Static-principles Of Momen

4.4 Principles of Moments  Also

known as Varignon’s Theorem “Moment of a force about a point is equal to the sum of the moments of the forces’ components about the point”  For F = F 1 + F 2, MO = r X F1 + r X F2 = r X (F 1 + F 2) =rXF

4.4 Principles of Moments  The

guy cable exerts a force F on the pole and creates a moment about the base at A MA = Fd  If the force is replaced by F x and F y at point B where the cable acts on the pole, the sum of moment about point A yields the same resultant moment

4.4 Principles of Moments  Fy

create zero moment about A M A = Fx h

 Apply

principle of transmissibility and slide the force where line of action intersects the ground at C, F x create zero moment about A M A = Fy b

4.4 Principles of Moments Example 4.6 A 200N force acts on the bracket. Determine the moment of the force about point A.

4.4 Principles of Moments Solution Method 1: From trigonometry using triangle BCD, CB = d = 100cos45° = 70.71mm = 0.07071m Thus, MA = Fd = 200N(0.07071m) = 14.1N.m (CCW) As a Cartesian vector, M A = {14.1k}N.m

4.4 Principles of Moments Solution Method 2:  Resolve 200N force into x and y components  Principle of Moments MA = ∑Fd MA = (200sin45°N)(0.20m) – (200cos45°)(0.10m) = 14.1 N.m (CCW) Thus, M A = {14.1k}N.m

4.4 Principles of Moments Example 4.7 The force F acts at the end of the angle bracket. Determine the moment of the force about point O.

4.4 Principles of Moments View Free Body Diagram

Solution Method 1 MO = 400sin30°N(0.2m)-400cos30°N(0.4m) = -98.6N.m = 98.6N.m (CCW) As a Cartesian vector, M O = {-98.6k}N.m

4.4 Principles of Moments Solution Method 2:  Express as Cartesian vector r = {0.4i – 0.2j}N F = {400sin30°i – 400cos30°j}N = {200.0i – 346.4j}N    For moment, i j k    M O = r XF = 0.4

{

}

− 0.2

0

200.0 − 346.4 0

 = − 98.6k N .m

4.5 Moment of a Force about a Specified Axis  For

moment of a force about a point, the moment and its axis is always perpendicular to the plane containing the force and the moment arm  A scalar or vector analysis is used to find the component of the moment along a specified axis that passes through the point

4.5 Moment of a Force about a Specified Axis Scalar Analysis Example Consider the pipe assembly that lies in the horizontal plane and is subjected to the vertical force of F = 20N applied at point A.  For magnitude of moment, MO = (20N)(0.5m) = 10N.m  For direction of moment, apply right hand rule

4.5 Moment of a Force about a Specified Axis Scalar Analysis  Moment

tends to turn pipe about the Ob axis  Determine the component of M O about the y axis, M y since this component tend to unscrew the pipe from the flange at O  For magnitude of M y, My = 3/5(10N.m) = 6N.m  For

direction of M y, apply right hand rule

4.5 Moment of a Force about a Specified Axis Scalar Analysis  If

the line of action of a force F is perpendicular to any specified axis aa, for the magnitude of the moment of F about the axis, Ma = Fda where da is the perpendicular or shortest distance from the force line of action to the axis

A

force will not contribute a moment about a specified axis if the force line of action is parallel or passes through the axis

4.5 Moment of a Force about a Specified Axis A

horizontal force F applied to the handle of the flex-headed wrench, tends to turn the socket at A about the z-axis  Effect is caused by the moment of F about the z-axis  Maximum moment occurs when the wrench is in the horizontal plane so that full leverage from the handle is achieved (MZ)max = Fd

4.5 Moment of a Force about a Specified Axis  For

handle not in the horizontal position, (MZ)max = Fd’ where d’ is the perpendicular distance from the force line of action to the axis  Otherwise, for moment, MA = Fd MZ = MAcosθ

4.5 Moment of a Force about a Specified Axis Vector Analysis Example M O = r A X F = (0.3i +0.4j) X (-20k) = {-8i + 6j}N.m Since unit vector for this axis is u a = j,

My = M O.u a = (-8i + 6j)·j = 6N.m

4.5 Moment of a Force about a Specified Axis Vector Analysis  Consider

body subjected to force F acting at point A  To determine moment, M a, - For moment of F about any arbitrary point O that lies on the aa’ axis MO = r X F where r is directed from O to A - M O acts along the moment axis bb’, so projected M O on the aa’ axis is M A

4.5 Moment of a Force about a Specified Axis Vector Analysis - For magnitude of M A,

MA = MOcosθ = M O·u a where u a is a unit vector that defines the direction of aa’ axis MA = u a·(r X F)  i

 j

 k

M a = (uax i + uay j + uaz k ) ⋅ rx Fx

ry Fy

rz Fz

- In determinant form,    

4.5 Moment of a Force about a Specified Axis Vector Analysis - Or expressed as,

uax

    M a = uax ⋅ (r XF ) = rx

Fx

uay

u az

ry

rz

Fy

Fz

where uax, uay, uaz represent the x, y, z components of the unit vector defining the direction of aa’ axis and rx, ry, rz represent that of the position vector drawn from any point O on the aa’ axis and Fx, Fy, Fz represent that of the force vector

4.5 Moment of a Force about a Specified Axis Vector Analysis  The

sign of the scalar indicates the direction of M a along the aa’ axis - If positive, M a has the same sense as u a - If negative, M a act opposite to u a  Express M a as a Cartesian vector, M a = Mau a = [u a·(r X F)] u a  For magnitude of M a, Ma = ∑[u a·(r X F)] = u a·∑(r X F)

4.5 Moment of a Force about a Specified Axis  Wind

blowing on the sign causes a resultant force F that tends to tip the sign over due to moment M A created about the aa axis MA = r X F  For magnitude of projection of moment along the axis whose direction is defined by unit vector u A is MA = u a·(r X F)

4.5 Moment of a Force about a Specified Axis Example 4.8 The force F = {-40i + 20j + 10k} N acts on the point A. Determine the moments of this force about the x and a axes.

4.5 Moment of a Force about a Specified Axis Solution Method 1

    rA = {−3i + 4 j + 6k }m   ux = i 1 0 0     M x = i ⋅ (rA XF ) = − 3 4 6 − 40 20 10 = −80 N .m

Negative sign indicates that sense of M x is opposite to i

4.5 Moment of a Force about a Specified Axis Solution We can also compute Ma using r A as r A extends from a point on the a axis to the force    3 4 uA = − i + j 5 5 −3 5     M a = u A ⋅ (rA XF ) = − 3 − 40 = −120 N .m

4

0 5 4 6 20 10

4.5 Moment of a Force about a Specified Axis Solution Method 2  Only 10N and 20N forces contribute moments about the x axis  Line of action of the 40N is parallel to this axis and thus, moment = 0  Using right hand rule Mx = (10N)(4m) – (20N)(6m) = -80N.m My = (10N)(3m) – (40N)(6m) = -210N.m Mz = (40N)(4m) – (20N)(3m) = 100N.m

4.5 Moment of a Force about a Specified Axis Solution

4.5 Moment of a Force about a Specified Axis Example 4.9 The rod is supported by two brackets at A and B. Determine the moment M AB produced by F = {-600i + 200j – 300k}N, which tends to rotate the rod about the AB axis.

4.5 Moment of a Force about a View Free Body Specified Axis Diagram Solution  Vector analysis chosen as moment arm from line of action of F to the AB axis is hard to determine  For unit vector defining direction of AB axis of the rod,     M AB = u B ⋅ (r XF )

 For

simplicity, choose r D

   rB 0.4i + 0.2 j  uB =  = rB (0.4) 2 + (0.2) 2   = 0.894i + 0.447 j

4.5 Moment of a Force about a Specified Axis Solution  For force,

    F = {−600i + 200 j − 300k }N

 In

determinant form,

    M AB = u B ⋅ ( rD XF ) =

0.894 0.447 0 0 0.2 0 − 600 200 − 300

= −53.67 N .m

Negative sign indicates M AB is opposite to u B

4.5 Moment of a Force about a Specified Axis Solution  In Cartesian form,

     M AB = M AB u B = (−53.67 N .m)(0.894i + 0.447 j )   = {−48.0i − 24.0 j }N .m

*Note: if axis AB is defined using unit vector directed from B towards A, the above formulation –u B should be used. M AB = MAB(-u B)