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20/04/1432 CHAPTER 13 : KINEMATICS OF A PARTICLE ‐ FORCE AND ACCELERATION Dr Mohamed AICHOUNI This presentation is an adapted version from : R. C. Hibbeler, « Engineering Mechanics – Dynamics », 12th Ed. 13.1 - Newton’s Law of Motion First Law: A particle originally at rest rest, or moving in a straight line with a constant velocity, will remain in this state provided the particle is not subjected to an unbalanced force. Second Law: A particle acted upon by an unbalanced force F experiences p an acceleration a that has the same direction as the force and a magnitude that is directly proportional to the force. 1 20/04/1432 Newton’s Law of Motion (13.1) Third Law: The mutual forces of action and reaction between two particles are equal, opposite and collinear. Equation of motion: F = ma Newton’s Law of Gravitational Attraction. A law governing the mutual attractive gravitational force acting between them. F =G m1m2 r2 G : universal constant of gravitation Newton’s Law of Motion (13.1) Mass and Weight. Mass is a property of matter by which we can compare the response of one body with that of another. It is an absolute quantity since the measurement can be made at any location. Weight of a body is not absolute since it is measured in a gravitational field, hence its magnitude depends p on the location measured. W = mg 2 20/04/1432 Newton’s Law of Motion (13.1) SI System of Units. The mass of the body is specified in kilograms (kg) and the weight must be calculated using the equation of motion, F = ma W = mg g ((N)) (g = 9.81 m/s2) 13.2 - The Equation of Motion The equation of motion may be written as ∑ F = ma Consider particle P of mass m and subjected to the action of two forces, F1 and F2. 3 20/04/1432 The Equation of Motion (13.2) From the free body diagram, the resultant of these forces produces the vector ma it magnitude and direction can be represented graphically on the kinetic ma, diagram. ∑ F = ma Note that if FR = ΣF = 0, then acceleration is zero, so that the particle will either remain at rest or moves along a straight line with a constant velocity. Such a condition is called static equilibrium, Newton’s First Law of Motion 13.3 - Equation of Motion for a System of Particles Consider a system of n particles isolated within an enclosed region in space. • Arbitrary ith particle having a mass of mi • is subjected to a system of internal forces which resultant force is represented by fi and a resultant external force Fi. 4 20/04/1432 Equation of Motion for a System of Particles (13.3) The free body diagram for the ith particle are shown. Applying equation of • motion yields Fi + fi = miai ΣF = ma; If equation of motion is applied to each of the other • particles, these equations can be added together vectorially, y, ΣFi + Σ fi = Σmiai Equation of Motion for a System of Particles • Since internal forces between particles all occur in equal but opposite collinear pairs the summation of these internal forces will equal zero pairs, zero. ΣFi = Σmiai • If rG is a position vector which locates the center of mass G of the particles, then mrG = Σmiri where m = Σmi is the total mass of all the particles • Differentiating twice w.r.t time yields • Therefore, maG = Σm Σ iai ΣF = maG • The sum of the external forces acting on the system of particles is equal to the total mass of the particles times the acceleration of its center of mass G. 5 20/04/1432 13.4 - Equation of Motion: Rectangular Coordinates • When a particle is moving relative to an inertial x, y, z frame of reference, the forces acting on the particle particle, and its acceleration may be expressed in term of their i, j, k components ΣF = ma ΣFxi + ΣFyj + ΣFzk = m(axi + ayj + azk) • We may write the following three scalar equations: ∑ Fx = max ∑ Fy = ma y ∑ Fz = maz Example 13.1 The 50-kg crate rests on a horizontal plane for which the coefficient of kinetic friction is μk = 0.3. If the crate is subjected to a 400-N towing force, determine the velocity of the crate in 3 s starting from rest. 6 20/04/1432 Example 13.1 - Solution Free-Body Diagram. The weight of the crate is W = mg = 50 (9.81) = 490.5 N. The frictional force has a magnitude F = μkNC and acts to the left, since it opposes the motion of the crate. The acceleration a is assumed to act horizontally, in the positive x direction. There are 2 unknowns, namely NC and a. Example 13.1 Equations of Motion. + → ∑ Fx = max ; 400 cos 30o − 0.3 NC = 50a + ↑ ∑ Fy = ma y ; NC − 490.5 + 400 sin 30o = 0 Solving for the two equations yields NC = 290.5 N a = 5.19m / s 2 7 20/04/1432 Example 13.1 Kinematics. Acceleration is constant, since the applied force P is constant. Initial velocity is zero, the velocity of the crate in 3 s is v = v0 + act = 0 + 5.19(3) = 15.6m / s → 13.5 - Equations of Motion: Normal and Tangential Coordinates When a particle moves over a curved path which is known, the equation of motion for the particle may be written in the tangential tangential, normal and binormal directions. We have ΣF = ma ΣFtut + ΣFnun + ΣFbub = mat +man Here ΣFt, ΣFn, ΣFb represent the sums of all the force components acting on the particle in the tangential, normal and binormal directions. 