Learning Goal:To gain insight into the independence of the scalartriple product from the point on the line chosen as thereference point of the calculation.The magnitude of a moment about a line segmentconnecting points P and Q due to a force F applied atpoint R (with R not on the line through P and Q) canbe calculated using the scalar triple product,MpQupq r × F,=where r is a position vector from any point on the linethrough P and Q to R and upq is the unit vector in thedirection of line segment PQ. The unit vector upq isthen multiplied by this magnitude to find the vectorrepresentation of the moment.As shown in the figure, the member is anchored at Aand section AB lies in the x-y plane. The dimensionsare x₁ = 1.2 m, y₁ = 1.8 m, and 2₁ = 1.6 m. Theforce applied at point C isF = [-170 i +105 j + 110 k] N.FigureB1 of 1Using 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.View Available Hint(s)Hint 1. Calculating the moment about a line segmentThe scalar triple product of the unit direction vector of the line segment, direction vector to the force, and the force vector givesthe magnitude of the moment. To ensure the moment vector lies along the line segment, multiply the magnitude by the unitdirection vector of the line segment.Hint 2. Find the unit vector in the direction of segment ABBecause segment AB lies in the x-y plane, there is no z component in the position vector AB. What is the unit vector from Ato B, UAB.Express the individual components to three significant figures, if necessary, separated by commas.► View Available Hint(s)UAB||Submit呃V—| ΑΣΦ | 1vecX?i, jHint 3. Find the position vector BCWhat is the position vector from B to C?Express the individual components to three significant figures, if necessary, separated by commas. Learning Goal:To gain insight into the independence of the scalartriple product from the point on the line chosen as thereference point of the calculation.The magnitude of a moment about a line segmentconnecting points P and Q due to a force F applied atpoint R (with R not on the line through P and Q) canbe calculated using the scalar triple product,MPQ= = upq r x F,where r is a position vector from any point on the linethrough P and Q to R and up is the unit vector in thedirection of line segment PQ. The unit vector upq isthen multiplied by this magnitude to find the vectorrepresentation of the moment.As shown in the figure, the member is anchored at Aand section AB lies in the x-y plane. The dimensionsare x₁ = 1.2 m, y₁ = 1.8 m, and z1₁ = 1.6 m. Theforce applied at point C isF = [-170 i +105 j + 110 k] N.FigureB1 of 1Using 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.View Available Hint(s)Hint 1. Calculating the moment about a line segmentThe scalar triple product of the unit direction vector of the line segment, direction vector to the force, and the force vector givesthe magnitude of the moment. To ensure the moment vector lies along the line segment, multiply the magnitude by the unitdirection vector of the line segment.Hint 2. Find the unit vector in the direction of segment ABBecause segment AB lies in the x-y plane, there is no z component in the position vector AB. What is the unit vector from Ato B, UAB?Express the individual components to three significant figures, if necessary, separated by commas.UABSubmitVG| ΑΣΦ |VORequest Answer↓↑ vec?i, jHint 3. Find the position vector ACWhat is the position vector from A to C?Express the individual components to three significant figures, if necessary, separated by commas.

Question

Accepted Answer

Given that,
x1=1.2 my1=1.8 mz1=1.6 mF=[ -170 i+105 j+110 k] N