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Transcript
3.2 Cross Products The cross product is only defined for vectors in three dimensions. For three dimensional vectors u and v that are specified by a magnitude and direction, the cross product uv is the vector defined as follows. 1. uv points in the direction perpendicular to u and v. Its direction is determined by the right hand rule. Position your right hand so the fingers lie along u and curl toward v. Your thumb should point in the direction of uv. 2. magnitude of uv = (magnitude of u) (magnitude of projection of v on the plane perpendicular to u) = | u | | v | sin = area of parallelogram with sides u and v 1 = 2 (area of triangle with sides u and v) where = angle between u and v Alternatively, to form uv from u and v do the following. 1. Project v on the plane perpendicular to u. 2. Rotate that vector counterclockwise by 90 in the plane perpendicular to u when viewed from the u direction. 3. Multiply that vector by | u |. Note that if u and v point in the same direction then uv is zero. Torque. One application of the cross product is to calculating the torque of a force acting on an object. Suppose we have an object that occupies more than a single point in space. Let C be the center of mass of the object (or some other designated point on the object). Suppose a force F acts on the object at some point on the object. Let r = vector from the C to where the force is applied. Then = the torque due to the force = rF 1-1 The torque turns out to be a good description of the rotating effect of the force. In fact, the force has two effects. 1. The center of mass of the object moves as if all the mass of the object were concentrated at the center of mass and the force acted at the center of mass. 2. The object tends to rotate about the axis through . The rotating power of the force depends on a. The distance | r | of the point of application of the force from the center of mass. b. The magnitude of the projection of the force on the plane perpendicular to r. In fact the rotating power of the force is equal to | r | | F | sin = magnitude of = | | Quantitatively one has d I dt = | | where I = moment of inertia of object about the axis through = angular velocity about the axis through r F In fact the above equation extends to a vector equation d I dt = where I = moment of inertia matrix of the object = angular velocity vector Example. (Problem #40 on p. 824 of the text) A certain object consists of two rods each 4 ft long and joined at a right angle at the one of the ends of each rod. The first rod extends from P 4 ft to the right to a point R. The second rod goes down 4 ft from R to a point Q. A force F of 36 lb acts at Q and points in the plane of the rod down and to the left at an angle of 30 with respect to the horizontal. Find the torque due to the force with respect to the point P. Solution. Let r = PQ be the directed line segment from P to Q. Then the torque = r F. One has | | = | r | | F | sin where = angle between r and F = 45 + 60 = 105. So | | = (4 2 ft) (36 lb) sin105 = 144 2 sin105 ft-lb = 144 (1.414) (0.9659) ft-lb = 196.7 ft-lb is directed perpendicular to the plane of the rod. 1-2 Algebraic properties of the cross product. The cross product has the following algebraic properties. In the following u, v and w are any vectors and t is any number. 1. uv = - vu (the cross product is antisymmetric) 2. uu = 0 3. (tu)v = t (uv) (the cross product is homogeneous) 4. (u + v)w = uv + vw (distributive) u (v + w) = uv + uw (distributive) 5. ii = 0 ij = k ik = - j ji = - k jj = 0 jk = i ki = j kj = - i kk = 0 Sketch of proof. (1) This follows from the fact that when you use the right hand rule to find the direction of vu, it points in the opposite direction as the direction of uv. (2) This follows from the fact that the angle between u and itself is zero. (3) Consider the case t > 0. Note that both tu and u have the same direction. Therefore (tu)v and uv have the same direction. So (tu)v and t(uv) have the same direction. Furthermore (tu)v and t (uv) have the same magnitude. (4) This is a little more subtle. It follows from the fact that projecting on the plane perpendicular to u, rotating about the line through u and multiplying by u all preserve sums. (5) The fact that ii = 0, jj = 0 and kk = 0 all follow from property (2). To see that ij = k, note that k is perpendicular to i and j and has the correct direction determined by the right hand rule. Also it has the correct length for ij. The other relations are proved in the same fashion. 1-3