An Overview of Mechanics

Mechanics: The study of how bodies react to forces acting on them. Dynamics: 1. Kinematics – concerned with the geometric aspects of motion 2. Kinetics –concerned with the forces causing the motion Statics: The study of bodies in equilibrium. An Overview of Mechanics

The displacementof the particle is defined as its change in position. Vector form:r=r’ - r Scalar form: s = s’ - s RECTILINEAR KINEMATICS: CONTINIOUS MOTION (Section 12.2) A particle travels along a straight-line path defined by the coordinate axis s. The position of the particle at any instant, relative to the origin, O, is defined by the position vector r, or the scalar s. Scalar s can be positive or negative. Typical units for r and s are meters (m) or feet (ft). The total distance traveled by the particle, sT, is a positive scalar that represents the total length of the path over which the particle travels.

The average velocityof a particle during a time interval t is vavg = r/ t The instantaneous velocityis the time-derivative of position.v = dr/ dt Speedis the magnitude of velocity: v = ds / dt VELOCITY Velocityis a measure of the rate of change in the position of a particle. It is a vectorquantity (it has bothmagnitude and direction). The magnitude of the velocity is called speed, with units of m/s or ft/s. Average speedis the total distance traveled divided by elapsed time: (vsp)avg = sT/ t

The instantaneous accelerationis the time derivative of velocity. Vector form: a= dv/ dt Scalar form: a = dv / dt = d2s / dt2 Acceleration can be positive (speed increasing) or negative (speed decreasing). ACCELERATION Accelerationis the rate of change in the velocity of a particle. It is a vectorquantity. Typical units are m/s2. As the text indicates, the derivative equations for velocity and acceleration can be manipulated to get a ds = v dv

Position: Velocity: v t v s s t ò ò ò ò ò ò = = = dv a dt or v dv a ds ds v dt v o v s s o o o o o SUMMARY OF KINEMATIC RELATIONS: RECTILINEAR MOTION • Differentiateposition to get velocity and acceleration. v = ds/dt ; a = dv/dt or a = v dv/ds • Integrateacceleration for velocity and position. •Note that so and vo represent the initial positionand velocity of the particle at t = 0.

CONSTANT ACCELERATION The three kinematic equations can be integrated for the special case when acceleration is constant (a = ac)to obtain very useful equations. A common example of constant acceleration is gravity; i.e., a body freely falling toward earth. In this case, ac = g = 9.81 m/s2 downward. These equations are: v t ò ò = = + dv a dt v v a t yields c o c v o o s t ò ò = = + + (1/2) a ds v dt s s v t t yields o o c 2 s o o v s ò ò = = + v dv a ds v (v ) 2a (s - s ) 2 yields 2 c o c o v s o o

GENERAL CURVILINEAR MOTION (Section 12.4) A particle moving along a curved path undergoes curvilinear motion. Since the motion is often three-dimensional, vectors are used to describe the motion. A particle moves along a curve defined by the path function, s. The position of the particle at any instant is designated by the vector r= r(t). Both the magnitude and direction of r may vary with time. If the particle moves a distance Ds along the curve during time interval Dt, the displacement is determined by vector subtraction: D r = r’ - r

VELOCITY Velocity represents the rate of change in the position of a particle. The average velocity of the particle during the time increment Dt is vavg = Dr/Dt . The instantaneous velocity is the time-derivative of position v = dr/dt . The velocity vector, v, is always tangent to the path of motion. The magnitude of v is called the speed. Since the arc length Ds approaches the magnitude of Dr as t→0, the speed can be obtained by differentiating the path function (v = ds/dt). Note that this is not a vector!

ACCELERATION Acceleration represents the rate of change in the velocity of a particle. If a particle’s velocity changes from v to v’over a time increment Dt, the average acceleration during that increment is: aavg = Dv/Dt = (v - v’)/Dt The instantaneous acceleration is the time-derivative of velocity: a = dv/dt = d2r/dt2 A plot of the locus of points defined by the arrowhead of the velocity vector is called a hodograph. The acceleration vector is tangent to the hodograph, but not, in general, tangent to the path function.

