ENGR-1100 Introduction to Engineering Analysis Pawel Keblinski Materials Science and Engineering MRC115 Office hours: Tuesday 1-3 phone: (518) 276 6858 email: keblip@rpi.edu

Mechanics -> Static Mechanics of rigid bodies Mechanics of deformable bodies Mechanics of fluids Static Kinematics Kinetics

Lecture outline Newton’s law. Law of gravitation. Basic concept of physical problem solving. Fundamentals of vectors

Newton’s Law of Motion Law #1: Body is in equilibrium SFi=0 Body will remain in rest or continue to move with same speed and direction unless unbalanced force is acted on the body or particle.

Newton’s Law of Motion F=ma Law #2: Body not in equilibrium SFi=0 Change of motion of a body is proportional to the net force imposed on the body in the direction of the net force Where: F is the external force acting on the body. m is the mass of the body. a is the acceleration of the body in the direction of the force. F=ma

Newton’s Law of Motion Law #3: The forces of action and reaction between bodies in contact have the same magnitude, same line of action, and opposite sense.

Law of gravitation M.m F=G F Two bodies of mass M and m are mutually attracted to each other with equal and opposite forces F and –F of magnitude F given by the formula: M.m r2 F=G Where r is the distance between the center of mass of the two bodies. And G is the universal gravitational constant G=3.439(10-8)ft3/(slug*s2) in the U.S customary system of units. G=6.673(10-11)m3/(kg*s2) in SI system of units M m r F

Mass and weight Me.m F=G =mg Me The mass m of a body is an absolute quantity. The weight W of a body is the gravitational attraction exerted on the body by the earth or by another massive body such as another planet. At the surface of the earth: Me.m re2 F=G =mg Where: Me is the mass of the earth. re is the mean radius of the earth Me re2 g=G At sea level and latitude 450 g=32.17 ft/s2 = 9.807 m/s2

Example 1 Determine the gravitational force, in kilonewtons, exerted by the earth on the moon. M m r Me.mm r2 F=G G= 6.673*10-11 m3/(kg*s2) Me=5.976*1024 kg mm=7.35*1022 kg r=3.844*108 m F=1.98*1017 kN

Example 2 At what distance from the surface of the earth, in kilometers, is the weight of the body equal to one-half of its weight on the earth’s surface? W2 Me.m r12 W1=G On earth Me.m r22 W2=G Above earth W1 D r1 earth The requirement: W2=1/2W1 W1/W2= r22 /r12=2 r2=√2 r1 D=r2-r1= (√2-1) r1= (√2-1)*6370 km = 2638km

Class assignment: Exercise set 1-1, 1-2, 1-17. please submit to TA at the end of the lecture 1-1) Calculate the mass m of a body that weighs 600 lb at the surface of the earth. 1-2) Calculate the weight W of a body at the surface of the earth if it has a mass m of 675 kg. 1-17) If a man weight 210 lb at sea level, determine the weight W of the man a) At the top of Mt. Everest (29,028 ft above sea level) b) In a satellite at an altitude of 200 mi. Solution: 1-1) M=18.65 slug 1-2) W=6.62 kN 1-17) a=209 lb b=190.3 lb

Units of measurement The U.S customary system of units (the British gravitational system) Base units are foot (ft) for length, pound (lb) for force, and second (s) for time. Pound is defined as the weight at sea level and altitude of 450 of a platinum standard. The international system of units (SI) Three class of units (1) base units (2) supplementary units. (3)derived units.

Base units Supplementary units Derived units Quantity unit Symbol Length meter m Mass kilogram kg time second s Supplementary units Quantity unit Symbol Plane angle radian rad Solid angle steradian sr Derived units Quantity unit Symbol Area Square meter m2 Volume Cubic meter m3 Linear velocity Meter per second m/s

SI/U.S. customary units conversion Quantity U.S. customary to SI SI to U.S. customary Length 1 ft = 0.3048 m 1 m = 3.281 ft Velocity 1 ft/s = 0.304 m/s 1 m/s = 3.281 ft/s Mass 1 slug = 14.59 kg 1 kg = 0.06854 slug

Dimensional homogeneity Equation does not depend on the units of measurement. F=mg is valid whether the force is measured in newtons or lb and the mass in slugs, kilograms or grams, provided g is measured in the same units of length, time and mass as m and F. The F=9.8m is not dimensional homogeneous since the equation applies only if the length is measured in meters and the time in second.

