CE 201 - Statics

Lecture 12

MOMENT OF A FORCE ABOUT A SPECIFIED AXIS

Scalar Analysis

Mo = (20 N) (0.5 m) = 10 N.m (F tends to turn about the Ob axis)How can we determine the component of Mo about the y-axis (My)?If is known, then My = Mo cos ()It is clear that in order to find MyDetermine MoResolve Mo along the y-axis

A

F = 20 N

0.5 m

O

MO

dy = 0.4 m

b

dx = 0.3 m

x

y

z

My

Generally, to find My directly, it is necessary to find the perpendicular distance from the line of action of the force to the y-axis (da), then:

Ma = F (da)where:a is the aa-axisda is the perpendicular distance from F line of action to aa-axis

Note: F will not have a moment about an axis if F is parallel to that axis or passes through it.

Vector Analysis

The two steps performed with scalar analysis can be performed with vector analysis, therefore:Mo = rA F = (0.3 i + 0.4 j) (-20 k) = (-8 i + 6 j) N.m

The component of Mo along the y-axis is determined from the dot product, then:

My = Mo . u = (-8 i + 6 j) . j = 6 N.m

A

F = (- 20 k) N

rOA

O

MO

0.4 m

b

0.3 m

x

y

z

My

u = j

Generally, if a body is subjected to force F acting at point (A), what is Ma??

A

F

rOA

O

MO = r F

b

a

b

a

Ma

(1) Calculate Mo ( Mo = rOA F )The moment axis is perpendicular to the plane containing F and r (say bb)

(2) Ma is the component along aa, where its magnitude is found from the dot product, ( Ma = Mo cos () = Mo . ua ), where ua is the unit vector defining the direction of aa, then:

Ma = (r F) . ua

Since the dot product is commutative (A . B = B . A), then:

Ma = ua . (r F)

The combination of the dot and cross products is called the triple scalar product.

i j k

Ma = (uax i + uay j + uaz k ) . rx ry rz

Fx Fy Fz

Or simply,

uax uay uaz

Ma = ua . (r F) = rx ry rz Fx Fy FzIf Ma is +ve, then it has the same direction as ua and if Ma is –ve, then it is in the opposite direction of ua.Whence Ma was calculated, Ma can be expressed in Cartesian vector form by:

+ Ma = Ma ua = [ ua . (r F) ] ua

The resultant moment of a system of forces about the axis is:Ma = [ ua . (r F) ] = ua . (r F)