Question

The line of action of force F passes through point B. Find the moment due to force F about axis AC, given:
  F̄ = <10, 70, -10> lbs,   A_X = 6 ft,   A_Y = 3 ft,   A_Z = 7 ft,   B_X = 7 ft,   B_Y= 5 ft,   B_Z = 9 ft,   C_X = 5 ft,   C_Y = 5 ft,   C_Z = 1 ft

M_AC =      lb·ft

M̄_AC= <   ,   ,   >  lb·ft

Find the moment about points A, B, C, D, and E, due to force F, given:

  F = 50 lbs,   L₁ = 6 ft,   L₂ = 7 ft,   L₃ = 4 ft

M_A =   lb·ft
  M_B =   lb·ft
  M_C =   lb·ft
  M_D =   lb·ft
  M_E =   lb·ft

3 ーF 1

Answer 1

Thus, the unit vector u along the aris Ac is given as 1, 2,-6 1, 2,6 Substitute, B,-7, By=5, B2-3

F 10, 0, -10 so, 10刊-10 10 2O8 -6 80-x-〈-1, 2, -6 80 41 41 30 and, MAC -80く1,-2,6 〉 [lbs. ft] So, MAc VT1 (b) MAFL 50x6 300 bs.Ft MB= MN FL1= 300 lbs. Ft - 50x 7 350 lbs ft MpF2

Since. the points and the force is planar, therefor, moment isbout point is given simply by is the perpendicular From the point to the line of action of the force and when the moment tend to yotate anticlockwise it is considered positive and negative otherwise Thot's why and the moment is positive because the force F tends to rotate anticloctwise about point A and same is the case with other points