Page 9-17
Thus, the result is θ = 122.891
o
. In RPN mode use the following:
[3,-5,6] ` [2,1,-3] ` DOT
[3,-5,6] ` ABS [2,1,-3] ` ABS *
/ ACOS NUM
Moment of a force
The moment exerted by a force F about a point O is defined as the cross-
product M = r×F, where r, also known as the arm of the force, is the position
vector based at O and pointing towards the point of application of the force.
Suppose that a force F = (2i+5j-6k) N has an arm r = (3i-5j+4k)m. To
determine the moment exerted by the force with that arm, we use function
CROSS as shown next:
Thus, M = (10i+26j+25k) m⋅N. We know that the magnitude of M is such
that |M| = |r||F|sin(θ), where θ is the angle between r and F. We can find
this angle as, θ = sin
-1
(|M| /|r||F|) by the following operations:
1 – ABS(ANS(1))/(ABS(ANS(2))*ABS(ANS(3)) calculates sin(θ)
2 – ASIN(ANS(1)), followed by NUM(ANS(1)) calculates θ
These operations are shown, in ALG mode, in the following screens:
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