To gain insight into the independence of the scalar triple product from the poin
ID: 1718362 • Letter: T
Question
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 magnitude of a moment about a line segment connecting points P and 0 due to a force F applied at point R (with R not on the line through P and O) can be calculated using the scalar triple product. where r is a position vector from any point on the line through P and Q to R and U_PQ is the unit vector m the direction of line segment PQ The unit vector U_PQ 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 AS lies in the x-y plane The dimensions are x_1 =1.6 m, y_1 = 1.7 m. and z_1 = 1.6 m. The force applied at point C is F = [-220 i+ 60 j+ 135 k] N. Calculating the moment about AB using the position vector AC Using the position vector from A to C. calculating the moment about segment AB due to force F Express the individual components to three significant figures. If necessary, separated by commas. Calculating the moment about AB using the position vector BC Using the position vector from 0 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.Explanation / Answer
>> Writing the co-ordinates:
>> A = (0,0,0)
>> B = (x1.y1.0) = (1.6,1.7,0)
>> C = ((x1, y1, z1) = (1.6,1.7,1.6)
>> As, Force, F = (- 220 i + 60 j + 135 k) {N}
>> As, Position Vector AC = (1.6 i + 1.7 j + 1.6 k) - (0)
=> AC = (1.6 i + 1.7 j + 1.6 k)
=> Moment of Force F about AB by AC = AC X F = (1.6 i + 1.7 j + 1.6 k) X (- 220 i + 60 j + 135 k)
=> Moment,M = i (1.7*135 - 60*1.6) - j (1.6*135 + 1.6*220) + k (1.6*60 + 220*1.7)
=> M = 133.5 i - 568 j + 470 k ........ANSWER...........PART A......
>> As, Position Vector BC = (1.6 i + 1.7 j + 1.6 k) - (1.6 i + 1.7 j + 0 k)
=> BC = (1.6 k)
=> Moment of Force F about AB by BC = BC X F = (1.6 k) X (- 220 i + 60 j + 135 k)
=> Moment,M = - 352 i - 96 j
=> M = - 352 i - 96 j ........ANSWER...........PART B......
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