Force, Moment, and Couple in Mechanics: A Comprehensive Guide with Examples, Summaries of Statics

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Metal Forming & Plasticity Lab.

Chapter 2

FORCE

SYSTEMS

2 - 2 Force (1) 1/

 Force

 is defined as the action of one body on another. (definition)  is a vector quantity, since its effect depends on the direction as well as on the magnitude.  The action of the cable tension P must be treated as a fixed vector because its effect on the bracket depends on the magnitude P , the angle

θ, and the location of the point of application A.

magnitude 크기 / angle 각 / point of application 작용점 / tension 장력

2 - 2 Force (3) 3/

 Force

 may be treated as a sliding vector in dealing with the mechanics of rigid bodies. (Why?)

 Principle of Transmissibility (전달성 원리)

 A force may be applied at any point on its given line of action without altering the resultant effects of the force external to the rigid body on which it acts.

rigid body 강체 / principle 원리 / transmissibility 전달성 / resultant 결과로서 생기는, 합성된

2 - 2 Force (4) 4/

 Force Classification

External force (외력): applied force (action) / reactive force (reaction) . Internal force (내력): results in internal stress and strain

Contact force (접촉력): generated through direct physical contact, applied over an area (= surface force)  Body force (물체력): applied by remote action (= non-contact force), applied over a volume, such as gravitational, electric and magnetic force

Distributed force (분포력): applied over a finite area or volume . Concentrated force (집중력): applied over a very small area or volume

stress 응력 / strain / surface force 표면력 / electric force 전기력 / magnetic force 자기력

2 - 2 Force (6) 6/

 Resultant of Two Forces

 Parallelogram law ^ Triangle law

parallelogram 평행사변형 / concurrent force 한 점에 동시에 작용하는 힘

R =F 1 +F 2

2 - 2 Force (7) 7/

 Vector Resolution

1 2 a b

F F R

F F R

 Addition of Parallel Forces

 Components :  Projections :

resolution 분해 / component 성분 / projection 정사영 / parallel 평행한

2 - 3 2 - D Rectangular Components (2) 9/

 Determining the Components of a Force

 The assignment of reference axes is a matter of arbitrary convenience.

reference axes 기준(좌표)축

2 - 3 2 - D Rectangular Components (3) 10/

 Resultant of Two Coplanar Forces

1 2 1 2

x x x y y y

R F F

R F F

x y

F

F

i j

i j

i j i j

R F F

x y

x x y y

x y x y

R R

F F F F

F F F F

1 2 1 2

1 1 2 2

1 2

coplanar forces 동일 평면 상의 힘들

2 - 3 2 - D Rectangular Components (5) 12/

 Sample Problem 2-2 : Single equivalent force

 Algebraic Solution

 In vector notation,

tan 6sin 60 0. 3 6 cos 60

α BD AD α

° ° °

R = P + T = 800 i + −( 600cos α i − 600sin α j ) = 346 i − 393 j

800 600 cos 346 N 0 600sin 393 N

x x x y y y

R = P +T α R = P +T α

(^2 2 2 )

1 1

346 ( 393) 524 N

tan tan 393 48. 346

x y y x

R = R + R

R

θ = R

− − °

equivalent 등가의 / algebraic solution 대수적 해

2 - 4 2 - D Moment (1) 13/

 Moment

 is the tendency of a force to rotate a body about an axis which neither intersects nor is parallel to the line of action of the force.  may be considered as a sliding vector with a line of action coinciding with the moment axis.  is also referred to as torque (= moment of force).  moment of inertia  moment of area  moment of momentum

2 - 4 2 - D Moment (3) 15/

 Varignon’s theorem

 The moments of a force about any point is equal to the sum of the moments of the components of the force about the same point.

R = P +Q M (^) O = r × R = r × ( P + Q ) = ( r × P ) + ( r ×Q)

• M (^) O = − Rd = pP − qQ

2 - 4 2 - D Moment (4) 16/

 Sample Problem 2-5 : Moment about the point O of the 600-N force.

(1) Moment arm

(2) Varignon’s theorem

(1) (2)

600 cos 40 460 600sin 40 386 460(4) 386(2) 2, 610

x y O

F N

F N

M N m

4 cos 40 2sin 40 4. O 600(4.35)^ 2, 610^ (CW)

d m M N m

2 - 5 2 - D Couple (1) 18/

 Couple (우력, 짝힘)

 The moment produced by two equal and opposite and non-collinear forces is called a couple.

 The moment expression contains no reference to O ( a, r A , r B ).  The moment of a couple has the same value for all moment centers.  Thus, we may represent M by a free vector.

M (^) O = F a + d ( ) - F a = F d M (^) O = r A (^) × F + r B (^) × − ( F ) = ( r A (^) − r B )× F = r ×F

2 - 5 2 - D Couple (2) 19/

 Sense of a Couple Vector in 2-D Analysis

 Equivalent Couples

 Four different configurations of the same couple M :

Counter-clockwise (CCW) Clockwise (CW)

sense 부호 / equivalent 등가의