Introduction to Statics and Dynamics P2 doc

2.1. Vector notation and vector addition 13 Graphical: a scalar multiplies an arrow. Indicate a vector’s direction by drawing an arrow with direction indicated by marked angles or slopes. The scalar multiple with a nearby scalar symbol, say F, as shown in figure 2.8b. This means F times a unit vector in the direction of the arrow. (Because F might be negative, signconfusioniscommonamongst beginners. Please see sample2.1.) Combined: graphical representation used to define a symbolic vector. The full symbolic notation can be used in a picture with the graphical information as a way of defining the symbol. For example if the arrow in fig. 2.8b were labeled with an  F instead of just F we would be showing that  F is a scalar multiplied by a unit vector in the direction shown. F 50 o (a) (b)  F Figure 2.8: Two different ways of draw- ing a vector (a) shows a labeled arrow. The magnitude and direction of the vector is given by the symbol  F , the drawn arrow has no quantitative information. (b) shows an arrow with clearly indicated orientation next to the scalar F. This means a unit vec- tor in the direction of the arrow multiplied by the scalar F. (Filename:tfigure1.d) The components of a vector A given vector, say  F , can be described as the sum of vectors each of which is parallel to a coordinate axis. Thus  F =  F x +  F y in 2D and  F =  F x +  F y +  F z in 3D. Each of these vectors can in turn be written as the product of a scalar and a unit vector along the positive axes, e.g.,  F x = F x ˆ ı (see fig. 2.9). So  F =  F x +  F y = F x ˆ ı + F y ˆ  (2D) or  F =  F x +  F y +  F z = F x ˆ ı + F y ˆ  + F z ˆ k. (3D) The scalars F x , F y , and F z are called the components of the vector with respect to the axes xyz. The components may also be thought of as the orthogonal projections (the shadows) of the vector onto the coordinate axes. Because the list of components is such a handy way to describe a vector we have a special notation for it. The bracketed expression [  F ] xyz stands for the list of components of  F presented as a horizontal or vertical array (depending on context), as shown below. [  F ] xyz = [F x , F y , F z ]or[  F ] xyz =   F x F y F z   . If we had an xy coordinate system with x pointing East and y pointing North we could write the components of a 5 N force pointed Northeast as [  F ] xy = [(5/ √ 2) N,(5/ √ 2) N]. x y ˆ ı ˆ   F  F  F F x F x F z F y F O O y x y z  F  F x  F y  F y  F x  F z Figure 2.9: A vector can be broken into a sum of vectors, each parallel to the axis of a coordinate system. Each of these is a componentmultiplied byaunitvectoralong the coordinate axis, e.g.,  F x = F x ˆ ı. (Filename:tfigure.vectproject) Note that the components of avectorinsomecrookedcoordinatesystem x  y  z  are differentthan the coordinatesforthesame vectorinthe coordinatesystem xyzbecause the projections are different. Even though  F =  F itisnottruethat [  F ] xyz = [  F ] x  y  z  (see fig. 2.19 on page 25). In mechanics we often make use of multiple coordinate systems. So to define a vector by its components the coordinate system used must be specified. Rather than using up letters to repeat the same concept we sometimes label the coordinate axes x 1 , x 2 and x 3 and the unit vectors along them ˆ e 1 , ˆ e 2 , and ˆ e 3 (thus freeing our minds of the silently pronounced letters y,z,j, and k). Manipulating vectors by manipulating components Because a vector can be represented by its components (once given a coordinate system) we should be able to relate our geometric understanding of vectors to their components. In practice, when push comes to shove, most calculations with vectors are done with components. 