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

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

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

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

The Characteristic of a Couple 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. z d F1 A B F2 O y x

Couple Transformation Translation to a parallel position Rotation of a couple

Changing the magnitude and distance provided the product F 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=C e 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

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

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

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

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

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