MASSACHUSETTS INSTITUTE OF TECHNOLOGY Depatment of hysics 801 Fall 2009 Review Module: Coss oduct We shall now intoduce ou second vecto opeation, called the coss poduct that takes any two vectos and geneates a new vecto The coss poduct is a type of multiplication law that tuns ou vecto space (law fo addition of vectos) into a vecto algeba (a vecto algeba is a vecto space with an additional ule fo multiplication of vectos) The fist application of the coss poduct will be the physical concept of toque about a point that can be descibed mathematically by the coss poduct between two vectos: a vecto fom to whee the foce acts, and the foce vecto Definition: Coss oduct Let A and B be two vectos Since any two non-paallel vectos fom a plane, we define the angle θ to be the angle between the vectos A and B as shown in Figue 1 The magnitude of the coss poduct A B of the vectos A and B is defined to be poduct of the magnitude of the vectos A and B with the sine of the angle θ between the two vectos, A B = ABsin(θ), (1) whee A and B denote the magnitudes of A and B, espectively The angle θ between the vectos is limited to the values 0 θ π insuing that sin( θ ) 0 Figue 1 Coss poduct geomety The diection of the coss poduct is defined as follows The vectos A and B fom a plane Conside the diection pependicula to this plane Thee ae two possibilities: we shall choose one of these two (shown in Figue 1) fo the diection of the coss poduct A B using a convention that is commonly called the ight-hand ule 10/24/2009 1
Right-hand Rule fo the Diection of Coss oduct The fist step is to edaw the vectos A and B so that thei tails ae touching Then daw an ac stating fom the vecto A and finishing on the vecto B Cul you ight finges the same way as the ac You ight thumb points in the diection of the coss poduct A B (Figue 2) Figue 2 Right-Hand Rule You should emembe that the diection of the coss poduct A B is pependicula to the plane fomed by A and B We can give a geometic intepetation to the magnitude of the coss poduct by witing the magnitude as A B = AB ( sin θ ) (2) The vectos A and B fom a paallelogam The aea of the paallelogam is equal to the height times the base, which is the magnitude of the coss poduct In Figue 3, two diffeent epesentations of the height and base of a paallelogam ae illustated As depicted in Figue 3(a), the tem Bsin( θ ) is the pojection of the vecto B in the diection pependicula to the vecto B We could also wite the magnitude of the coss poduct as A B = ( Asin( θ )) B (3) Now the tem Asin( θ ) is the pojection of the vecto A in the diection pependicula to the vecto B as shown in Figue 3(b) 10/24/2009 2
Figue 3(a) and 3(b) ojection of vectos and the coss poduct The coss poduct of two vectos that ae paallel (o anti-paallel) to each othe is zeo since the angle between the vectos is 0 (o π ) and sin(0) = 0 (osin( π ) = 0 ) Geometically, two paallel vectos do not have a unique component pependicula to thei common diection opeties of the Coss oduct (1) The coss poduct is anti-commutative since changing the ode of the vectos coss poduct changes the diection of the coss poduct vecto by the ight hand ule: A B= B A (4) (2) The coss poduct between a vecto c A whee c is a scala and a vecto B is ca B= c( A B) (5) Similaly, A cb= c( A B) (6) (3) The coss poduct between the sum of two vectos A and B with a vecto C is ( A+ B) C= A C+ B C (7) Similaly, A ( B+ C) = A B+ A C (8) Vecto Decomposition and the Coss oduct We fist calculate that the magnitude of coss poduct of the unit vecto î with ĵ : 10/24/2009 3