14 CHAPTER 2. Vectors for mechanics Adding and subtracting with components Because a vector can be broken into a sum of orthogonal vectors, because addition is associative, and because each orthogonal vector can be written as a component times a unit vector we get the addition rule: [  A +  B] xyz = [(A x + B x ), (A y + B y ), (A z + B z )] which can be described by the tricky words ‘the componentsof the sum oftwovectors are given by the sums of the corresponding components.’ Similarly, [  A −  B] xyz = [(A x − B x ), (A y − B y ), (A z − B z )] Multiplying a vector by a scalar using components The vector  A can be decomposed into the sum of three orthogonal vectors. If  A is multiplied by 7 than so must be each of the component vectors. Thus [c  A] xyz = [cA x , cA y , cA z ]. The components of a scaled vector are the corresponding scaled components. Magnitude of a vector using components The Pythagorean theorem for right triangles (‘A 2 + B 2 = C 2 ’) tells us that |  F |=  F 2 x + F 2 y , (2D) |  F |=  F 2 x + F 2 y + F 2 z . (3D) To get the result in 3D the 2D Pythagorean theorem needs to be applied twice suc- cessively, first to get the magnitude of the sum  F x +  F y and once more to add in  F z . 2.1. Vector notation and vector addition 15 SAMPLE 2.1 Drawing a vector from its components: Draw the vector  r = 3ft ˆ ı − 2ft ˆ  using its components. A x y 3 ft 2 ft  r Figure 2.10: A vector  r = 3ft ˆ ı −2ft ˆ  is drawn by locating its end point which is 3 units away along the x-axis and 2 units away along the negative y-axis. (Filename:sfig1.2.4a) Solution To draw  r using its components, we first draw the axes and measure 3 units (any units that we choose on the ruler) along the x-axis and 2 units along the negative y-axis. We mark this point as A (say) on the paper and draw a line from the origin to the point A. We write the dimensions ‘3ft’ and ‘2ft’ on the figure. Finally, we put an arrowhead on this line pointing towards A. SAMPLE 2.2 Drawing a vector from its length and direction: A vector  r is 3.6ft long and is directed 33.7 o from the x-axis towards the negative y-axis. Draw  r . A x y  r 33.7 o 3.6 units Figure 2.11: A vector  r with a given length (3.6ft ) and direction (slope angle θ =−33.7 o ) is drawn by measuring its length along a line drawn at angle θ from the positive x-axis. (Filename:sfig1.2.4b) Solution We first draw the x and y axes and then draw  r as a line from the origin at an angle −33.7 o from the x-axis (minus sign means measuring clockwise), measure 3.6 units (magnitude of  r ) along this line and finally put an arrowhead pointing away from the origin. Comments: Note that this is the same vector as in Sample 2.1. In fact, you can easily verify that r x = r cos θ = 3.6ft·cos(−33.7 o ) = 3ft and r y = r sin θ = 3.6ft·sin(−33.7 o ) =−2ft. Thus  r = r x ˆ ı +r y ˆ  = (3ft) ˆ ı − (2ft) ˆ  as given in Sample 2.1. 16 CHAPTER 2. Vectors for mechanics SAMPLE 2.3 Various ways of representing a vector: A vector  F = 3N ˆ ı + 3N ˆ  is represented in various ways, some incorrect, in the following figures. The base vectors used are shown first. Comment on each representation, whether it is correct or incorrect, and why. 45 o (a) 3 √ 2N 3N 3N (b) (c) -3 √ 2N 45 o (d) (i) (j) 3 √ 2N ˆ ı  3 √ 2N ˆ ı  3 √ 2N  (e) 3N ˆ ı + ˆ 3Nj ˆ  3N 3N (f) 3N (g) 45 o (h) 3 √ 2N ˆ ı 45 o 45 o ˆ ı ˆ  ˆ ı  ˆ   Figure 2.12: (Filename:sfig2.vectors.rep) Solution The given vector is a force with components of 3N each in the positive ˆ ı and ˆ  directions using the unit vectors ˆ ı and ˆ  shown in the box above. The unit vectors ˆ ı  , and ˆ   are also shown. a) Correct: 3 √ 2Nˆı  . From the picture defining ˆ ı  , you can see that ˆ ı  is a unit vector with equal components in the ˆ ı and ˆ  directions; i.e., it is parallel to  F .So  F is given by its magnitude  (3N) 2 + (3N) 2 times a unit vector in its direction, in this case ˆ ı  . It is the same vector. b) Correct: Here two vectors are shown: one with magnitude 3 N in the direction of the horizontal arrow ˆ ı, and one with magnitude 3N in the direction of the vertical arrow ˆ . When two forces act on an object at a point, their effect is additive. So the net vector is the sum of the vectors shown. That is, 3N ˆ ı +3N ˆ . It is the same vector. c) Correct: Here we have a scalar 3 √ 2 N next to an arrow. The vector described is the scalar multiplied by a unit vector in the direction of the arrow. Since the arrow’s direction is marked as the same direction as ˆ ı  , which we already know is parallel to  F , this vector represents the same vector  F . It is the same vector. d) Correct: The scalar −3 √ 2 N is multiplied by a unit vector in the direction indicated, − ˆ ı  . So we get (−3 √ 2N)(− ˆ ı  ) which is 3 √ 2N ˆ ı  as before. It is the same vector. e) Incorrect: 3 √ 2N ˆ   . The magnitude is right, but the direction is off by 90 degrees. It is a different vector. f) Incorrect: 3N ˆ ı − 3N ˆ . The ˆ ı component of the vector is correct but the ˆ  component is in the opposite direction. The vector is in the wrong direction by 90 degrees. It is a different vector. 2.1. Vector notation and vector addition 17 g) Incorrect: Right direction but the magnitude is off by a factor of √ 2. h) Incorrect: The magnitude is right. The direction indicated is right. But, the algebraic symbol 3 √ 2N ˆ ı takes precedence and it is in the wrong direction ( ˆ ı instead of ˆ ı  ). It is a different vector. i) Correct: A labeled arrow. The arrow is only schematic. The algebraic symbols 3 √ 2N ˆ ı  define the vector. We draw the arrow to remind us that there is a vector to represent. The tip or tail of the arrow would be drawn at the point of the force application. In this case, the arrow is drawn in the direction of  F but it need not. j) Correct: Like (i) above, the directional and magnitude information is in the algebraic symbols 3 N ˆ ı + 3N ˆ . The arrow is there to indicate a vector. In this case, it points in the wrong direction so is not ideally communicative. But (j) still correctly represents the given vector. It is the same vector. 18 CHAPTER 2. Vectors for mechanics SAMPLE 2.4 Adding vectors: Three forces,  F 1 = 2N ˆ ı +3N ˆ ,  F 2 =−10 N ˆ , and  F 3 = 3N ˆ ı + 1N ˆ  −5N ˆ k, act on a particle. Find the net force on the particle. Solution The net force on the particle is the vector sum of all the forces, i.e.,  F net =  F 1 +  F 2 +  F 3 = (2N ˆ ı + 3N ˆ ) + (−10 N ˆ ) + (3N ˆ ı + 1N ˆ  −5N ˆ k) = 2N ˆ ı + 3N ˆ  + 0 ˆ k + 0 ˆ ı − 10 N ˆ  + 0 ˆ k + 3N ˆ ı + 1N ˆ  − 5 ˆ k = (2N+3N) ˆ ı + (3N− 10 N + 1N) ˆ  +(−5N) ˆ k = 5N ˆ ı − 6N ˆ  −5N ˆ k.  F net = 5N ˆ ı − 6N ˆ  −5N ˆ k Comments: In general, we do not need to write the summation so elaborately. Once you feel comfortable with the idea of summing only similar components in a vector sum, you can do the calculation in two lines. SAMPLE 2.5 Subtracting vectors: Two forces  F 1 and  F 2 act on a body. The net force on the body is  F net = 2N ˆ ı.If  F 1 = 10N ˆ ı − 10 N ˆ , find the other force  F 2 . Solution  F net =  F 1 +  F 2 ⇒  F 2 =  F net −  F 1 = 2N ˆ ı − (10 N ˆ ı − 10 N ˆ ) = (2N−10N) ˆ ı − (−10 N) ˆ  =−8N ˆ ı + 10 N ˆ .  