CURVILINEAR MOTION: RECTANGULAR COMPONENTS (Section 12.5) It is often convenient to describe the motion of a particle in terms of its x, y, z or rectangular components, relative to a fixed frame of reference. The position of the particle can be defined at any instant by the position vector r = x i+ y j+ z k . The x, y, z components may all be functions of time, i.e., x = x(t), y = y(t), and z = z(t) . The magnitude of the position vector is: r = (x2 + y2 + z2)0.5 The direction of ris defined by the unit vector: ur = (1/r)r

The velocity vector is the time derivative of the position vector: v= dr/dt = d(xi)/dt + d(yj)/dt + d(zk)/dt Since the unit vectorsi, j, k are constant in magnitude and direction, this equation reduces to v = vxi + vyj + vzk where vx = = dx/dt, vy = = dy/dt, vz = = dz/dt • • • z y x RECTANGULAR COMPONENTS: VELOCITY The magnitude of the velocity vector is v = [(vx)2 + (vy)2 + (vz)2]0.5 The direction of vis tangent to the path of motion.

RECTANGULAR COMPONENTS: ACCELERATION The acceleration vector is the time derivative of the velocity vector (second derivative of the position vector): a= dv/dt = d2r/dt2 = ax i + ay j+ azk where ax = = = dvx/dt, ay = = = dvy/dt, az = = = dvz/dt •• • •• • y x vx vy •• • z vz The magnitude of the acceleration vector is a = [(ax)2 + (ay)2 + (az)2 ]0.5 The direction of a is usually not tangent to the path of the particle.

NORMAL AND TANGENTIAL COMPONENTS (Section 12.7) When a particle moves along a curved path, it is sometimes convenient to describe its motion using coordinates other than Cartesian. When the path of motion is known, normal (n) and tangential (t) coordinates are often used. In the n-t coordinate system, the origin is located on the particle (the origin moves with the particle). The t-axis is tangent to the path (curve) at the instant considered, positive in the direction of the particle’s motion. The n-axis is perpendicular to the t-axis with the positive direction toward the center of curvature of the curve.

NORMAL AND TANGENTIAL COMPONENTS (continued) The positive n and t directions are defined by the unit vectorsun and ut, respectively. The center of curvature, O’, always lies on the concave side of the curve. The radius of curvature, r, is defined as the perpendicular distance from the curve to the center of curvature at that point. The position of the particle at any instant is defined by the distance, s, along the curve from a fixed reference point.

The magnitude is determined by taking the time derivative of the path function, s(t). v = v ut where v = s = ds/dt . VELOCITY IN THE n-t COORDINATE SYSTEM The velocity vector is always tangent to the path of motion (t-direction). Here v defines the magnitude of the velocity (speed) and utdefines the direction of the velocity vector.

Acceleration is the time rate of change of velocity: a= dv/dt = d(vut)/dt = vut + vut . . Here v represents the change in the magnitude of velocity andut represents the rate of change in the direction of ut. . After mathematical manipulation, the acceleration vector can be expressed as: . a = vut + (v2/r) un = at ut + an un. ACCELERATION IN THE n-t COORDINATE SYSTEM .

The tangential component is tangent to the curve and in the direction of increasing or decreasing velocity. • at = v or atds = v dv . ACCELERATION IN THE n-t COORDINATE SYSTEM (continued) So, there are two components to the acceleration vector: a = at ut + an un • The normal or centripetal component is always directed toward the center of curvature of the curve. an = v2/r • The magnitude of the acceleration vector is • a = [(at)2 + (an)2]0.5

1) The particle moves along a straight line. r => an = v2/r = 0 => a = at = v The tangential component represents the time rate of change in the magnitude of the velocity. 2) The particle moves along a curve at constant speed. at = v = 0 => a = an = v2/r . The normal component represents the time rate of change in the direction of the velocity. SPECIAL CASES OF MOTION There are some special cases of motion to consider. .

4) The particle moves along a path expressed as y = f(x). The radius of curvature, r, at any point on the path can be calculated from r= ________________ [ 1 + (dy/dx)2 ]3/2 d2y/dx 2 SPECIAL CASES OF MOTION (continued) 3) The tangential component of acceleration is constant, at = (at)c. In this case, s = so + vot + (1/2) (at)c t2 v = vo + (at)c t v2 = (vo)2 + 2 (at)c (s – so) As before, so and vo are the initial position and velocity of the particle at t = 0. How are these equations related to projectile motion equations? Why?

CYLINDRICAL COMPONENTS (Section 12.8) We can express the location of P in polar coordinates as r = r ur. Note that the radial direction, r, extends outward from the fixed origin, O, and the transverse coordinate, q, is measured counter-clockwise (CCW) from the horizontal.