ˆi ˆj = ˆi ˆj sin( π /2) = 1, (9) since the unit vecto has magnitude ˆ i = ˆ j = 1 and sin( π / 2) = 1 By the ight hand ule, the diection of ˆ i ˆ j is in the +k ˆ as shown in Figue 4 Thus ˆ i ˆ j = k ˆ Figue 4 Coss poduct of ˆ i ˆ j We note that the same ule applies fo the unit vectos in the y and z diections, ˆj kˆ = ˆi, kˆ ˆi = ˆj (10) Note that by the anti-commutatively popety (1) of the coss poduct, ˆj ˆi = kˆ, ˆi kˆ = ˆj (11) The coss poduct of the unit vecto to each othe, (sin(0) = 0), î with itself is zeo because the two unit vectos ae paallel ˆi ˆi = ˆi ˆi sin(0) = 0 (12) The coss poduct of the unit vecto ĵ with itself and the unit vecto ˆk with itself ae also zeo fo the same eason ˆj ˆj = 0, kˆ k ˆ = 0 (13) With these popeties in mind we can now develop an algebaic expession fo the coss poduct in tems of components Let s choose a Catesian coodinate system with the vecto B pointing along the positive x-axis with positive x-component B x Then the vectos A and B can be witten as and A= A ˆi+ A ˆj + A kˆ x y z (14) 10/24/2009 4
B= B x ˆi, (15) espectively The coss poduct in vecto components is A B = ( A ˆi+ A ˆj + A k) ˆ B ˆi (16) x y z x This becomes, A B= ( A ˆi B ˆi) + ( A ˆj B ˆi) + ( A kˆ B ˆi) x x y x z x = AB ( ˆi ˆi) + AB ( ˆj ˆi) + AB ( kˆ ˆi) x x y x z x = AB kˆ + AB y x z x ˆj (17) The vecto component expession fo the coss poduct easily genealizes fo abitay vectos and A= A ˆi+ A ˆj + A kˆ x y z (18) B= B ˆi+ B ˆj + B kˆ, (19) x y z to yield A B= ( AB AB ) ˆi+ ( AB AB) ˆj + ( AB AB) kˆ (20) y z z y z x x z x y y x Example 1: Angula Momentum fo a point paticle an example of coss poduct Conside a point paticle of mass m moving with a velocity v (Figue 5) Figue 5 A point paticle and its position and velocity vectos The linea momentum of the paticle is p= m v Conside a point S located anywhee in space Let Sm, denote the vecto fom the point S to the location of the object Definition: The angula momentum L S about the point S of a point paticle is defined to be the vecto coss poduct of the vecto fom the point S to the location of the object with the momentum of the paticle, L = p (21) S S, m 10/24/2009 5
Magnitude: The magnitude of the angula momentum about S is given by LS = S, m p sinθ, (22) whee θ is the angle between the vectos Sm, and p, and lies within the ange [0 θ π ] The 2 1 SI units fo angula momentum ae kg m s Diection of the Angula Momentum Remembe that, as in all coss poducts, the diection of the angula momentum is pependicula to the plane fomed by Sm, and p, and is given by the ight hand ule Figue 6 The Right Hand Rule Thee ae two ways to visualize the magnitude of the angula momentum Figue 7 Vecto diagam fo angula momentum Define the moment am,, as the pependicula distance fom the point S to the line defined by the diection of the momentum Then = Sm, sinθ (23) 10/24/2009 6
Hence the magnitude of the angula momentum is the poduct of the moment am with the magnitude of the momentum L S = p (24) Altenatively, define the pependicula momentum, p, to be the magnitude of the component of the momentum pependicula to the line defined by the diection of the vecto Sm, Thus p = p sinθ (25) Thus the magnitude of the angula momentum is the poduct of the distance fom S to the point object with the pependicula momentum, L S = S, m p (26) Example 2: Toque as a Coss oduct Let a foce F with magnitude F = F act at a point Let S, be the vecto fom the point S to a point, with magnitude = S, Let F = Fsinθ be the component of the foce F that is pependicula to the line passing fom the point S to The angle between the vectos S, and F is θ with [0 θ π ] (Figue 8) Figue 8 Toque about a point S Definition: The toque τ S about a point S, due to a foce F vecto coss poduct of the vectos S, and F, S S,, acting at the point, is the τ = F (27) Toque is a vecto quantity with magnitude and diection given as follows: (1) Magnitude: The magnitude of the toque about S is τ S = F = F sinθ (28) 10/24/2009 7
