vector equation. You may only need two equations if the motion is in 2D.
9 9 / 40 PROCEDURE FOR ANALYSIS (continued) Kinematics The second law only provides solutions for forces and accelerations. If velocity or position have to be found, kinematics equations are used once the acceleration is found from the equation of motion. Any of the kinematics tools learned in Chapter 12 may be needed to solve a problem. Make sure you use consistent positive coordinate directions as used in the equation of motion part of the problem!
10 10 / 40 EXAMPLE Given: The 200lb mine car is hoisted up the incline. The motor M pulls in the cable with an acceleration of 4 ft/s 2. Find: The acceleration of the mine car and the tension in the cable. Plan: 1 Draw the freebody and kinetic diagrams of the car. 2 Using a dependent motion equation, determine an acceleration relationship between cable and mine car. 3 Apply the equation of motion to determine the cable tension.
11 11 / 40 Solution EXAMPLE (solution) 1. Draw the freebody and kinetic diagrams of the mine car: Since the motion is up the incline, rotate the x y axes. Motion occurs only in the xdirection. We are also neglecting any friction in the wheel bearings, etc., on the cart.
12 12 / The cable equation results in s p + 2s c = l t Taking the derivative twice yields a p + 2a c = 0 (1) The relative acceleration is a p = a c + a p/c. As the motor is mounted on the car, a p/c = 4 ft/s 2. So, a p = a c + 4 ft/s 2 (2) Solving equations (1) and (2), yields a C = ft/s 2
13 3. Apply the equation of motion in the xdirection: m = 200/32.2 = slug (+ ) F x = ma x 3T mg(sin 30 ) = ma x 3T (200)(sin 30 ) = 6.211(1.333) T = 36.1 lb 13 / 40
14 14 / 40 CHECK YOUR UNDERSTANDING QUIZ 1. If the cable has a tension of 3 N, determine the acceleration of block B. (a) 4.26 m/s 2 (b) 4.26 m/s 2 (c) 8.31 m/s 2 (d) 8.31 m/s 2 ANS: (d) 2. Determine the acceleration of the block. (a) 2.20 m/s 2 (b) 3.17 m/s 2 (c) 11.0 m/s 2 (d) 4.26 m/s 2 ANS: (a)
15 15 / 40 GROUP PROBLEM SOLVING Given: W A = 10 lb, W B = 20 lb, v oa = 2 ft/s( ), and µ k = 0.2 Find: v A when A has moved 4 feet to the right. Plan: Since both forces and velocity are involved, this problem requires both the equation of motion and kinematics. First, draw free body diagrams of A and B. Apply the equation of motion to each. Using dependent motion equations, derive a relationship between a A and a B and use with the equation of motion formulas.
16 GROUP PROBLEM SOLVING (Solution) Freebody and kinetic diagrams of B: Apply the equation of motion to B: + ΣF y = ma y W B 2T = m B a B 20 2T = a B (3) Freebody and kinetic diagrams of A: Apply the equations of motion to A: (+ ) ΣF y = ma y = 0 (+ ) ΣF x = ma x N = W A = 10 lb f = µ k N = 2 lb f T = m A a A 2 T = a A (4) 16 / 40
17 17 / 40 Now consider the kinematics. Constraint equation: Therefore s A + 2s B = constant v A + 2v B = 0 a A + 2a B = 0 a A = 2a B (5) (Notice a A is considered positive to the left and a B is positive downward.)
