Couples. Resolution of a force into a force and a couple.

Couples. • Resolution of a force into a force and a couple.

F F A Couple A system of forces whose resultant force is zero but the resultant moment about a point is not zero.

A system of forces whose resultant force is zero but the resultant moment about a point is not zero. z d F1 A B F2 O x A Couple MA=|F2|d MB=|F1|d If |F1|= | F2| : MA= MB=Fd y

z Since F2 equals -F1: d F1 M0= r1 X F1+ r2 X (-F1) A rAB r1 F2 B r2 O Therefore: M0= rABX F1 = F1d en x The sum of the moment of two forces about any point O is: M0= r1 X F1+ r2 X F2 =(r1–r2)X F1= rABX F1

1) The magnitude of the moment of the couple. 2) The sense (direction of rotation) of the couple. 3) The orientation of the moment of the couple. The Characteristic of a Couple z d F1 A F2 B O y x

Translation to a parallel position Rotation of a couple Couple Transformation

Changing the magnitude and distance provided the product F.d remains constant.

The Resultant Couple C= SCx+ SCy+ SCz=SCx i + SCy j + SCz k The magnitude of the couple: |C|= SCx2 +SCy2 +SCz2 The couple can also be written as: C=Ce Where: e=cos (qx) i+ cos (qy) j +cos (qz) k The direction: qx=cos-1(Cx/C); qy=cos-1(Cy/C); qz=cos-1(Cz/C);

Example Determine the moment of the couple shown in Fig. P4-82 and the perpendicular distance between the two forces 760 N A 760 N 200 mm B 350 100 mm

|MB|= Mx2 + My2 + Mz2 = 168 Nm 760 N A 760 N 200 mm B i j k -0.10.2 0 -622 -435.90 350 100 mm MB= = 168 k Nm d=M/F=168/760=0.22 m Solution FA = -760 cos(350) - 760 sin(350) =-622 i – 435.9 jN rBA = -0.1 i + 0.2 jm

Determine the moment of the couple shown in Fig.P4-81 and the perpendicular distance between the two forces. Solution: MA= 3030 k in lb d=8.66 in

Two parallel forces of opposite sense, F1 = (-70i - 120j - 80k) lb and F2 = (70i + 120j + 80k) lb, act at points B and A of a body as shown in Fig. P4-83. Determine the moment couple and the perpendicular distance between the two forces. Solution: MA= 320 i -920 j +1100 k ft lb d=9.17 ft

-F O M0  O  F d F F Resolution of a force into force and a couple

Example Replace the 350 N force shown in Fig. P4-104 by a force at point B and a couple. Express your answer in Cartesian form. C

-FC rBC rBC = 0.1 i + 0.25 jm FC C i j k 0.10.25 0 268.1 -2250 CB C= = -89.5 k Nm FC Solution C FC = 350 cos(400) - 350 sin(400) = 268.1 i – 225 jN

Couples. Resolution of a force into a force and a couple.