The SI units fo toque ae [N m] (2) Diection: The diection of the toque is pependicula to the plane fomed by the vectos S, and F (fo [0 < θ < π ] ), and by definition points in the diection of the unit nomal vecto to the plane ˆn 1 as shown in Figue 9 Figue 9 Diection of toque about a point S The magnitude of the coss poduct is the aea of the paallelogam defined by vectos S, F, the height times the base Figue 10 shows the two diffeent ways of defining height and base, Aea = τ S = F = F (29) and Figue 10 Aea of the toque paallelogam The diection of the coss poduct is defined by the ight hand ule and so the coss poduct of S, and F points out of the page in Figue 10 10/24/2009 8
oblems: oblem 1 Given two vectos, A= 2ˆi+ 3ˆj + 7kˆ and B= 5ˆi+ ˆj + 2kˆ, find A B Solution: A B= ( AB AB ) ˆi+ ( AB AB) ˆj+ ( AB AB) kˆ y z z y z x x z x y y x = (( 3)(2) (7)(1)) ˆi + ((7)(5) (2)(2)) ˆj + ((2)(1) ( 3)(5)) kˆ = 13 ˆi+ 31 ˆj+ 17 kˆ oblem 2 ove the law of sines using the coss poduct (Hint: Conside the aea of a tiangle fomed by thee vectos A, B, and C, whee A+ B+ C= 0 ) Solution: Since A+ B+ C= 0, we have that 0 = A ( A+ B+ C) = A B+ A C Thus A B= A C o A B = A C Fom the figue we see that A B = A B sinγ and A C = A C sin β Theefoe ABsinγ = ACsin β, and hence B / sin β = C / sinγ A simila agument shows that B /sin β = A /sinα poving the law of sines 10/24/2009 9
oblem 3 Find a unit vecto pependicula to A= ˆi+ ˆj kˆ and B= 2ˆi ˆj + 3kˆ Solution: The coss poduct A B is pependicula to both A and B Theefoe the unit vectos nˆ =± A B/ A B ae pependicula to both A and B We fist calculate A B= ( AB AB ) ˆi+ ( AB AB) ˆj+ ( AB AB) kˆ y z z y z x x z x y y x = ((1)(3) ( 1)( 1)) ˆi + (( 1)(2) (1)(3)) ˆj + ((1)( 1) (1)(2)) kˆ = 2ˆi 5ˆj 3kˆ We now calculate the magnitude A B = 2 + 5 + 3 = (38) ( ) 1/2 2 2 2 1/2 Theefoe the pependicula unit vectos ae nˆ =± A B/ A B =± 2i 5j 3 k /(38) ( ˆ ˆ ˆ 1/2 ) oblem 4 Show that the volume of a paallelpiped with edges fomed by the vectos A, B, and C is given by A ( B C) Solution: The volume of a paallelpiped is given by aea of the base times height If the base is fomed by the vectos B and C, then the aea of the base is given by the magnitude of B C The vecto B C= B Cnˆ whee ˆn is a unit vecto pependicula to the base The pojection of the vecto A along the diection ˆn gives the height of the paallelpiped This pojection is given by taking the dot poduct of A with a unit vecto and is equal to An ˆ = height Theefoe A ( B C) = A B C nˆ = B C A nˆ = ( aea)( height) = ( volume) ( ) ( ) 10/24/2009 10
oblem 5 Let A be an abitay vecto and let ˆn be a unit vecto in some fixed diection A= A nˆ nˆ + nˆ A nˆ Show that ( ) ( ) Solution: Let A= A nˆ + eˆ n A whee A is the component A in the diection of ˆn, ê is the n diection of the pojection of A in a plane pependicula to ˆn, and A is the component of A in the diection of ê Since en ˆ ˆ = 0, An ˆ = An Note that nˆ A= nˆ A nˆ + A eˆ = nˆ A eˆ = A ( nˆ eˆ) ( ) n The unit vecto nˆ e ˆ lies in the plane pependicula to ˆn and is also pependicula to ê Theefoe ( nˆ eˆ) n ˆ is also a unit vecto that is must be paallel to ê (by the ight hand ule So nˆ A nˆ = eˆ Thus ( ) A A= A nˆ nˆ + nˆ A nˆ = A nˆ + eˆ ( ) ( ) oblem 6 Angula Momentum of a oint-like aticle A paticle of mass m = 20 kg moves as shown in the sketch with a unifom velocity 1 ˆ 1 v= 30 m s i+ 30 m s ˆj At time t, the paticle passes though the point = ˆ ˆ 0, m 20m i+ 30m j Find the diection and the magnitude of the angula momentum about the oigin at time t n A Solution: Choose Catesian coodinates with unit vectos shown in the figue above The angula momentum vecto L 0 of the paticle about the oigin is given by: 10/24/2009 11