The instantaneous velocity is defined as: v = dr/dt = d(rur)/dt . dur v = rur+ r dt Using the chain rule: dur/dt = (dur/dq)(dq/dt) We can prove that dur/dq = uθ so dur/dt = quθ Therefore: v = rur + rquθ . . . Thus, the velocity vector has two components: r, called the radial component, and rq called the transverse component. The speed of the particle at any given instant is the sum of the squares of both components or v =(r q)2 + ( r )2 . . . . VELOCITY in POLAR COORDINATES)

The instantaneous acceleration is defined as: a = dv/dt = (d/dt)(rur + rquθ) . . After manipulation, the acceleration can be expressed as a = (r – rq2)ur + (rq + 2rq)uθ . .. . . .. . .. The term (r – rq2)is the radial acceleration or ar. The term (rq + 2rq) is the transverse acceleration or aq. . .. . .. . . .. . The magnitude of acceleration is a = (r – rq2)2 + (rq + 2rq)2 ACCELERATION (POLAR COORDINATES)

. . . Velocity: vP =rur +rquθ + zuz Acceleration: aP =(r – rq2)ur+ (rq+ 2rq)uθ+ zuz . .. . .. . .. CYLINDRICAL COORDINATES If the particle P moves along a space curve, its position can be written as rP = rur + zuz Taking time derivatives and using the chain rule:

NEWTON’S LAWS OF MOTION (Section 13.1) The motion of a particle is governed by Newton’s three laws of motion. First Law: A particle originally at rest, or moving in a straight line at constant velocity, will remain in this state if the resultant force acting on the particle is zero. Second Law: If the resultant force on the particle is not zero, the particle experiences an acceleration in the same direction as the resultant force. This acceleration has a magnitude proportional to the resultant force. Third Law: Mutual forces of action and reaction between two particles are equal, opposite, and collinear.

NEWTON’S LAWS OF MOTION (continued) The first and third laws were used in developing the concepts of statics. Newton’s second law forms the basis of the study of dynamics. Mathematically, Newton’s second law of motion can be written F = ma where F is the resultant unbalanced force acting on the particle, and a is the acceleration of the particle. The positive scalar m is called the mass of the particle. Newton’s second law cannot be used when the particle’s speed approaches the speed of light, or if the size of the particle is extremely small (~ size of an atom).

NEWTON’S LAW OF GRAVITATIONAL ATTRACTION Any two particles or bodies have a mutually attractive gravitational force acting between them. Newton postulated the law governing this gravitational force as F = G(m1m2/r2) where F = force of attraction between the two bodies, G = universal constant of gravitation , m1, m2 = mass of each body, and r = distance between centers of the two bodies. When near the surface of the earth, the only gravitational force having any sizable magnitude is that between the earth and the body. This force is called the weight of the body.

MASS AND WEIGHT It is important to understand the difference between the mass and weight of a body! Mass is an absolute property of a body. It is independent of the gravitational field in which it is measured. The mass provides a measure of the resistance of a body to a change in velocity, as defined by Newton’s second law of motion (m = F/a). The weight of a body is not absolute, since it depends on the gravitational field in which it is measured. Weight is defined as W = mg where g is the acceleration due to gravity.

UNITS: SI SYSTEM VS. FPS SYSTEM SI system:In the SI system of units, mass is a base unit and weight is a derived unit. Typically, mass is specified in kilograms (kg), and weight is calculated from W = mg. If the gravitational acceleration (g) is specified in units of m/s2, then the weight is expressed in newtons (N). On the earth’s surface, g can be taken as g = 9.81 m/s2. W (N) = m (kg) g (m/s2) => N = kg·m/s2

EQUATION OF MOTION (Section 13.2) The motion of a particle is governed by Newton’s second law, relating the unbalanced forces on a particle to its acceleration. If more than one force acts on the particle, the equation of motion can be written F = FR = ma where FR is the resultant force, which is a vector summation of all the forces. To illustrate the equation, consider a particle acted on by two forces. First, draw the particle’s free-body diagram, showing all forces acting on the particle. Next, draw the kinetic diagram, showing the inertial force ma acting in the same direction as the resultant force FR.

INERTIAL FRAME OF REFERENCE This equation of motion is only valid if the acceleration is measured in a Newtonian or inertial frame of reference. What does this mean? For problems concerned with motions at or near the earth’s surface, we typically assume our “inertial frame” to be fixed to the earth. We neglect any acceleration effects from the earth’s rotation. For problems involving satellites or rockets, the inertial frame of reference is often fixed to the stars.

