Physics Tutoial V1 2D Vectos 1 Resolving Vectos & Addition of Vectos A vecto quantity has both magnitude and diection. Thee ae two ways commonly used to mathematically descibe a vecto. y (a) The pola fom:, yj (b) The Catesian fom: xi+yj, whee the unit vectos i and j indicate the diections of xi x the x and y axes espectively. Conveting between the two foms equies the use of Pythagoas ( = x2 + y 2 ) and basic tigonomety (x = cos, y = sin ) 1.1 Expess each of the vectos shown below in Catesian fom. (a) (b) 8 132 (c) (d) 228 312 1
1.2 Expess each of the vectos shown below in pola fom. (a) 2i + 3j (b) 2i + 3j 3j 3j 2i 2i (c) 2i 3j (d) 2i 3j 2i 2i 3j 3j 1.3 (a) Expess each of the vectos A, B, C and D in Catesian fom: 60 y B E A 30 A = B = 6 C = 5 D = 5 x 5 C 36.9 D (b) Calculate the vecto A + B + C + D, and expess you answe in both Catesian and pola fom. (c) Calculate the vecto E, again expessing you answe in both Catesian and pola fom. 2
1. Given the two vectos a = 3i + 5j and b = 5i + 2j : (a) Calculate s = a + b and t = a b (b) Expess each of the vectos s and t in pola fom, and hence detemine the angle between these vectos. (c) Detemine the Catesian fom of the unit vecto ŝ and use this to find (again in Catesian fom) the vecto u which has the same magnitude as t and same diection as s. 1.5 An object is located at the point with Catesian co-odinates (2,3) befoe moving to a new position with co-odinates (5,7). (a) Detemine the initial and final position vectos ( i and f ) of the object. (b) Calculate the displacement vecto s = f i fo the paticle s motion, in both Catesian and pola foms. 1.6 Two foces ae acting on an object. The fist foce F 1 is 5 N in the x diection and the second, F 2 is 10 N at an angle of 60 above the x-axis. Calculate the magnitude and diection of the esultant foce F 1 + F 2. 1.7 A beam AB of length m is hinged to a wall at the end A and suppoted by a ope attached to the othe end B as seen below ight: The weight foce W of magnitude 60 N acts downwad at the cente of the beam, and the ope applies a tension foce T with magnitude 50 N at the end B. The wall pushes on the beam at point A with the eaction foce R. 3 m R T If the beam is not moving (so the sum of the foces acting on it is zeo), it is in static equilibium. Calculate the eaction foce R equied fo this to be tue. Expess you answe in both Catesian and pola foms. A m W B 3
2 Scala Poduct The scala (o dot) poduct of two vectos a and b is defined by: a b = ab cos whee is the angle between the two vectos. Note that a b is a SCALAR (i.e. NOT a VECTOR) Fom this definition it should be clea that fo any vecto a: a a = a 2 and that if a b = 0, then the two vectos ae pependicula to each othe. (Assuming of couse that neithe vecto is the zeo vecto!) Fo 2-dimensional Catesian vectos: a b = (a x i + a y j) (b x i + b y j) = a x b x + a y b y and thus: a = a 2 x + a 2 y Reaanging the definition of the scala poduct enables us to calculate the angle between two vectos a and b: ( ) a b = cos 1 ab 2.1 A fomula we will see soon that uses the dot poduct is fo the wok W (SI units: J) done by a foce F (SI: N) acting though a displacement s (SI: m): W = F s Calculate the wok W done when F = 10 N, s = 5 m and the angle between these vectos is : (a) 37 (b) 13 (c) 90 (d) 180
2.2 Calculate the wok done by the foce F = 20i 30j N when it acts though a displacement s = 2i 5j m. Also find the angle between these two vectos. 2.3 Use the dot poduct to find the angle between the vectos s and t peviously calculated in Q1.. 2. Which of the following vectos ae pependicula to each othe? (a) 2i 3j (b) 3i + j (c) 1.5i + j (d) 2 3 i 1 2 j 2.5 Given the vectos: u = 3i + j and v = 2i + Xj, whee X is an unknown numbe, detemine the value of X fo which the two vectos will be: (a) paallel (b) pependicula 2.6 Use the scala poduct to pove that the angle between the diagonals of the ectangle at ight is given by: ( ) a = cos 1 2 b 2 a 2 + b 2 a b Answes 1.1 (a) 2.7i + 3.0j (b) 2.7i + 3.0j (c) 2.7i 3.0j (d) 2.7i 3.0j 1.2 (a) = 3.6, = 56 (b) = 3.6, = 12 (c) = 3.6, = 236 (d) = 3.6, = 30 1.3 (a) A = 3.6i+2.00j; B = 3.00i+5.20j; C = 3.5i 3.5j; D =.00i 3.00j (b) 0.92i + 0.66j; 1.13 at 36 (c) E = B A = 6.6i + 3.20j; 7.21 at 153.6 1. (a) s = 2i + 7j; t = 8i + 3j (b) s = 7.28 at 106 ; t = 8.5 at 21 ; 85 (c) ŝ = 0.27i + 0.96j; u = tŝ = 2.31i + 8.20j 1.5 i = 2i + 3j; f = 5i + 7j; s = 3i + j = 5 at 53 1.6 13.2 N at 1 1.7 R = 0i + 30j N = 50 N at 37 2.1 (a) 39.9 J (b) -39.9 J (c) 0 (d) -50 J 2.2 190 J, 11.9 2.3 85 2. (a) and (c) ae pependicula, also (b) and (d) 2.5 (a) 8 (b) -1.5 3 5