Question

Learning Goal: To gain insight into the independence of the scalar triple product from the point on the line chosen as the reference point of the calculation. The general process (not referenced to Figure 1) to calculate the moment of a force about a specified axis is as follows: The magnitude of a moment about a line segment connecting points P and Q due to a force F applied at point R (with R not on the line through P and Q) can be calculated using the scalar triple product, MPQ=uPQ⋅r×F, where r is a position vector from any point on the line through P and Q to R and uPQ is the unit vector in the direction of line segment PQ. The unit vector uPQ is then multiplied by this magnitude to find the vector representation of the moment. As shown in the figure, the member is anchored at A and section AB lies in the x–y plane. The dimensions are x1 = 1.6 m, y1 = 1.7 m, and z1 = 1.5 m. The force applied at point C is F=[−165 i+80 j+130 k] N.

Part B - Calculating the moment about AB using the position vector AC Using the position vector from A to C, calculate the moment about segment AB due to force F. Express the individual components to three significant figures, if necessary, separated by commas.

Part C - Calculating the moment about AB using the position vector BC Using the position vector from B to C, calculate the moment about segment AB due to force F. Express the individual components to three significant figures, if necessary, separated by commas.

Answer 1

Thank you.please like the answer .ASAP

Here ' ② given recter AB = {X2, 42,227 = <1.6,1.7, 1.5). m. Ê = [-16sî, +80 § + 130 û] N The moment of AB = A x ² = î (101) - ġ (455.5) + (408.5) 1.6 | 165 1.7 80 1.5 130 Hence, Required moment = 101 î – 455.5 Î + 408-5ý τ © due Ā to force Nole moment about AB segment Ton - AB XF = 101 î - 455.5 1 + 408.5 ☆ Iltne = 620.1237 è 620.124 ·

Answer 2

on givensi=1.6m, y = 1.1m, 21=1.5m. F = [-1651 +80j +130 kN. The Coordinates of various points A, B and Coordinates of A 10,0,0) Coordinates of B,16,17,0) Coordinates of Cicl 6, 17, 15). to the equation for the position vectory AB, TAB -rg-ra = (1.6-0) 1+(1.7-6)j + (0-0) Calculate the magnitude of vector YAB IYABI = 21.6²4 1.72 = 2.334 The value of unito cedor the UAB= AO = 1.61 +12) VAB - 2.33 2.33 = 0-6851 +0.729 The position of rector of AC (1-6-6)+(1:7-6)j +0.5-byk YA- 16)+1.75 +1.sk

6) The aquation for the calculation of moment about point MA Tmc x = (1+73+5)x145+8° +13 ok) 11:7 15- 1-165 180 130 = 1011 - 455.5j tk (1525) (408.5) - 1011 - 455.5j + 4085k The moment about AG MAB = MAGMA C06851 +0.7283) Coli-455.5) +4685) = 69.185-331604 =-262.8419 Nm moment in anti clerk wodnech The equation in vector form white MAB = MAB UAB = -22.419 (0.6851+0.28) M -179751-90)

Imag= -179.751 - 1915 to pult) The position roctor BC Vec = (1.6-1.671 +(1.7-1.7)) + (1500- | |-5k MB = PBC XF = 1.5k(-1651 +80)+bok). - 165 80 130 - 1201 - 247.5 The moment about the point AB MAB= MABMB - (0-6851+07283) (1201-3475) = 872-180-18 = -97,98 The moment in the vector form MAB = MABUAB =(-97-98) (06859 +.7283) Mag=-67 11631-71.329 /nych The voctor