Question

(2.2 Coaching Activity Three-dimensional Force Systems 4 of 5 > II Review Four forces are shown: F1, F2, F3, and F4. (Figure 1) Classify the x, y, and z components of each vector as being less than zero. equal to zero, or greater than zero. Drag the appropriate items to their respective bins. Learning Goal: To use the conditions of equilibrium to solve for unknown forces in a three-dimensional force system. For particle equilibrium, the sum of all forces must be zero: F =0 If the forces are resolved into their i, j, and k components, the condition for equilibrium becomes Reset Help Fi+FJ+ Fik=0 which results in the following three scalar component equations that can be used to solve for unknowns: 000000000000 Fx=0 Fy=0 Fo=0 Less than zero Equal to zero Greater than zero Submit Previous Answers Correct When using vectors, pay attention to the location of the vector's head and tail. F2 is pointing toward the origin rather than away from the origin. If a vector is defined by two arbitrary points A WAYA,ZA) and B2B, YB, 2B), where point A is the head and point is the tall, the vector's resulting components arer = ( R)i + (WA W ) + (24 ) Part B As shown, a system of cables supports a bucket. (Figure 2) The dimensions in the figure are as follows: 21 = 4.35 ft. 23 = 1.40 ft. = 1.60 ft. W2 = 3.50 ft. 21 = 2.55 ft, and 2= 3.00 ft. If the bucket and its contents have a combined weight of W = 25.5 lb, determine TA,TB, and Tc. the tensions in cable segments DA, DB, and DC, respectively. Express your answers numerically in pounds to three significant figures separated by commas. View Available Hint(s) Figure 1 of 2 > YO AED Ivec O ? TATRIC 10.9,6.92,18.7 Submit Previous Answers * Incorrect; Try Again Part C Complete previous part(s) Provide Feedback Next >

Answer 1

IDA1 = The woydinales are: A (4:35,0,2055) 8 (1.4,0,0)(C033.533) P(1.4,1601 DA = OA-OD = (4.35i+gj+ a-55k) -(1:41+1:6j+ok) = 2.951 -1.6j+2:55 * • * 12.9546-16):55? - Yol ft UDA DA = 2.951-16)+2.55K = 0 +1 -0138j+0.65 TDA =TAD (0.7 -0.38j +0.605K) = 0 IT DAI-0:38TDAJ to GOSTOAK PB - (08-00) = (141+0jtok)-(1-41 +16jtok)= 01-1:6;7012 IDBI= $1.62 = 16 UDB= oi- 16jtok IDAL 421 1.6 - oijtok TOB = - TbBj DC = OC-OD= Coitasj+36)-(1.41 +1:6jtok) - =-41 +19j+3K. Loc=743719232 - 3.8171 Upc = -147+19j+3K -0.367i+ 0.498j+0.784 K 3181 TDC = -0.367 TOC 1+0.498 Tojito 787 TDCK W = -Wk = - 25.5

TOA+TDBT TOctwo 0.4TDAT -038 TORJ +0.60STDAK - T08j +(-0-367Toci to.498 Tocj+0787TOCK = 0 Equate i, j&K components, 0 ITDA - 0.367Tbc = 0 0 - 0.38 TDA - TOB+0.498 TDC=0 @ 0.605 TDA +0.78770c -25.5 =0.- Solve o, & TDA - 12.11b TOB - 6.89 16 TDC - 23.1 lb. TA, TB, TC = ( 1211, 6.89, 23.1|1b.