3장 1/16

Chapter 3

질점의평형

(Equilibrium of a Particle)

3장 2/16

3.1질점의평형조건

(Condition for the Equilibrium of a Particle)

뉴턴 운동법칙에서 질점이 일정한 속도를 유지하기 위한 필요충분조건은 질점

에작용하는힘의합력

∑F = 0이며,

이러한상태를 정적평형(static equilibrium)이라한다. 보통, 속도가 0인 정적평형상

태즉질점이정지해있는평형상태를중시한다. (왜그럴까?)

3장 3/16

3. 2 자유물체도 (The Free-Body Diagram)자유물체도는 대상물체만을 주위환경과 분리하여 그린 후 주위환경으로부터

질점에작용하는모든힘을표시하는하나의스케치이다.

질점의평형문제에서자주나오는두가지형태의연결방식

①스프링

스프링상수(spring constant) 혹은 강성(stiffness)이 k인 선형탄성스프링 (linear elastic spring)이힘을받지않는위치(unstretched position)에서 s 만큼 변형(인장 또는 압축)하도록 스프링에 작용하는 힘은 F = ks 이다.

3장 4/16

②케이블(cable, 줄)과풀리(pulley, 도르레)

Fig. 3-2에서와 같이 케이블(inextensible)은 인장

력(tension)만 받으며 풀리(frictionless)를 감은 각도

와무관하게일정한인장력 T를받는다.

자유물체도를그리는순서

①물체의개략적인모양을그린다.

②물체에작용하는힘들을표시한다.

③각힘에대한정보(크기와방향)를표시한다. –모를땐미지수로.

3장 5/16

예제 3-1 : Example 3-1

The sphere in Fig. 3-3a has a mass of 6kg and is

supported as shown. Draw a free-body diagram of the

sphere, the cord CE, and the knot at C.

(Sol.)

3장 6/16

3.3평면력계 (Coplanar Force Systems)

∑F = ∑Fxi + ∑Fyj = 0

∑Fx= 0, ∑Fy= 0 (3-3)

풀이과정

Coplanar force equilibrium problems for a particle can be solved using the following

procedure.

3장 7/16

자유물체도자유물체도

• Establish the x, y axes in any suitable orientation.

• Label all the known and unknown force magnitudes and directions on the diagram.

•The sense of a force having an unknown magnitude can be assumed.

평형방정식평형방정식

• Apply the equations of equilibrium ∑Fx= 0 and ∑Fy= 0.

• Components are positive if they are directed along a positive axis, and negative if

they are directed along a negative axis.

• If more than two unknowns exist and the problem involves a spring, apply F = ks to

relate the spring force to the deformation s of the spring.

• If the solution yields a negative result, this indicates the sense of the force is the

reverse of that shown on the free-body diagram.

3장 8/16

예제 3-2 : Problem 3-36

Determine the mass of each of the two

cylinders if they cause a sag of s = 0.5 m

when suspended from the rings at A and B.

Note that s = 0 when the cylinders are

removed.

(Sol.)

실린더가제거되었을때

5.1

2

5.25.12 22 =+

3장 9/16

2

2°45

828.22222 22 ==+

ABTA

mg

ACT

실린더가설치되었을때

TAC = ks = 100(2.828 − 2.5)

= 32.8 N

점 A의평형을위하여

∑Fy = TAC sin 45° - mg = 0

∴ kg36.222

81.98.3245sin =×=°=

gT

m AC

좌우대칭이므로 두 물체의 질량은 같다.주의

3장 10/16

예제 3-3 : Problem 3-37

The ring of negligible size is subjected to

a vertical of 200 lb. Determine the required

length l of cord AC such that the tension

acting in AC is 160 lb. Also what is the

force in cord AB? Hint: Use the equilibrium

condition to determine the required angle θ

for attachment, that determine l using

trigonometry applied to triangle ABC.

(Sol.)

점 A의평형을위해

∑Fx = 0; FABcos40° − 160cosθ = 0

∑Fy = 0; FABsin40° + 160sinθ − 200 = 0

θ °40

ABF

lb200

lb160

3장 11/16

두식에서 FAB를소거하면,

sinθ + 0.8391cosθ = 1.25

양변을제곱하면

1 − cos2θ = 1.252 − 2(1.25)(0.8391)cosθ + 0.83912cos2θ

cosθ에대한 2차방정식을풀면

θ = 33.25° or 66.75°

① θ = 33.25°일때 FAB = 175 lb

∴ l = 2.34 ft

② θ=66.75°일때 FAB = 82.4 lb

∴ l = 1.40 ft

θθθ 222 cos8391.025.1cos1sin −=−=

°=

° 40sin25.33sin2 l

°=

° 40sin75.66sin2 l

θ °40l ft2

3장 12/16

3.4 3차원힘계 (Three-Dimensional Force Systems)3차원힘계가작용하는질점의평형을위해선

∑F = ∑Fxi+∑Fyj+∑Fzk = 0

즉 ∑Fx = 0, ∑Fy = 0, ∑Fz = 0 (3-5)

를만족해야한다.

예제 3-4 : Problem 3-59

If the maximum allowable tension in

cable AB and AC is 500 lb, determine the

maximum height z to which the 200-lb crate

can be lifted. What horizontal force F must

be applied? Take y = 8 ft.

3장 13/16

(Sol.)

기하학적대칭에 의해 FAC = FAB가 성립하며 이 두 인장력이

최대허용인장에도달할때의상태를생각하면

FAC = FAB = 500 lb 이다.

점 A의평형을위해

222 )4(85)4(85z

zFACAC−++

−+−=

kjiFACF

ABF

F

A

∑Fy = 0 ; (1)

∑Fz = 0 ; (2)

0)4(85

85002222

=+⎥⎥⎦

⎤

⎢⎢⎣

⎡

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

−++

− Fz

020045002222

=−⎥⎥⎤

⎢⎢⎡

⎟⎟⎞

⎜⎜⎛

−++

− z)4(85 ⎦⎣ ⎠⎝ z

3장 14/16

식 (2)를식 (1)로나누면

이결과를식 (1)에대입하면

양변을제곱하여 F와 z를구하면

F = 831 lb, z = 2.07 ft

Fz 200

84

=−

Fz 16004 =−

222 1600858000

⎟⎠⎞

⎜⎝⎛++=

FF

왜 z는 4보다작아야하나?

∑Fx = 0는왜적용하지않는

가?

주의

3장 15/16

예제 3-5 : Problem 3-61

The bucket has a weight of 80 lb and is being hoisted

using three springs, each having an unstretched length of

l0 = 1.5 ft and stiffness of k = 50 lb/ft. Determine the

vertical distance d from the rim to point A for equilibrium.

(Sol.)

x

y

zA

C

rC – rA : FAC 와같은방향

22)5.1(5.1

ddFF AC

AC

ACACAC

+

−=

−−

=kj

rrrr

F

3장 16/16

점 A의평형을위해

∑Fz = 0 : (1)

(2)

(2) → (1)

d 4 − 1.067d 3 + 0.284d 2 − 2.4d + 0.64 = 0 d = 1.64 ft

kF ⋅= ACACZF

kFFFkFkFkF ⋅++=⋅+⋅+⋅ )()( ADACABADACAB

0)5.1(

)()3(8022=

+

−+

ddFAC

( )5.1)5.1(50)( 220 −+=−== dllkksFAC

기하학적대칭에의해세스프링에작용하는인장력은동일하며이인

장력들의수직성분도동일하다는점이이용되었다. FAC대신에 FAB나 FAD

를사용하여해석해보라.

주의