8 20/04/1432 Equations of Motion: Normal and Tangential Coordinates Since the particle is constrained to move along the path, there is no motion in the binormal direction ∑ Ft = mat ∑ Fn = man ∑ Fb = 0 • at (=dv/dt) represents the time rate of change in the magnitude of velocity • Therefore if ΣFt acts in the direction of motion, the particle’s speed will increase. If it acts in the opposite direction, the particle will slow down. • an (=v2/ρ) represents the time rate of change in the velocity’s direction. Equations of Motion: Normal and Tangential Coordinates Since this vector always acts in the positive n direction, i.e. toward the path’s • center of curvature curvature, then ΣFn, which causes an, also act in this direction. direction 9 20/04/1432 Example 13.6 Determine the banking θ for the race track so that the wheels of the racing cars will not have to depend upon friction to prevent any car from sliding up or down the track. Assume the cars have negligible size a mass m, and travel around the curve of radius ρ with a speed v. Example 13.6 Free-Body Diagram. There is no frictional force acting on the car car. Here NC represents the resultant of the ground on all four wheels. Since an can be calculated, the unknown are NC and θ. 10 20/04/1432 Example 13.6 Equations of Motion. + Using the n, b axes shown, v2 → ∑ Fn = man ; NC sin θ = m +↑ ∑ Fb = 0; NC cosθ − mg = 0 ρ Solving g the 2 equations, q eliminating g NC and m, v2 tan θ = gρ −1⎛⎜ v 2 ⎞⎟ θ = tan ⎜ ⎟ ⎝ gρ ⎠ 13.6 - Equations of Motion: Cylindrical Coordinates When all forces acting on a particle are resolved into cylindrical components components, ii.e. e along the unit unit-vector vector directions ur, uθ, uz, the equation of motion may be expressed as: ΣF = ma ΣFrur + ΣFθuθ + ΣFzuz = marur +maθuθ+mazuz 11 20/04/1432 Equations of Motion: Cylindrical Coordinates We may write the following three scalar equations of motion: ∑ Fr = mar ∑ Fθ = maθ ∑ Fz = maz Equations of Motion: Cylindrical Coordinates Tangential and Normal Forces. • Determination of the resultant force components ΣFr, ΣFθ, ΣFz causing a particle to move with a known acceleration. • If acceleration is not specified at given instant, directions or magnitudes of the forces acting on the particle must be known or computed to solve. • Consider the force P that causes the particle to move along a path r = f(θ) 12 20/04/1432 Equations of Motion: Cylindrical Coordinates • The normal force N which the path exerts on the particle is always perpendicular to the tangent of the path path. • Frictional force F always acts along the tangent in the opposite direction of motion. Equations of Motion: Cylindrical Coordinates • The directions of N and F can be specified relative to the radial coordinate by using the angle ψ, ψ which is defined between the extended radial line and the tangent to the curve. tanψ = 13 r dr / dθ 20/04/1432 Equations of Motion: Cylindrical Coordinates • If ψ is positive, it is measured from the extended radial line to the tangent in a CCW sense or in the positive direction θ • If it is negative, it is measured in the opposite direction to positive θ Example 13.10 The 2-kg block moves on a smooth horizontal track such that its path is specfied in polar coordinates by the parametric equations r = (3t2) m and θ = (0.5t) (0 5t) rad where t is in seconds. Determine the magnitude of the tangential force F causing the motion at the instant t = 1 s. 14 20/04/1432 Example 13.10 Free-Body Diagram. ψ The normal force N, and the tangential force F are located at an angle from the r and θ axes. By expressing r = f(θ), we yield r = 12θ2. When t = 1 s, θ = 0.5 rad. Example 13.10 tanψ = r 12θ 2 = dr / dθ 12(2θ ) = 0.25 θ = 0.5rad ψ = 14.04o ψ Because is a p positive q quantity, y it is measured counterclockwise from the r axis to the tangent (same direction as θ). There are four unknowns: F, N, ar and aθ 15 20/04/1432 Example 13.10 Equations of Motion. + ↓ ∑ Fr = mar ; F cos14.04o − N sin 14.04o = 2ar ↑ + ∑ Fθ = maθ ; F sin 14.04o + N cos14.04o = 2aθ Kinematics. r = 3t 2 t =1s = 3m θ = 0.5t t =1s = 0.5rad r& t =1s = 6m / s, &r& t =1s = 6m / s 2 , θ& = 0.5rad / s, θ&& = 0 Example 13.10 ar = &r& − rθ& 2 = 6 − 3(0.5) 2 = 5.25m / s 2 aθ = rθ&& + 2r&θ& = 3(0) + 2(6)(0.5) = 6m / s 2 Substituting into the two equations of motion and solving, F = 13.10 13 10 N N = 9.22 N 16 20/04/1432 Example 13.11 A smooth 2-kg cylinder C has a peg P through its center which passes through the slot in arm arm. If the arm rotates in the vertical plane at a constant rate , determine the force that the arm exerts on the peg at the instant θ = 60° θ& = 0.5rad / s Example 13.11 Free-Body Diagram. The force on the peg, Fp, acts perpendicular to the slot in the arm. arm ar and aθ are assumed to act in the directions of positive r and θ respectively. 17 20/04/1432 Example 13.11 Equations of Motion. + + ∑ Fr = mar ; ∑ Fθ = maθ ; Kinematics. 19.62 sin θ − NC sin θ = 2ar (1) 19.62 cosθ + FP − NC cosθ = 2aθ (2) From the FBD, r can be related to θ by the equation, r= 0.4 = 0.4 cscθ sin θ Example 13.11 θ& = 0.5 θ&& = 0 r = 0.4 cscθ r& = −0.4(cscθ cot θ )θ& = −0.2 cscθ cot θ &r& = −0.2( − cscθ cot θ )(θ&) cot θ − 0.2 cscθ (− csc2 θ )θ& = 0.1cscθ (cot 2 θ + csc2 θ ) 18 20/04/1432 Example 13.11 Evaluating the formulas at θ = 60°, we get, θ& = 0.5 θ&& = 0 r = 0.462 r& = −0.133 &r& = 0.192 ar = &r& − rθ& 2 = 0.0770 aθ = rθ&& + 2r&θ& = −0.133 Substituting these results into Eqs. 1 and 2 with θ = 60°, NC = 19.4 N 19 FP = -0.356 N