18 18 / 40 Now combine equations (3), (4), and (5) Constraint equation: T = 22 = 7.33 lb 3 a A = ft/s 2 = ft/s 2 ( ) Now use the kinematic equation: (v A ) 2 = (v 0A ) 2 + 2a A (s A s 0A ) (v A ) 2 = (2) 2 + 2( 17.16)( 4) v A = 11.9 ft/s( )
19 19 / 40 ATTENTION QUIZ 1. Determine the tension in the cable when the 400 kg box is moving upward with a 4 m/s 2 acceleration. (a) 2265 N (b) 3365 N (c) 5524 N (d) 6543 N ANS: (c) T 60 a = 4 m/s
20 20 / 40 ATTENTION QUIZ 2. A 10 lb particle has forces of F 1 = (3i + 5j) lb and F 2 = ( 7i + 9j) lb acting on it. Determine the acceleration of the particle. (a) ( 0.4i + 1.4j) ft/s 2 (b) ( 0.4i + 14j) ft/s 2 (c) ( 12.9i + 45j) ft/s 2 (d) (13i + 4j) ft/s 2 ANS: (c)
21 21 / 40 EQUATIONS OF MOTION: NORMAL AND TANGENTIAL COORDINATES Today s objectives: Students will be able to 1 Apply the equation of motion using normal and tangential coordinates. Inclass activities: Reading Quiz Applications Equation of Motion in n t Coordinates Concept Quiz Group Problem Solving Attention Quiz
22 22 / 40 READING QUIZ 1 The normal component of the equation of motion is written as ΣF n = ma n, where ΣF n is referred to as the? (a) impulse (b) centripetal force (c) tangential force (d) inertia force ANS: (b) 2 The positive n direction of the normal and tangential coordinates is. (a) normal to the tangential component (b) always directed toward the center of curvature (c) normal to the binormal component (d) All of the above. ANS: (d)
23 23 / 40 APPLICATIONS Race tracks are often banked in the turns to reduce the frictional forces required to keep the cars from sliding up to the outer rail at high speeds. If the car s maximum velocity and a minimum coefficient of friction between the tires and track are specified, how can we determine the minimum banking angle (θ) required to prevent the car from sliding up the track?
24 APPLICATIONS(continued) The picture shows a ride at the amusement park. The hydraulicallypowered arms turn at a constant rate, which creates a centrifugal force on the riders. We need to determine the smallest angular velocity of the cars A and B so that the passengers do not loose contact with the seat. What parameters do we need for this calculation? 24 / 40
25 25 / 40 APPLICATIONS(continued) Satellites are held in orbit around the earth by using the earth s gravitational pull as the centripetal force  the force acting to change the direction of the satellite s velocity. Knowing the radius of orbit of the satellite, we need to determine the required speed of the satellite to maintain this orbit. What equation governs this situation?
26 26 / 40 NORMAL & TANGENTIAL COORDINATES (Sec. 13.5) When a particle moves along a curved path, it may be more convenient to write the equation of motion in terms of normal and tangential coordinates. The normal direction (n) always points toward the path s center of curvature. In a circle, the center of curvature is the center of the circle. The tangential direction (t) is tangent to the path, usually set as positive in the direction of motion of the particle.
27 27 / 40 EQUATIONS OF MOTION Since the equation of motion is a vector equation, ΣF = ma, it may be written in terms of the n and t coordinates as ΣF t u t + ΣF n u n + ΣF b u b = ma t + ma n F t and F n are the sums of the force components acting in the t and n directions, respectively. This vector equation will be satisfied provided the individual components on each side of the equation are equal, resulting in the two scalar equations: Ft = ma t and Fn = ma n Since there is no motion in the binormal (b) direction, we can also write F b = 0.
28 28 / 40 NORMAL AND TANGENTIAL ACCERLERATIONS The tangential acceleration, a t = dv/dt, represents the time rate of change in the magnitude of the velocity. Depending on the direction of ΣF t, the particle s speed will either be increasing or decreasing. The normal acceleration, a n = v 2 /ρ, represents the time rate of change in the direction of the velocity vector. Remember, a n always acts toward the path s center of curvature. Thus, ΣF n will always be directed toward the center of the path. Recall, if the path of motion is defined as y = f(x), the radius of curvature at any point can be obtained from ρ = dy [1 + ( d2 y dx 2 dx )2 ] 3/2
29 29 / 40 SOLVING PROBLEMS WITH n t COORDINATES Use n t coordinates when a particle is moving along a known, curved path. Establish the n t coordinate system on the particle. Draw freebody and kinetic diagrams of the particle. The normal acceleration (a n ) always acts inward (the positive ndirection). The tangential acceleration (a t ) may act in either the positive or negative t direction. Apply the equations of motion in scalar form and solve. It may be necessary to employ the kinematic relations: a t = dv dt = v dv ds a n = v2 ρ (6) (7)
30 EXAMPLE Given: At the instant θ = 45, the boy with a mass of 75 kg, moves a speed of 6 m/s, which is increasing at 0.5 m/s 2. Neglect his size and the mass of the seat and cords. The seat is pin connected to the frame BC. Find: Horizontal and vertical reactions of the seat on the boy. Plan: 1 Since the problem involves a curved path and requires finding the force perpendicular to the path, use n t coordinates. 2 Draw the boy s freebody and kinetic diagrams. 3 Apply the equation of motion in the n t directions. 30 / 40
31 31 / 40 EXAMPLE (Solution) 1. The n t coordinate system can be established on the boy at angle θ = 45. Approximating the boy and seat together as a particle, the freebody and kinetic diagrams can be drawn.