Method of problem solving Statement of the problem Identification of physical principles Sketch and tabulation Mathematical modeling Solution Checking

Class assignment: Exercise set 1-25 please submit to TA at the end of the lecture Determine the weight W, in U.S. Customary units, of an 85-kg steel bar under standard conditions (sea level at a latitude of 45 degrees). Solution: W=187.4 lb

Scalar and vectors A scalar quantity is completely described with only a magnitude (a number). - Examples: mass, density, length, speed, time, temperature. A vector quantity has both magnitude and direction and obeys the parallelogram law of addition. - Examples: force, moment, velocity, acceleration.

Vector B A Terminal point Initial point Direction of arrow direction of vector Length of arrow magnitude of vector

The sum of two vectors – geometrical representation Two vectors can be vectorially added using the parallelogram law. R F1 F2 Position vector F1 so that its initial point coincides with the terminal point of F2. The vector F1+F2 is represented by the vector R.

Vectors in rectangular coordinate systems- two dimensional x y (v1,v2) v (v1,v2) are the terminal point of vector v V = Vx i + Vy j

The sum of two vectors – analytic representation (two dimensional ) x y (v1+w1,v2+w2) v2 w1 (w1,w2) w2 w v (v1,v2) v1 v+w=(v1+w1,v2+w2) v +w = (v1 + w1 )i + (v2 + w2 ) j

The sum of two vectors – rectangular components (Three dimensional ) z x y (a1,a2,a3) (b1,b2,b3) a b a+b=(a1 +b1,a2+b2, a3 +b3) a +b = (a1 + b1 )i + (a2 + b2 ) j+ (a3 + b3 ) k

Vectors with initial point not on the origin P1(x1 ,y1 ,z1) y z v w x P2(x2 ,y2 ,z2)

Example Find the components of the vector having initial point P1 and terminal point P2 P1(-1,0,2), P2(0,-1,0) Solution: v= (0+1,-1-0,0-2)=(1,-1,-2)

Class assignment: Exercise set 3.1-3a-3c,3e,6a

Vector arithmetic If u,v,w are vectors in 2- or 3-space and k and l are scalar, then the following relationship holds: u+v=v+u u+0=0+u=u k(lu)=(kl)u (k+l)u=ku+lu (u+v)+w=u+(v+w) u+(-u)=0 k(u+v)= ku+ kv 1u=u

The norm of a vector (the length of a vector) x y (v1,v2) v v2 v1 From the Theorem of Pythagoras it follows that the length of a vector is : v = v12 + v22

Three dimensional space z (a1,a2,a3) a a3 y a1 a2 x a = a12 + a22 + a32

Example Let u=(2,-2,3), v=(1,-3,4), w=(3,6,-4). Evaluate the following expression: 3u-5v+w Solution 3u-5v+w =3(2,-2,3)-5(1,-3,4)+(3,6,-4) =(6,-6,9)-(5,-15,20)+(3,6,-4)= 3u-5v+w =(4,15,-25) 3u-5v+w = 42+152+(-15)2 =19.9

Class assignment: Exercise set 3.2-3e

Unit vector Any vector can be written as a product of its magnitude and a unit vector in the direction of the given vector. A vector F in the positive n direction can be written as follows: F=|F|en=Fen Where en is the unit vector and can be written as follows: en= = i + j + k F Fx Fy Fz F F F F

Class assignment: Find the magnitude of any unit vector

Hanchen Stopped here on Aug 28, 2004

Right-hand system The Cartesian coordinates axes are arranged as a right-hand system.

Multiplication of Cartesian vectors The scalar (or dot) product. The vector (or cross) product.

