Question
I don't get why Tmin=T(2/3i-1/3j+2/3k)
how can I know T vector is 2/3i-1/3j+2/3k
A 450-lb load hangs from the comer C of a rigid piece of pipe ABCD which has been bent as shown. The pipe is supported by the ball-and-socket joints A and D. which are fastened, respectively, to the floor and to a vertical wall, and by a cable attached at the midpoint E of the portion BC of the pipe and at a point G on the wall. Determine (a) where G should be located if the tension in the cable is to be minimum. (b) the corresponding minimum value of the tension. The free-body diagram of the pipe includes the load W = (- 450 lb)j, the reactions at A and D. and the force T exerted by the cable. To eliminate the reactions at A and D from the computations, we express that the sum of the moments of the forces about AD is zero. Denoting by lambda the unit vector along AD, we write The second term in Eq. (1) can be computed as follows: Substituting the value obtained into Eq. (1), we write Minimum Value of Tension. Recalling the commutative property for mixed triple products, we rewrite Eq. (2) in the form which shows that the projection of T on the vector lambda times is a constant. It follows that T is minimum when parallel to the vector Since the corresponding unit vector is , we write Substituting for T and lambda times in Eq. (3) and computing the dot products, we obtain 6T = -1800 and. thus, T = -300. Carrying this value into (4), we obtain Tmtn = -200i + 100j - 200k Tmtn = 300 lb Location of G. Since the vector and the force Tmtn have the same direction, their components must be proportional. Denoting the coordinates of G by x, y, 0, we write x - 6/-200 = y - 12/+100 = 0 - 6/-200 x = 0 y = 15ft
Explanation / Answer
It is clear...as it said...For T to be min...It must be parallel to the cross product on lambda and vector AE...i. e. ( lambda X AE).......... Now...we need to calculate the vector...(lambda x AE)...which comes out to be...4i - 2j + 4k.... We know that a unit vector represents the direction of any vector... I hope u know how to find a unit vector of a given vector....we divide the given vector by the sqrt of ( sum of squares of coeffiecients )... Here the unit vector will be obtained by dividing the vector (lambda x AE) by SQRT ( 4^2 + 2^2 + 4^2 )...this comes out to be 6. The unit vector in the direction of (lambda x AE)...i.e...in the direction of T...will be...2/3i-1/3j+2/3k....... Now...it is assumed that the magnitude of vector T is "T"....so to find the tension vector...we multiply "T" with the unit vector in that direction....i.e.... "T" times (2/3i-1/3j+2/3k)
