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

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

3. 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

4. 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

5. 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

6. Couple Transformation
Translation to a parallel position
Rotation of a couple

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

8. 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);

9. 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

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

11. 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

12. 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. P Determine the moment couple and the perpendicular distance between the two forces.
Solution:
MA= 320 i -920 j k ft lb
d=9.17 ft

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

14. 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

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