The scalar (or dot) product The scalar product of two intersecting vectors is defined as the product of the magnitudes of the vectors and the cosine of the angle between them q A B A•B=B•A=AB cos(q) 0< q <1800 Finding the rectangular scalar component of vector A along the x-axis Ax=A•i=A cos(qx) Along any direction An=A•en=A cos(qn) A n y q en et At=A-An

The scalar product of the two vectors written in Cartesian form are: A•B = (Ax i + Ay j + Az k) • (Bx i + By j + Bz k) = Ax Bx (i•i) + Ax By (i•j) + Ax Bz (i•k)+ Ay Bx (j•i) + Ay By (j•j) + Ay Bz (j•k)+ Az Bx (k•i) + Az By (k•j) + Az Bz (k•k) Since i, j, k are orthogonal: i•j= j•k= k•j=(1)*(1)*cos(900)=0 i•i= j•j= k•k=(1)*(1)*cos(00)=1 Therefore: A•B = Ax Bx + Ay By + Az Bz

Example Determine the angle q between the following vectors: A=3i +0j +4k and B=2i -2j +5k A•B=AB cos(q) cos(q)= A•B/ AB A = 32+42 = 5 B = 22 +22 +52 = 5.74 AB= 28.7 A•B=3*2+0*(-2)+4*5=26 cos(q)= 26/28.7 q= 25.10

Class assignment: Exercise set 2-63 Two forces are applied to an eye bolt as shown in Fig. P2-63. Determine the x,y, and z scalar components of vector F1. Express vector F1 in Cartesian vector form. Determine the angle a between vectors F1 and F2. x y z 3 ft 4 ft 6 ft F2=700 lb F1=900 lb Fig. P2-63

Solution +72 d1 = d1 = x12+ y12+ z12 (-6)2 +32 =9.7 ft F2=700 lb F1=900 lb +72 d1 = d1 = x12+ y12+ z12 (-6)2 +32 =9.7 ft F1x = F1 cos(qx) =900 *{(–6)/9.7}=-557 lb F1y = F1 cos(qy) =900 *{3/9.7}=278.5 lb F1z = F1 cos(qz) =900 *{7/9.7}=649.8 lb b) Express vector F1 in Cartesian vector form. F1 = -557 i + 278.5j+ 649.8 k lb

c) Determine the angle a between vectors F1 and F2. cos(a)= F1•e2 / F1 e2 x y z 3 ft 4 ft 6 ft F2=700 lb F1=900 lb F1 = -557 i + 278.5j+ 649.8 k lb d2 = x22+ y22+ z22 d2 = (-6)2 +62 +32 =9 ft F1•e2 = (-557)*(-2/3)+278.5*2/3+649.8*0.33=771.4lb e2 = -6/9 i + 6/9j + 3/9 k = -0.67 i + 0.67 j + 0.33 k cos(a)=771.4/900 a=310

Orthogonal (perpendicular) vectors (w1 ,w2 ,w3) z w (v1 ,v2 ,v3) v y x w and v are orthogonal if and only if w·v=0

Properties of the dot product If u , v, and w are vectors in 2- or 3- space and k is a scalar, then: u·v= v·u u·(v+w)= u ·v + u·w k(u·v)= (ku) ·v = u·(kv) v·v> 0 if v=0, and v·v= 0 if v=0

Characteristics of a Force A force is characterizes by the following: (1) Magnitude, (2) Direction, (3) Point of application. F B F B F A

Force Representation Two dimensions Three dimensions y y F F x x x y z q x 5m x Three dimensions x y z qx qy qz F x y z F 7 m 5 m 4 m

Principle of Transmissibility The external effect of a force on a rigid body is the same for all points of application of the force along its line of action. Push Pull Line of action

Forces Classification Contacting or surface forces For example: push or pull Body forces For example: gravitational forces, magnetic forces.

Concurrent forces A force system is said to be concurrent if the action lines of all forces intersect at a common point

Two Concurrent Forces Any two concurrent forces F1 and F2 acting on a body can be replaced by a single force, called the resultant R. F1+F2=R The two forces can be vectorially added using the parallelogram law. F1 R F2

F2 R F1 a b c The law of sines: Sin a sin b sin g = = = = b = sin-1(F2 sin f/R) The law of cosines: c2 = a2 + b2 - 2ab cos g R2 = F12 + F22 - 2F1F2 cos g R2 = F12 + F22 + 2F1F2 cos f

Example 300 450 A barge is pulled by two tugboats. If the resultant of the forces exerted by the tugboat is a 5000 lb force directed along the axis of the barge, determine the tension in the ropes.

