Question

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,

M_PQ=u_PQ · r × F

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 in the direction of line segment P Q. 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 A B lies in the x-y plane. The dimensions are x₁=1.6 ~m, y₁=1.8 m, and z₁=1.6 m. The force applied at point C is F=[-165 i+105 j+180 k] N

Part B - Calculating the moment about A B using the position vector AC

Using the position vector from A to C, calculate the moment about segment A B due to force F.

Part C - Calculating the moment about A B using the position vector BC

Using the position vector from B to C, calculate the moment about segment A B due to force F.

Answer 1

ftns :- Given that, x,= 1.6m, y,= l.8m and , = 1.Gm F 165i t105j + 180k (N) where Y is rom A to C B UAB. YXF MAR Y1.6+t.8i+t.6K (m) & UpR = 1.6 it1-83 t 6k N.G1-8t0 j k YXF (180 X18-105 X -6) i -(180x1-6t t65x1-6)j .6 .8 1-165 o5 t80 +(180X1.g+ \05X1-6) K YXF 156i-552j +U65 K (N-m). MAB (6-6644i+0.44구4j +oK). (156i-552jtu65 K). MAB li) (N m) 0.6644 (156) 103.65 MAe ) 0.474 (-55a) = -412-5t (N.M) MAB CK)= o(465) = 0 (N m) MA8 = ( 103.65 ,-412.57, 0)N-m Y is from 8 to c Y= 01toj +t.6K (%,,9i,0) to (x, ,1,2) & F=-165i + 105j 180K t(0,6,2) [72-,2-,2-2 K. YXf = 0 106 165 to5 180 YXF= 0-(165 XI.6)i (t65 Xl-6)3 + Co-0)K (N.m. TXF= 168 i - 264j +0K (N m) MAB uAB. (YXF) & uaR= 0.6644i to.744jtok (0.66u4i +0.tutujtok). (-168i - asujtok) (N.m) MAB ci)=-t68 Co.6644) =t11.62 N.M MAG () N.m MAA CK) =oto) t61- = (ERE O ) H9C- N-M MAB (-tti.62,-191-31, 0) N.m