F 2 =−8N ˆ ı + 10 N ˆ  SAMPLE 2.6 Scalar times a vector: Two forces acting on a particle are  F 1 = 100 N ˆ ı − 20 N ˆ  and  F 2 = 40N ˆ .If  F 1 is doubled, does the net force double? Solution  F net =  F 1 +  F 2 = (100N ˆ ı − 20 N ˆ ) + (40 N ˆ ) = 100 N ˆ ı + 20 N ˆ  After  F 1 is doubled, the new net force  F (net) 2 is  F (net) 2 = 2  F 1 +  F 2 = 2(100N ˆ ı − 20 N ˆ ) + (40 N ˆ ) = 200 N ˆ ı − 40 N ˆ  +40 N ˆ  = 200 N ˆ ı = 2(100N ˆ ı + 20 N ˆ )     F net No, the net force does not double. 2.1. Vector notation and vector addition 19 SAMPLE 2.7 Magnitude and direction of a vector: The velocity of a car is given by  v = (30 ˆ ı + 40 ˆ ) mph. (a) Find the speed (magnitude of  v) of the car. (b) Find a unit vector in the direction of  v. (c) Write the velocity vector as a product of its magnitude and the unit vector. Solution (a) Magnitude of  v: The magnitude of a vector is the length of the vector. It is a scalar quantity, usually represented by the same letter as the vector but without the vector notation (i.e. no bold face, no underbar). It is also represented by the modulus of the vector (the vector written between two vertical lines). The length of a vector is the square root of the sum of squares of its components. Therefore, for  v = 30mph ˆ ı + 40 mph ˆ , v =|  v|=  v 2 x + v 2 y =  (30 mph) 2 + (40 mph) 2 = 50 mph which is the speed of the car. speed = 50 mph (b) Direction of  v as a unit vector along  v: The direction of a vector can be spec- ified by specifying a unit vector along the given vector. In many applications you will encounter in dynamics, this concept is useful. The unit vector along a given vector is found by dividing the given vector with its magnitude. Let ˆ λ v be the unit vector along  v. Then, ˆ λ v =  v |  v| = 30 mph ˆ ı + 40 mph ˆ  50 mph = 0.6 ˆ ı + 0.8 ˆ . (unit vectors have no units!) ˆ λ v = 0.6 ˆ ı + 0.8 ˆ  (c)  v as a product of its magnitude and the unit vector ˆ λ v : A vector can be written in terms of its components, as given in this problem, or as a product of its magnitude and direction (given by a unit vector). Thus we may write,  v =|  v| ˆ λ v = 50mph(0.6 ˆ ı + 0.8 ˆ ) which, of course, is the same vector as given in the problem.  v = 50 mph(0.6 ˆ ı + 0.8 ˆ ) 20 CHAPTER 2. Vectors for mechanics SAMPLE 2.8 Position vector from the origin: In the xyz coordinate system, a particle is located at the coordinate (3m, 2m, 1m). Find the position vector of the particle. Solution The position vector of the particle at P is a vector drawn from the origin y x z 2m 3m 1m  r (3m,2m,1m) Figure 2.13: The position vector of the particle is a vector drawn from the origin of the coordinate system to the position of the particle. (Filename:sfig2.vec1.6) of the coordinate system to the position P of the particle. See Fig. 2.13. We can write this vector as  r P = (3m) ˆ ı + (2m) ˆ  +(1m) ˆ k or  r P = (3 ˆ ı + 2 ˆ  + ˆ k) m.  r P = 3m ˆ ı + 2m ˆ  +1m ˆ k SAMPLE 2.9 Relative position vector: Let A (2m, 1m, 0) and B (0, 3m, 2m) be two points in the xyzcoordinate system. Find the position vector of point B with respect to point A, i.e., find  r AB (or  r B/A ). Solution From the geometry of the position vectors shown in Fig. 2.14 and the rules y x z A (2,1,0) B (0,3,2) 2 1 3 2  r A  r AB  r B Figure 2.14: The position vector of B with respect to A is found from  r AB =  r B −  r A . (Filename:sfig2.vec1.7) of vector sums, we can write,  r B =  r A +  r AB ⇒  r AB =  r B −  r A = (0 ˆ ı + 3m ˆ  +2m ˆ k) − (2m ˆ ı + 1m ˆ  +0 ˆ k) =−2m ˆ ı + 2m ˆ  +2m ˆ k.  r AB ≡  r B/A =−2m ˆ ı + 2m ˆ  +2m ˆ k 2.1. Vector notation and vector addition 21 SAMPLE 2.10 Finding a force vector given its magnitude and line of action: A x y z A B 1m 0.6 m 0.5 m 0.2m 0.2 m  F Figure 2.15: (Filename:sfig1.2.2) string is pulled with a force F = 100 N as shown in the Fig. 2.15. Write F as a vector. Solution A vector can be written, as wejust showed in the previous sample problem, as the product of its magnitude and a unit vector along the given vector. Here, the magnitude of the force is given and we know it acts along AB. Therefore, we may write  F = F ˆ λ AB where ˆ λ AB is a unit vector along AB. So now we need to find ˆ λ AB . We can easily x y z A B 1m 0.6 m 0.5 m 0.2m 0.2 m  r A  r B  r AB Figure 2.16:  r AB =  r B −  r A . (Filename:sfig1.2.2b) find ˆ λ AB if we know vector AB. Let us denote vector AB by  r AB (sometimes we will also write it as  r B/ A to represent the position of B with respect to A as a vector). Then, ˆ λ AB =  r AB |  r AB | . To find  r AB , we note that (see Fig. 2.16)  r A +  r AB =  r B where  r A and  r B are the position vectors of point A and point B respectively. Hence,  r B/ A =  r AB =  r B −  r A = (0.2m ˆ ı + 0.6m ˆ  +0.2m ˆ k) − (0.5m ˆ ı + 1.0m ˆ k) =−0.3m ˆ ı + 0.6m ˆ  −0.8m ˆ k. Therefore, ˆ λ AB = −0.3m ˆ ı + 0.6m ˆ  −0.8m ˆ k  (−0.3) 2 + (0.6) 2 + (−0.8) 2 m =−0.29 ˆ ı + 0.57 ˆ  −0.77 ˆ k, and, finally,  F = ( F  100 N) ˆ λ AB =−29 N ˆ ı + 57 N ˆ  −77 N ˆ k.  F =−29 N ˆ ı + 57 N ˆ  −77 N ˆ k 22 CHAPTER 2. Vectors for mechanics SAMPLE2.11 Addingvectorsoncomputers: The followingsixforces actatdifferent points of a structure.  F 1 =−3N ˆ ,  F 2 = 20 N ˆ ı − 10N ˆ ,  F 3 = 1N ˆ ı + 20N ˆ  − 5N ˆ k,  F 4 = 10N ˆ ı,  F 5 = 5N( ˆ ı + ˆ  + ˆ k),  F 6 =−10 N ˆ ı − 10 N ˆ  +2N ˆ k. (a) Write all the force vectors in column form. (b) Find the net force by hand calculation. (c) Write acomputerprogramtosum n vectors,eachoflength3. Useyourprogram to compute the net force. Solution (a) The 3-D vector  F = F x ˆ ı + F y ˆ  + F z ˆ k is represented as a column (or a row) as follows: [  F ] =   F x F y F z   xyz Following this convention, we write the given forces as [  F 1 ] =   0 −3N 0   xyz , [  F 2 ] =   20 N −10 N 0   xyz , ···, [  F 6 ] =   −10 N −10 N 2N   xyz (b) The net force,  F net =  F 1 +  F 2 +  F 3 +  F 4 +  F 5 +  F 6 or [  F net ] =   0201105−10 −3 +−10 + 20 + 0 + 5 +−10 00−505 2   xyz N =   26 2 2   xyz N (c) The steps to do this addition on computers are as follows. • Enter the vectors as rows or columns: F1=[0-30] F2 = [20 -10 0] F3 = [1 20 -5] F4 = [10 0 0] F5=[5 5 5] F6 = [-10 -10 2] • Sum the vectors, using a summing operation that automatically does ele- ment by element addition of vectors: Fnet=F1+F2+F3+F4+F5+F6 • The computer generated answer is: Fnet = [26 2 2].  F net = 26N ˆ ı + 2N ˆ  +2N ˆ k [...]... vector in a given direction, to reduce a vector to components, to reduce vector equations to scalar equations, to define work and power, and to help solve geometry problems The dot product of two vectors A and B is written A · B (pronounced ‘A dot B’) The dot product of A and B is the product of the magnitudes of the two vectors times a number that expresses the degree to which A and B are parallel: cos... have seen how a vector can be broken down into a sum of components each parallel to one of the orthogonal base vectors Another useful decomposition is this: ˆ Given any vector A and a unit vector λ the vector A can be written as the sum of two parts, ⊥ A=A +A A A ⊥ A || ˆ λ ˆ Figure 2.20: For any A and λ, A can be ˆ decomposed into a part parallel to λ and a ˆ part perpendicular to λ (Filename:tfigure.Graham1)... base vectors F = Fx i + Fy j or F = Fx i + Fy j + Fz k in three dimensions Here the base vectors marked with primes, i , j and k , are unit vectors parallel to some mutually orthogonal x , y , and z axes These x , y , and z axes