Solution T1 T2 T2 T1 T1 T2 5000 sin 450 sin 300 sin 1050 = = 5000 lb 450 300 1050 450 300 5000 lb T2 T1 T1 T2 5000 sin 450 sin 300 sin 1050 = = T1=3660 lb T2=2590 lb

Example – 2-16 Determine the magnitude of the resultant R and the angle q between the x-axis and the line of action on the resultant for the following problem:

f tan(l)=65/120 l=28.440 tan(d)=70/120 d=30.260 f=180-l-d=121.30 Using We get R2 = F12 + F22 + 2F1F2 cos f R2 = 8252 + 7402 + 2*825*740 cos(121.30) R= 771 N

b = sin-1(F2 sin f/R) Using We get b = sin-1(740sin(121.3)/771)=55.140 l q b = sin-1(F2 sin f/R) Using We get b = sin-1(740sin(121.3)/771)=55.140 q=b+l=55.140+28.440=83.580

Class Assignment: Exercise set 2-13 please submit to TA at the end of the lecture Use the law of sines and the law of cosines to determine the magnitude of the resultant R and the angle q between the x-axis and the line of action of the resultant. You could use : b = sin-1(F2 sin f/R) R2 = F12 + F22 + 2F1F2 cos f Solution R= 980.5 lb q=46.230

Solution 600 lb 500 lb f l g b R Using the cosine equation: R2 = F12 + F22 + 2F1F2 cos f tan(l)=2/5 tan(g)=4/1 l=21.80 g=75.960 f=g-l=54.16 R2 = 6002 + 5002 + 2*600*500 cos(54.160) R= 980.5 lb

b q R f l g b R b = sin-1(F2 sin f/R) Using We get 500 lb b 600 lb 980.5 lb q R 600 lb 500 lb f l g b R b = sin-1(F2 sin f/R) Using We get b = sin-1(500sin(54.16)/980.5) b=24.420 q=l+b=24.420 +21.80=46.230

Three or More Concurrent Forces The previous two concurrent forces can be extended to cover three or more forces. F1 +F2 +…+Fn =R F1 F2 F3 F4 F34 F12 F12 F3 F4 R F12 F34

Example Use the laws of sines and cosines to determine the magnitude of the resultant R and the angle q between the x-axis and the line of action of the resultant.

Solution 250 lb consider first two forces (the 250 lb and 350 lb forces ) b 350 lb f12 a g Using the cosine equation: R122 = F12 + F22 + 2F1F2 cos f12 tan(a)=1/3 tan(g)=4/3 a=18.40 g=53.130 f12=g-a=34.730 R122 = 3502 + 2502 + 2*350*250 cos(34.730) R12= 573.4 lb

b b f a q12 g b = sin-1(F2 sin f12/R) Using We get 250 lb 250 lb 573.4 lb b 350 lb b f a q12 350 lb g b = sin-1(F2 sin f12/R) Using We get b = sin-1(250sin(34.73)/573.4) b=14.380 q12=a+b= 18.40 +14.380 =32.780

Using the cosine equation: R2 = R122 + F32 + 2R12F3 cos f12,3 573.4 lb 500 lb 32.780 450 f12,3 Using the cosine equation: R2 = R122 + F32 + 2R12F3 cos f12,3 f12,3= 32.780 +450 =77.780 R2 = 573.42 + 5002 + 2*573.4 *500cos(77.780) R= 837 lb

Solution: q123 f12,3 b3 b3 = sin-1(F3 sin f12,3/R) Using We get 573.4 lb q123 f12,3 450 b3 500 lb b3 = sin-1(F3 sin f12,3/R) Using We get b3 = sin-1(573.4sin(77.78)/837) b3 =42.030 q123=450 -42.030 =2.970 Solution: R= 837 lb

Class Assignment: Exercise set 2-22 please submit to TA at the end of the lecture Use the law of sines and the law of cosines to determine the magnitude of the resultant R and the angle q between the x-axis and the line of action of the resultant. Solution: R=468; q=35.1