32 EXAMPLE (continued) .//01234#35,6.,+# 75"/89#35,6.,+#! '#! +,! # $ % # ()*# +, " # 2. Apply the equations of motion in the n t directions. a. ma n = R x cos θ R y sin θ + W sin θ Using a n = v 2 /ρ = 6 2 /10, W = 75(9.81) N, and m = 75 kg, we get: R x cos 45 R y sin = (75)(6 2 /10) b. ma t = R x sin θ + R y cos θ W cos θ we get: R x sin 45 + R y cos = (75)(0.5) Using above two equations, solve for R x, R y. R x = 217 N and R y = 572 N. "# $ & # "# 32 / 40
33 CONCEPT QUIZ &"!"#"$"%" 1. A 10 kg sack slides down a smooth surface. If the normal force on the surface at the flat spot, A, is 98.1 N ( ), the radius of curvature is (a) 0.2 m (b) 0.4 m (c) 1.0 m (d) None of the above ANS: (d) 2. A 20 lb block is moving along a smooth surface. If the normal force on the surface at A is 10 lb, the velocity is. (a) 7.6 ft/s (b) 9.6 ft/s (c) 10.6 ft/s (d) 12.6 ft/s ANS: (c) 33 / 40
34 GROUP PROBLEM SOLVING Given: A 800 kg car is traveling over the hill having the shape of a parabola. When it is at point A, it is traveling at 9 m/s and increasing its speed at 3 m/s 2. Find: The resultant normal force and resultant frictional force exerted on the road at point A. Plan: 1 Treat the car as a particle. 2 Draw the freebody and kinetic diagrams. 3 Apply the equations of motion in the n t directions. 4 Use calculus to determine the slope and radius of curvature of the path at point A. 34 / 40
35 1. The n t coordinate system can be established on the car at point A. Treat the car as a particle and draw the freebody and kinetic diagrams: W = mg = weight of car N = resultant normal force on road F = resultant friction force on road 35 / 40
36 36 / Apply the equations of motion in the n t directions: W cos θ N = ma n Using W = mg and a n = v 2 /ρ = (9) 2 /ρ (800)(9.81) cos θ N = (800)(81/ρ) N = 7848 cos θ 64800/ρ (8) W sin θ F = ma t Using W = mg and a t = 3m/s 2 (given) (800)(9.81) sin θ F = (800)(3) F = 7848 sin θ 2400 (9)
37 37 / Determine ρ by differentiating y = f(x) at x = 80 m: y = 20(1 x 2 /6400) dy/dx = ( 40)x/6400 d 2 y/dx 2 = ( 40)/640 (10) dy [1 + ( dx ρ x=80 = )2 ] 3/2 [1 + ( 0.5) 2 ] 3/2 d2 y = = meter dx 2 Determine θ from the slope of the curve at A: tan θ = dy/dx x=80 = 40x 6400 = 0.5 x=80 θ = tan 1 ( dy dx ) = tan 1 ( 0.5) = 26.6
38 38 / 40 From Eq. (8): N = 7848 cos θ 64800/ρ = 7848 cos(26.6 ) 64800/223.6 = 6728 N (11) From Eq.(9) : F = 7848 sin θ 2400 = 7848 sin(26.6 o ) 2400 = 1114 N (12)
39 39 / 40 ATTENTION QUIZ 1 The tangential acceleration of an object (a) represents the rate of change of the velocity vector s direction. (b) represents the rate of change in the magnitude of the velocity. (c) is a function of the radius of curvature. (d) Both (b) and (c). ANS: (b)
40 40 / 40 ATTENTION QUIZ 1 The block has a mass of 20 kg and a speed of v = 30 m/s at the instant it is at its lowest point. Determine the tension in the cord at this instant. (a) 1596 N (b) 1796 N (c) 1996 N (d) 2196 N #$"%"!" &"'"($"%)*" ANS: (c)
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