EQUATION OF MOTION FOR A SYSTEM OF PARTICLES (Section 13.3) The equation of motion can be extended to include systems of particles. This includes the motion of solids, liquids, or gas systems. As in statics, there areinternal forces and external forces acting onthe system. What is the difference between them? Using the definitions of m = mi as the total mass of all particles and aG as the acceleration of the center of mass G of the particles, then m aG =mi ai. The text shows the details, but for a system of particles: F = m aGwhere F is the sum of the external forces acting on the entire system.

KEY POINTS 1) Newton’s second law is a “law of nature”-- experimentally proven, not the result of an analytical proof. 2) Mass (property of an object) is a measure of the resistance to a change in velocity of the object. 3) Weight (a force) depends on the local gravitational field. Calculating the weight of an object is an application of F = m a, i.e., W = m g. 4) Unbalanced forces cause the acceleration of objects. This condition is fundamental to all dynamics problems!

PROCEDURE FOR THE APPLICATION OF THE EQUATION OF MOTION 1) Select a convenient inertial coordinate system. Rectangular, normal/tangential, or cylindrical coordinates may be used. 2) Draw a free-body diagram showing all external forces applied to the particle. Resolve forces into their appropriate components. 3) Draw the kinetic diagram, showing the particle’s inertial force, ma. Resolve this vector into its appropriate components. 4) Apply the equations of motion in their scalar component form and solve these equations for the unknowns. 5) It may be necessary to apply the proper kinematic relations to generate additional equations.

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 F = maorFxi + Fyj +Fzk = m(axi+ay j +azk) or, as scalar equations, Fx= max,Fy= may, and Fz= maz. RECTANGULAR COORDINATES (Section 13.4) The equation of motion, F = m a, is best used when the problem requires finding forces (especially forces perpendicular to the path), accelerations, velocities, or mass. Remember, unbalanced forces cause acceleration!

PROCEDURE FOR ANALYSIS • Free-Body Diagram (always critical!!) 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 ‘k s’ should be included on the FBD.

PROCEDURE FOR ANALYSIS (continued) • Equations of Motion If the forces can be resolved directly from the free-body diagram (often the case in 2-D problems), use the scalar form of the equation of motion. In more complex cases (usually 3-D), a Cartesian vector is written for every force and a vector analysis is often best. A Cartesian vector formulation of the second law is F = maor Fxi + Fyj+Fzk= m(ax i+ayj+azk) Three scalar equations can be written from this vector equation. You may only need two equations if the motion is in 2-D.

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!

NORMAL & TANGENTIAL COORDINATES (Section 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.

EQUATIONS OF MOTION Since the equation of motion is a vector equation,F = ma, it may be written in terms of the n & t coordinates as Ftut + Fnun+ Fbub = mat+man Here Ft & Fn are the sums of the force components acting in the t & 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 = matand Fn = man. Since there is no motion in the binormal (b) direction, we can also writeFb = 0.

NORMAL AND TANGENTIAL ACCERLERATIONS The tangential acceleration, at = dv/dt, represents the time rate of change in the magnitude of the velocity. Depending on the direction of Ft, the particle’s speed will either be increasing or decreasing. The normal acceleration, an = v2/r, represents the time rate of change in the direction of the velocity vector. Remember, analways acts toward the path’s center of curvature. Thus, Fnwill always be directed toward the center of the path. Recall, if the path of motion is defined as y = f(x), the radius of curvatureat any point can be obtained from

SOLVING PROBLEMS WITH n-t COORDINATES • Use n-t coordinates when a particle is moving along a known, curved path. • Establish then-t coordinate system on the particle. • Draw free-body and kinetic diagrams of the particle. The normal acceleration (an) always acts “inward” (the positive n-direction). The tangential acceleration (at) 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: • at = dv/dt = v dv/ds an = v2/r

Equilibrium equations or “Equations of Motion” in cylindrical coordinates (using r, q , and z coordinates) may be expressed in scalar form as:  Fr = mar = m (r – r q 2)  Fq= maq= m (r q– 2 r q)  Fz = maz = m z . .. .. . . .. CYLINDRICAL COORDINATES (Section 13.6) This approach to solving problems has some external similarity to the normal & tangential method just studied. However, the path may be more complex or the problem may have other attributes that make it desirable to use cylindrical coordinates.

If the particle is constrained to move only in the r – qplane (i.e., the z coordinate is constant), then only the first two equations are used (as shown below). The coordinate system in such a case becomes a polar coordinate system. In this case, the path is only a function of q.  Fr = mar = m(r – rq2)  Fq= maq= m(rq– 2rq ) . .. .. . . CYLINDRICAL COORDINATES (continued) Note that a fixed coordinate system is used, not a “body-centered” system as used in the n – t approach.

An Overview of Mechanics