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**ENGR-1100 Introduction to Engineering Analysis**

Lecture 10

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**Previous Lecture Outline**

Vector representation of a moment: - Moment of a force about a point. -Moment of a force about a line.

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

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

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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

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**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

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**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

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**Couple Transformation**

Translation to a parallel position Rotation of a couple

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**Changing the magnitude and distance provided the product F**

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

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**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);

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Example P4-82 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

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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

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**Class Assignment: Exercise set 4-81**

please submit to TA at the end of the lecture 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

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**Class Assignment: Exercise set 4-83**

please submit to TA at the end of the lecture 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

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**Resolution of a force into force and a couple**

d O F O F -F M0 F

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Example P4-104 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

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**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

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**Class Assignment: Exercise set 4-102**

please submit to TA at the end of the lecture Replace the 600-N force shown in Fig. P-102 by a force at point A and a couple. Express your answer in Cartesian vector form. Solution: CA= k Nm

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