L = p= v 0 0, m 0, m m 1 1 ( 20m ˆi 30m ˆj ) (2kg)( 30m s ˆi 30m s ˆj ) = + + 2 1 ˆ 2 1 = 0+ 12kg m s k 18kg m s ( kˆ) + 0 2 1 = 6kg m s kˆ In the above, the elations i j = k, j i = k, i i = j j= 0 wee used oblem 7 Angula Momentum of a oint aticle Undegoing Cicula Motion Conside a point paticle of mass m moving in a cicle of adius R with velocity v = Rω ˆ θ, with angula speed ω > 0 Find the diection and magnitude of the angula momentum about the cente of a cicle in tems of the adius of the cicle R, the mass of the moving object m, and the angula speed ω Solution: The velocity of the paticle is given by v = Rω ˆ θ The vecto fom the cente of the cicle (the point S) to the object is given by ˆ, = R The angula momentum about the cente of the cicle is the coss poduct Sm L p v k k k ˆ ˆ 2 ˆ S = S, m = S, m m = Rmv = RmRω = mr ω The magnitude is 2 L = mr ω, and the diection is in the + kˆ -diection S 10/24/2009 12
oblem 8 Toque and Static Equilibium: the Ankle A peson of mass m = 75 kg is couching with thei weight evenly distibuted on both tiptoes The foces on the skeletal pat of the foot ae shown in the diagam In this position, the tibia acts on the foot with a foce F of an unknown magnitude F = F and makes an unknown angle β with the vetical This foce acts on the ankle a hoizontal distance s = 48 cm fom the point whee the foot contacts the floo The Achilles tendon also acts on the foot and is unde consideable tension with magnitude T T 0 and acts at an angle α = 37 with the hoizontal as shown in the figue The tendon acts on the ankle a hoizontal distance b = 60 cm fom the -2 point whee the tibia acts on the foot You may ignoe the weight of the foot Let g = 98 m s be the gavitational constant Compute the toque about the point of action of the tibia on the ankle due to (i) the nomal foce of the floo; (ii) the tendon foce; (iii) the foce of the tibia If the toque about the point of action of the tibia on the ankle is zeo, what is the magnitude of the tendon foce? Solution: We shall fist calculate the toque due to the foce of the Achilles tendon on the ankle The tendon foce has the vecto decomposition T= Tcosαˆi+ Tsinαˆj 10/24/2009 13
The vecto fom the point S to the point of action of the foce is given by ST, = bˆi Theefoe the toque due to the foce of the tendon T on the ankle about the point whee the tibia exets a foce on the foot is then τ T ˆi ( cosαˆi sin αˆj ) sinαkˆ ST, = ST, = b T + T = bt The toque diagam fo the nomal foce is shown in the figue below; The vecto fom the point S to the point whee the nomal foce acts on the foot is given by ˆ ˆ SN, = ( si hj) Since the weight is evenly distibuted on the two feet, the nomal foce on one foot is equal to half the weight, o N ( 12) = mg The nomal foce is theefoe given by N = N ˆj= (1/ 2) mgˆj Theefoe the toque of the nomal foce about the point S is τ = ˆj = ( ˆi ˆj) ˆj= kˆ = (1/2) kˆ SN, SN, N s h N sn smg The foce F that the tibia exets on the ankle will make no contibution to the toque about this point S since the tibia foce acts at the point S and theefoe the vecto 0, If the toque about the point S is zeo then the two vecto contibutions to the toque about the point S add up to zeo SF = btsin αkˆ + (1/ 2) smgkˆ = 0 We can use this toque condition to solve fo the tension in the tendon, T 2 2 (1/ 2) smg (1/ 2)(48 10 m)(75 kg)(98 m s ) 2 49 10 N 2 o = = = bsin α (60 10 m)sin(37 ) 10/24/2009 14