Question

a) Using the position vector from A to C, calculate the moment about segment AB due to force F

b) Using the position vector from B to C, calculate the moment about segment AB due to force F

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 u_{P Q} is the unit vector in the direction of line segment P Q. The unit vector u_{P Q} 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 A B lies in the x-y plane. The dimensions are x1=1.6 m, y1=1.8 m, and z1=1.5 m. . The force applied at point C is F=[-165 i+95 j+190 k] N

Answer 1

"Vector F = -165 i + 95 j + 190 k N position vector: vector r_c/A = 1.6 i + 1.8 j + 1.5 k m also, vector AB = 1.6 i + 1.8 j therefore unit vector, U_AB = 1.6 i + 1.8 j/ squareroot 1.6^2 + 1.8^2 Rightarrow U_AB = 0.664 i + 0.747 j moment about AB will be: M_AB = u_AB .(vector r_CA times vector F) Rightarrow M_AB = (0.664 i + 0.747 j) .[(1.6i + 1.8j + 1.5 k) times (- 165i + 95j + 190 k)] =