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4.4 Principles of Moments Also

known as Varignons 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 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

Apply

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

4.4 Principles of MomentsSolution 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 MomentsSolution Method 2: Resolve 200N force into x and y components Principle of Moments MA = Fd MA = (200sin45N)(0.20m) (200cos45)(0.10m) = 14.1 N.m (CCW) Thus, M A = {14.1k}N.m

4.4 Principles of MomentsExample 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 MomentsView Free Body Diagram

Solution Method 1 MO = 400sin30N(0.2m)-400cos30N(0.4m) = -98.6N.m = 98.6N.m (CCW) As a Cartesian vector, M O = {-98.6k}N.m

4.4 Principles of MomentsSolutionMethod 2: Express as Cartesian vector r = {0.4i 0.2j}N F = {400sin30i 400cos30j}N = {200.0i 346.4j}N For moment, i j k M O = r XF = 0.4 0.2 0 = 98.6k N .m

{

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200.0 346.4 0

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

A

4.5 Moment of a Force about a Specified AxisA

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 = MAcos

4.5 Moment of a Force about a Specified AxisVector AnalysisExample 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 AxisVector 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 AxisVector Analysis- For magnitude of M A,

MA = MOcos = M Ou awhere u a is a unit vector that defines the direction of aa axis MA = u a(r X F) - In determinant form, i j k M a = (uax i + uay j + uaz k ) rx Fx ry Fy rz Fz

4.5 Moment of a Force about a Specified AxisVector Analysis- Or expressed as, M a = uax (r XF ) = rx uax Fx uay ry Fy u az rz 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 AxisVector 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 AxisExample 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 AxisSolution Method 1 rA = {3i + 4 j + 6k }m ux = i 1 0 0 M x = i (rA XF ) = 3 4 6 40 20 10

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

= 80 N .m

4.5 Moment of a Force about a Specified AxisSolutionWe can also compute Ma using r A as r A extends from a point on the a axis to the force 3 i +4 j uA = 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 AxisSolution 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 AxisSolution

4.5 Moment of a Force about a Specified AxisExample 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 DiagramSolution 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

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

simplicity, choose r D

4.5 Moment of a Force about a Specified AxisSolution For force, In

F = {600i + 200 j 300k }N

determinant form,0.894 0.447 0 0 0.2 0 600 200 300

M AB = u B ( rD XF ) = = 53.67 N .m

Negative sign indicates M AB is opposite to u B

4.5 Moment of a Force about a Specified AxisSolution 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)

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