60
B
60
100 mm
150 mm
300 mmn
Given
Force magnitude:
F=450N
From the diagram, direction of F:
cos 60° in x
• cOs 60° in y
• COs 45° in z
So:
F= -450cos 60° i + 450 cos 60°j+ 450 cos 45° k
F=-225 i+ 225j + 318.2 k
Position vector from A toB
From diagram:
•a=-0.15, m
•y = 0.30, m
•z= 0.10, m
r'AB =-0.15i+0.3j +0.1k
Moment about x-axis
We use:
M, = u, · (IAB x F)
Since u, =ithis becomes:
M, = i-component of (rAB x F)
Compute only i-component
(shortcut)
M, = (0.3) (318.2) - (0.1)(225)
M, = 95.46 – 22.5
M, =72.96 ~ 73
VFinal Answer
M, =73 N-m