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A lesson from engr210 - spring 2005 course taught by ahmed abdel-rahim. It explains the concept of couples, which are two parallel forces with the same magnitude but opposite directions separated by a perpendicular distance. The moment of a couple, the effect of moving a force, and finding an equivalent force-couple system. It also includes an example of finding the equivalent resultant force and couple moment acting at a point.
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Engr210 – Spring 2005 Instructor: Ahmed Abdel-Rahim Lesson # 11: Couples/Force Couple Systems Page 1 of 2
Today’s Objectives:
1. Define a couple and determine **the moment of a couple
Concept
A couple is two parallel forces with the same magnitude but opposite in direction separated by a perpendicular distance d.
M O = F d (using a scalar analysis) M O = r x F (using a vector analysis).
Here r is any position vector from the line of action of –F to the line of action of F.
The net external effect of a couple is that the net force equals zero and the magnitude of the net moment equals F d
Since the moment of a couple depends only on the distance between the forces, the moment of a couple is a free vector. It can be moved anywhere on the body and have the same external effect on the body.
Moments due to couples can be added using the same rules as adding any vectors.
Two force and couple systems are called equivalent systems if they have the same external effect on the body.
Moving a force from A to O, when both points are on the vectors’ line of action, does not change the external effect. Hence, a force vector is called a sliding vector.
Moving a force from point A to O (as shown above) requires creating an additional couple moment. Since this new couple moment is a “free” vector, it can be applied at any point P on the body.
Engr210 – Spring 2005 Instructor: Ahmed Abdel-Rahim Lesson # 11: Couples/Force Couple Systems Page 2 of 2
Finding the resultant of a force and couple system
Reducing a force-moment to a single force
If FR and MRO are perpendicular to each other, then the system can be further reduced to a single force, FR , by simply moving FR from O to P.
Example
Find: The equivalent resultant force and couple moment acting at A.
Solution Plan
→ ΣFx = (4/5) 150 lb + 50 lb sin 30° = 145 lb
↑ ΣFy = (3/5) 150 lb + 50 lb cos 30° = 133.3 lb
FRA = (145 2 + 133.3 2)1/2 = 197 lb and θ = tan-1 (133.3/145) = 42.6 °