Moment of a force – HP 48gII User Manual

Page 292

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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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