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
= (2
i
+5
j
-6
k
) N has an arm
r
= (3
i
-5
j
+4
k
)m. To
determine the moment exerted by the force with that arm, we use function
CROSS as shown next:
Thus,
M
= (10
i
+26
j
+25
k
) 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:
