The moment of the force about joint B is given by
M_{B}
= r_{BQ} ×> F
The cross product in determinant form is
Expanding the equation gives
M_{B} = (r_{BQy} F_{z}
 r_{BQz} F_{y})i + (r_{BQz} F_{x}
 r_{BQx} F_{z})j
+ (r_{BQx} F_{y}  r_{BQy} F_{x})k
Substituting the appropriate values and simplifing to determine the
moment at joint B gives
M_{B} = (2,400 + 3,600 sinα)i
+
(3,600 cosα)j + (3,600 sinα)k
If the magnitude of M_{B} is plotted as a function of
the angle α, then the moment is a maximum
at
α = 90°, with a value of 6,997 Ncm or about 70 Nm. This
is approximately 62% greater than the moment at α
= 0°.