may be crooked in relation to the x, y, and z axis That is, the x axis need not be parallel to the x axis, the y not parallel to the y axis, and the z axis not parallel to the... both sides with a common vector All the distributive, associative, and commutative laws of ordinary addition and multiplication hold 1 1 Caution: But you cannot divide a vector ˆ by a vector or a scalar by a vector: 7/ı andA/C are nonsense expressions And it does not make sense to add a vector and a scalar, 7 + A is a nonsense expression 28 CHAPTER 2 Vectors for mechanics Vector calculations on the computer... will define the moment of a force intuitively, geometrically, and finally using vector algebra We will do this first in 2 dimensions and then in 3 The main mathematical tool here is the vector cross product, a second way of multiplying vectors together The cross product is used to define (and calculate) moment and to calculate various quantities in dynamics The cross product also sometimes helps solve three-dimensional... we call our coordinate x1 , x2 , and x3 ; and our unit base ˆ ˆ ˆ ˆ ˆ ˆ vectors e1 ,e2 , and e3 we would have A = A1 e1 + A2 e2 + A3 e3 ˆ ˆ ˆ and B = B1 e1 + B2 e2 + B3 e3 and the dot product has the tidy 3 form: A · B = A1 B1 + A2 B2 + A3 B3 = Ai Bi i=1 2.2 The dot product of two vectors 25 ˆ assume that the dot products between the standard base vectors and the vector  ˆ ˆ ˆ ˆ (i.e., ı ·  , ... adds a force that is directed towards the hinge it causes no tendency to rotate Because any force can be decomposed into a sum of forces, one perpendicular to the teeter totter and the other towards the hinge, and because we assume that the affect of the sum of these forces is the sum of the affects of each separately, and because the force towards the hinge has no tendency to rotate, we have deduced:... example, if vector A is the same as the vector B, A = B For any scalar a and any vector C, we then A + C = B + C, a A = a B, and A·C = B · C, because performing the same operation on equal quantities maintains the equality The vectors A, B, and C might themselves be expressions involving other vectors The equations above show the allowable manipulations of vector equations: adding a common term to both sides,... unit vector λ 2.2 The dot product of two vectors 27 Sum of orthogonal component vectors F = F x + F y is a sum of two vectors parallel to the x and y axis, respectively In three dimensions, F = F x + F y + F z ˆ ˆ ˆ ˆ ˆ Components times unit base vectors F = Fx ı + Fy  or F = Fx ı + Fy  + Fz k ˆ in three dimensions One way to think of this sum is to realize that F x = Fx ı , ˆ ˆ F y = Fy  and F z... unit vectors and they are perpendicular to each other In math language they are ‘orthonormal.’ The standard crooked base vectors are orthonormal The identities above lead to the following equivalent ways of expressing the dot product of A and B (see box 2.2 on page 24 to see how the component formula follows from the geometric definition above): B θ A os The dot product is used to project a vector in . of two vectors 23 2.2 The dot product of two vectors The dot product is used to project a vector in a given direction, to reduce a vector to components, to reduce vector equations to scalar equations,. main mathematical tool here is the vector cross product, a second way of multiplying vectors together. The cross product is used to define (and calculate) moment and to calculate various quantities in dynamics. . vector can be broken down into a sum of components each parallel to one of the orthogonal base vectors. Another useful decomposition is this: Given any vector  A and a unit vector ˆ λ the vector  A