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EML 3004C CHAPTER 9 Statics, Dynamics, and Mechanical Engineering Namas Chandra Introduction to Mechanical engineering Chapter 9-1 EML 3004C Objectives Understand the concept of a Vector Write a vector in component form Understand the concepts of a couple and moment Construct a free-body diagram of a physical system Sections 9.1 9.2 9.3 9.4 9.5 9.6 Introduction The Concept of a Vector Forces, Couples, and Moments Equilibrium and Free-body Diagrams Frictional Forces Motion of a Rigid Body Namas Chandra Introduction to Mechanical engineering Chapter 9-2 EML 3004C 9.1 Introduction Statics deals with bodies (Solid, Liquid, Gases) that are in equilibrium with applied forces. At static equilibrium, a body is at rest or in constant motion (no acceleration). Newton’s Law First: A body is at rest or in uniform motion (constant velocity) along a straight-line unless acted upon by some external (unbalanced) forces. Second: Every body accelerates in the direction of external forces. The acceleration is proportional to the magnitude of the force. Third: For every action, there is an equal and opposite reaction. Statics is critical in the study of bodies under forces in the design. Free body diagram isolates forces/bodies and is a critical tool. Namas Chandra Introduction to Mechanical engineering Chapter 9-3 EML 3004C 9.2 Concept of a Vector While scalar has only magnitude, e.g.: mass, temperature -vector has magnitude and direction, e.g. velocity, force, moment Head Magnitude of the vector Tail The above is a geometric representation of the magnitude of the vector is unity, it is called a unit vector Namas Chandra Introduction to Mechanical engineering Chapter 9-4 EML 3004C 9.2.1Component of Vector F Fx Fy =Fx ˆi+Fˆ y ˆj (2-D) =Fx ˆi+Fy ˆj+Fz kˆ (3-D) ˆi, ˆj, kˆ are unit vectors Magnitude of F F Fx Fy + Fz 2 Namas Chandra Introduction to Mechanical engineering 2 2 Chapter 9-5 EML 3004C 9.2.2Component of Vector If F=22 N is being pulled at 30, then Fx = F sin = 22 sin = 11N Fy = F cos = 22 cos = 11 3N Namas Chandra Introduction to Mechanical engineering Chapter 9-6 EML 3004C 9.2.3Direction Cosines of Vector Let , , be the angle of vector F has with coordinate axis projection of F on x-axis cos = magnitude of F Fx = F Fy Fz cos = ; cos = F F Fx ˆ Fy ˆ F Unit vector c i+ j + z kˆ = cos ˆi+cos ˆj+cos kˆ F F F Note that cos 2 cos 2 cos 2 1 Namas Chandra Introduction to Mechanical engineering Chapter 9-7 EML 3004C 9.2.4Addition/Subtraction of Vectors Since vectors have magnitude and directions only components can be added For example, adding 2iˆ to 3jˆ directly as (2+3) makes no sense So always resolve components and then add Example: F 3iˆ + 5jˆ +8kˆ Fˆ 2iˆ + 4jˆ + 5kˆ 2 Fˆ3 Fˆ1 + Fˆ2 = ˆi + 9jˆ + 13k̂ Fˆ4 Fˆ1 - Fˆ2 = 5iˆ + ˆj + 3 kˆ Namas Chandra Introduction to Mechanical engineering Chapter 9-8 EML 3004C 9.3 Forces, Couples, and Moments Magnitude of a couple: M = Fd Moment of a force M = Force x Lever arm The direction of the moment is normal to the plane containing the force and the arm. Namas Chandra Introduction to Mechanical engineering Chapter 9-9 EML 3004C 9.4 Equilibrium and Free-Body Diagram For static equilibrium, using Newton’s Second Law F 0 and M = 0 Free body diagrams show all the forces and boundary conditions on a given body. Use the following steps Isolate the body from surroundings and draw a simple sketch Pick the right coordinate system (2D, 3D, Cartesian, Cylindrical) Add all extreme forces (right direction) and internal forces Label all known quantities (magnitude/direction) Namas Chandra Introduction to Mechanical engineering Chapter 9-10 EML 3004C 9.5 Frictional forces A body on a surface (horizontal or sloped) is acted upon by weight acting vertically down and a reaction normal to the surface Using Coloumb’s Law for Static case fs s N f s = frictional force s Static coefficient of friction N = Normal force For bodies in motion, fk k N k Kinetic frictional coefficient, usually lower than s s and k do not depend on the force, only on the pair of surfaces Namas Chandra Introduction to Mechanical engineering Chapter 9-11 EML 3004C 9.5.2 Frictional Forces Given 15 Kg crate up 30 degree slope with s 0.3 Find: Force required to push the 15 kg crate up the hill Solution: F 0 M 0 Draw the free body diagram For sliding f = s N = .3 N F m a P-m g sin 30 + f =0 F m a N -m g cos 30 =0 Now M 0 f (0.1)+N (x) = 0; x= 0.03 x x y y 0 Namas Chandra Introduction to Mechanical engineering Chapter 9-12 EML 3004C 9.5.2 Frictional Forces..2 Given 15 Kg crate up 30 degree slope with s 0.3 m N=m g cos 30 = 15 kg 9.81 2 cos 30 = 127.4 N s P=m g sin 30 + f P=111.8 N Namas Chandra Introduction to Mechanical engineering Chapter 9-13 EML 3004C 9.6 Motion of Rigid Body If F 0, then there is unbalanced force leading the acceleration and hence motion Now F m a M I I - moment of inertia (kg m 2 or slug ft 2 ) rad - angular acceleration 2 s Namas Chandra Introduction to Mechanical engineering Chapter 9-14 EML 3004C 9.6.1 Crate Problem Revisited If the friction =0.3 compute the dynamics of motion? f = s N 0.3 N Let s 0.3 F y N - m g cos 0 N=m g cos F x -m g sin + f = m a x f= N = 0.3N = 0.3 (m g cos 30) Thus, a x = -g sin 30 + (0.3) g cos 30 = g (0.3 cos 30-sin 30) m (0.3 cos 30-sin 30) sec 2 m =2.36 2 sec a x = 9.81 Namas Chandra Introduction to Mechanical engineering Chapter 9-15 EML 3004C CHAPTER 9…concludes Statics, Dynamics, and Mechanical Engineering Namas Chandra Introduction to Mechanical engineering Chapter 9-16