1

MEM202 Engineering Mechanics - Statics MEM

Chapter 3Statics of Particles

(Equilibrium of Concurrent Force Systems)

0=++==

++=++=

∑∑∑∑ kFjFiFF

kRjRiRRRRR

zyx

zyxzyxrrrr

rrrrrrr

000 ====== ∑∑∑ zzyyxx FRFRFR

2

MEM202 Engineering Mechanics - Statics MEM

3.2 Free-Body Diagrams

( ) 021 ∑∑ =++= LifFFFrrrr

04321 =+++=∑ FFFFFrrrrr

1Fr

2Fr

3Fr

4Fr

1Fr

2Fr

( )Lifr

3Fr

4Fr

( )Rifr

( ) 043 ∑∑ =++= RifFFFrrrr

021 =++=∑ LFFFFrrrr

043 =++=∑ RFFFFrrrr

1Fr

2Fr

( )∑= LiL fFrr 3F

r

4Fr

( )∑= RiR fFrr

( ) ( )Ri

Li ff

rr−= RL FF

rr−=

If the body is in a state of equilibrium, then

Any sub-set of the body should also be in a state of equilibrium

3

MEM202 Engineering Mechanics - Statics MEM

3.2 Free-Body Diagrams

4

MEM202 Engineering Mechanics - Statics MEM

3.2 Free-Body Diagrams

F

Action/reaction on a smooth contact surface are always normal to the surface.

Force in a flexible cable is always tensile and directed along the axis of the cable.

Add a friction force F for a rough surfaces.

5

MEM202 Engineering Mechanics - Statics MEM

3.3 Equilibrium of A Particle2-D Example

FBD

( )( ) 022020sin25sin

020cos25cos

=−+=↑+

=−=→+

∑∑

oo

oo

ABy

ABx

FFF

FFFEquilibrium

lb 292lb 282 == BA FF

6

MEM202 Engineering Mechanics - Statics MEM

kg 50=m

3.3 Equilibrium of A Particle2-D Example

( )( ) 060sin45sin

060cos45cos

=−+=↑+

=−=→+

∑∑

mgFFF

FFF

BAy

BAx

oo

oo

lb 359lb 254 == BA FF

7

MEM202 Engineering Mechanics - Statics MEM

3.3 Equilibrium of A Particle2-D Example

( )

( )

