Question

4-47. Determine the magnitude of the moment of each of the three forces about the axis AB. Solve the problem (a) using a Cartesian vector approach and (b) using a scalar approach. IF 0N 25 N F -45 N SOLUTION a) Vector Analysis Position Vector and Force Vector: 1- {-1.5j} m - - 0 F:- {-60k}N F. - {8Si) N Unir Vector Along AB Aris: (2-0)i + (0 - 1.5) * V(2 - 0) + (0 - 15) F - 45j}N -0.8i - 0.6 Moment of Each Force About AB Axis: Applying Eq.4-11, we have (MAB)- UAB (XF) 0.8 -0.6 0 | - O -1. 5 0 0 0 -60 -0.8 (-1.5)(-60) - 0] -0+ 0 - 72.0N·m (MAD) - 48" (r; XB) 0.8 -0. 60 - 0 0 0 - 0 85 0 0 (MAB) - :48(XF) 0.8 -0. 6 0 - 0 0 0 - 0 0 450 b) Scalar Analysis: Since moment arm from force F, and F3 is equal to zero, (MAB)2-(MB) - 0 Moment arm d from force F to axis AB is d - 1.5 sin 53.13 - 1.20 m, (MAB) - Fd - 60(1.20) - 72.0 N·m

Answer 1

SOLUTION: a) using a cartesian vector approach Moment of each force about AB axis is (Mae): = UAB (V2XF) (MAB) 2 = UAB (82 XF2) (MAB)3 = UAB. (73XFZ) where is unit vector along AB axis. Jis position vector Fir Force vector r = (-.55)m, 72 =Y2 = 0 F2 = (-60K)N; F2 = (85.0N, Fz=(453) ſ Upe= (-o)i + (0-1.5); (2-0)2+(o-5) 2 = 0.8i - 0.6j Now (Mae)? = UAB: (x,x Ft) = (0.8 i-065). (-15J XC-50K)} - | 0 - 6 2 1 10 -115 0. 10 0 -60 = 0.8[6-1,5)(-60) - ]-o to (MAB)s = 720 Nimt due to force fe (MAB)2 = Une: ( 72 x F2) = 68.8.4-0.63). (o *C853)) = ' 0.8 -0.6 :01 | 85 0 0 (MRB)2 = ole Due to Force Fa

(MAB), = UAB (m3 * F) (0.84-0.65) · (ox (451) 1 0.8 -0.6 .0 45 0 Inez = 0 Due to Fasce Fz using a scalar apporo act te) Fo=G0N 2m t.5m Since arm Force F2 t F3 is equal to zero (MAB)2 = (F2x.de)=0 Due to Force Fe I (MRB) 3= (F3xd3)=0 * Due to force Fa For Force FL arm distance da= 1.5 sin(53.13) = 1.20 m. (Mne), = Fxd7 = 60 x 1.20 Mae) = 72.0 Nim 1 Due to Force Fe