059.216.335.0866.0165sin236sin30sin60sin

sinsinsinsin

0659.937.22866.05.0165cos236cos30cos60cos

coscoscoscos

21

4321

44332211

4321

21

4321

44332211

4321

=+−+=+++=

+++=

+++=↑+

=−−+=+++=

+++=

+++=→+

∑

∑

FFFFFF

FFFF

FFFFF

FFFFFF

FFFF

FFFFF

yyyyy

xxxxx

oooo

oooo

θθθθ

θθθθ

kip 9.24 kip 9.20 21 == FFooo

ooo

o

ooo

16515180

23656180

30

603030

4

3

2

1

=−=

=+=

=

=+=

θ

θ

θ

θ

Determine F1 and F2 so that the particle is in equilibrium

8

MEM202 Engineering Mechanics - Statics MEM

3.3 Equilibrium of A Particle2-D Example

Alternative Approach

kip 9.20

045sin1026sin4030sin

1

1431

=⇒

=+−=++=∑F

FFFFF nnnnooo

2Fen rr⊥

kip 9.240165cos10236cos4030cos60cos9.20

2

24321

=⇒

=+++=+++=∑F

FFFFFF xxxxxoooo

Determine F1 and F2 so that the particle is in equilibrium

9

MEM202 Engineering Mechanics - Statics MEM

3.3 Equilibrium of A Particle2-D Example

Method 1 Method 2

?=T

?=N

( )( ) 050035sin15cos

035cos15sin

=−+=↑+

=−=→+

∑∑

oo

oo

NTF

NTF

y

x

055sin50020cos

055cos50020sin

=−=

=+−−=

∑∑

oo

oo

TF

TNF

y

x

lb 7.137lb 436 == NTlb 7.137lb 436 == NT

10

MEM202 Engineering Mechanics - Statics MEM

N 2835010cos

N 2879050010sin

=⇒

=−=

=⇒

=−=

∑

∑

A

CAx

C

Cy

TTTF

TTF

o

o

N 7789020cos

N 8289020sin

=⇒

=−=

=⇒

=−=

∑

∑

D

DBy

B

ABx

TTTF

TTTF

o

o

kg 794

N 77890

==⇒

=⇒

=−=∑

gWm

WWTF

AA

A

ADy

?????

=====

A

D

C

B

A

mTTTT

3.3 Equilibrium of A Particle2-D Example (Multiple FBDs)

11

MEM202 Engineering Mechanics - Statics MEM

3.3 Equilibrium of A Particle3-D Example

lb 200001111 kjikFjFiFF zyx

rrrrrrr−+=++=

( ) ( )lb 02.2371.79

01032501012502222

kji

kjikFjFiFF zyxrrr

rrrrrrr

++=

++=++=

lb 9.1290.750

30cos15060cos15003333

kji

kjikFjFiFF zyxrrr

rrrrrrroo

++=

++=++=

kFjFiFF zyx

rrrr4444 ++=

0=++=++= ∑∑∑ kFjFiFkRjRiRR zyxzyx

rrrrrrr

lb 1.70lb 2.312

lb 1.79

09.129020000.752.2370

001.790

4

4

4

4

4

4

=−=−=

⇒=+++−===+++==

=+++==

∑∑∑

z

y

x

zzz

yyy

xxx

FFF

FFRFFR

FFR

lb 33024

24

244 =++= zyx FFFF

o

o

o

7.77cos

2.161cos

9.103cos

4

414

4

414

4

414

==

==

==

−

−

−

FFFFFF

zz

yy

xx

θ

θ

θ

lb 1.702.3121.794 kjiFrrrr

+−−=

Determine F4 for equilibrium

12

MEM202 Engineering Mechanics - Statics MEM

3.3 Equilibrium of A Particle3-D Example

Determine TA, TB, and TC

( ) ( ) ( )( )lb 7682.03841.05121.0

634

634222

kjiT

kjiTT

A

AA

rrr

rrrr

+−=

⎥⎥⎦

⎤

⎢⎢⎣

⎡

+−+

+−=

( ) ( ) ( )( )lb 4851.07276.04851.0

464

464222

kjiT

kjiTT

B

BB

rrr

rrrr

+−−=

⎥⎥⎦

⎤

⎢⎢⎣

⎡

+−+−

+−−=

lb jTT cC

rr=

lb 3000kWrr

−=

W = 3000 lb

lb 2700lb 2474lb 2343

030004851.07682.0

07276.03841.004851.05121.0

===

⇒=−+==+−−=

=−=

∑∑∑

C

B

A

BAz

cBAy

BAx

TTT

TTFTTTF

TTF

13

MEM202 Engineering Mechanics - Statics MEM

3.3 Equilibrium of A Particle3-D Example

Determine TA, TB, TC, and TD

( ) ( ) ( )N

4.25.12.1

4.25.12.1222 ⎥⎥⎦

⎤

⎢⎢⎣

⎡

+−+

+−=

kjiTT AA

rrrr

( ) ( ) ( )N

1.21.28.1

1.21.28.1222 ⎥⎥⎦

⎤

⎢⎢⎣

⎡

+−+−

+−−=

kjiTT BB

rrrr

N 500kTD

rr−=

N 317N 4.32N 459

000

===

⇒===

∑∑∑

C

B

A

z

y

x

TTT

FFF

N 500=DT

( ) ( ) ( )N

9.08.12.1

9.08.12.1222 ⎥⎥⎦

⎤

⎢⎢⎣

⎡

++−

++−=

kjiTT CC

rrrr