problems (equilibrium of particles) 1. -...

1. The 40-kg block rests on the rough surface. Length of the spring is 180 mm in the position

shown. Unstretched length of the spring is 200 mm. Determine the coefficient of friction

required for the equilibrium. ( m = ? )

Problems (Equilibrium of Particles)

FBD of Block

20o

x

yW=mg

For spring: l=180 mm, lo=200 mm (spring is compressed)

NlkFspring 5018.02.02500

NNNmgFy 74.36820cos81.940020cos:0

NFmgFFF frictionfrictionspringx 21.18420sin81.94050020sin:0

5.0N

FNF

friction

friction mm

Equations of equilibrium

2. 4-kg sphere rests on the smooth parabolic surface. Determine the reaction of the

surface on the sphere and the mass mB required to hold it in the equilibrium position.

A

mB

x

y

0.4 m

60o

y=2.5x2


A

x

y

0.4 m

60o

y=2.5x2

FBD of Sphere

q=63.43o

N

W=mg T

Tangent to parabola

o

.x

.xdx

dytan 436325

40

qq

N.T

.sinNcosT

Fx

791

0436360

0

Equations of equilibrium

N.TN.N

..cosNsinT

F

N.

y

12356219

08194436360

0

791

FBD of Block B

T

mBg

kg.m

.mT

F

B

B

.

y

583

0819

0

1235

3. Determine the stretch in springs AC and AB for equilibrium of the 2 kg block. The springs

are shown in the equilibrium position.


SOLUTION:

FBD of A

From equilibrium of A :

45o

FAC

AFAB

36.87o

x

y

W=mg=2(9.81) N

045cos87.36cos:0 ACABx FFF

2

1

081.9245sin87.36sin:0 ACABy FFF

mxNF ACAC 793.020

86.1586.15

mxNF ABAB 467.030

01.1401.14

4. A continuous cable of total length 4 m is wrapped

around the small pulleys at A, B, C and D. If each spring is

stretched 300 mm, determine the mass m of each block

for equilibrium. Neglect the weight of the pulleys and

cords. The springs are unstretched when d=2 m.


FBD of Pulley B

Tqq

C

T

Fspring

NlkFspring 1503.0500

qq

sin

750sin2150

0

TT

Fy

md 290 qC'

A'

A

C

Bq

1 m0.7 m

NT 1.1074.441

7.0arcsin

q

q

T

Tq

Bmg

FBD of Pulley C

kgm

gmT

Fx

6.15

0cos2

0

81.9

q

5. The pail and its contents have a mass of 10 kg. If the length of cable ABC is 7 m, determine

the horizontal distance x for equilibrium. Also find the tension in the cable. The mass of the

small pulley at B is small and can be neglected.

1 m

4 mx

A

B C


L=7 m, x = ? (for equilibrium). T=?

1 m

4 mx

A

B C

x

yA

B

C

yy 1

q q'

FBD of B

T

W=mg

q q'

T From equilibrium of B :

qqqq 0coscos:0 TTFx

0sin2:0 WTFy q

From geometry: xaaxaaxxa

x

a

x

4

747

7

4cos

q

15.557

4

4

7cos qq

x

x

a

xxy

x

y44.144.1tan q

mxx

x

x

x

x

y

x

y347.2

4

144.144.1

4

1

(*)

(*) NTT 817.59081.91015.55sin2

B

x4x

6. The cylinder of mass 1 kg having a

very small diameter is held against a

semi-cylinder with a much larger

diameter by two identical springs,

which are fixed to points C and C on

the ground. The springs are

unstretched when at point A. Knowing

that the small cylinder is in

equilibrium at point B, what is the

spring constant?

r=200 mm

h=120 mm

A

B

h r

C

C

C,C


7. Two bodies weighing 150 N and 200 N, respectively, rest on a cylinder and are

connected by a rope as shown. Find the reactions of the cylinder on the bodies, the

tension in the rope and the angle q. Assume all surfaces to be smooth.

ropeW1=200 N

W2=150 N

q90q


W1=200 Nq

90q

qN1

T

q


FBD of Particle 1

y

x

q

qqq

qqq

q

qq

q

qqq

sinT

sinsinTcosT

sinTcossin

cosT

sinTcosN

F

sin

cosTNcosTsinN

F

y

x

200

200

0200

0200

0

0

0

22

1

11

SOLUTION:

8. A force of F=100 N holds the 400-N crate in equilibrium. Determine the coordinates(0, y, z) of point A if the tension in cords AC and AB is 700 N each.


5 m

5 m

4 m

SOLUTION:FBD of A

FAC=700 NA

FAB=700 N

W=400 N

F=100 N

Coordinates:

𝐴 0, 𝑦, 𝑧𝐵 −5, 0,4𝐶 5, 0,4

222 45

45700

zy

kzjyiFAC

222 45

45700

zy

kzjyiFAB

jF

100 kW

400


222 45

45700

zy

kzjyiFAC

222 45

45700

zy

kzjyiFAB

jF

100 kW

400

0

45

3500

45

35000

222222

zyzyFx

0100

45

700

45

7000

222222

zy

y

zy

yFy

0400

45

4700

45

47000

222222

zy

z

zy

zFz

1

2

3

22 4251001400 zyy 2'

22 42540041400 zyz 3'

2'

3' zy

z

y

44

4

1

4

2' 22222 16251416251001400 yyyyyy

mzmy 51.23737.0

9. The shear leg derrick is used to haul the 200 kg net of fish onto the dock. Determine the

compressive force along each of the legs AB and CB and the tension in the winch cable DB.

Assume the force in each leg acts along its axis.


SOLUTION: FBD of B

𝑭BD

B

W=200(9.81) N𝑭AB

𝑭CB

6

442 kjiFF ABAB

kW

1962

Coordinates:

𝐴 2,0,0 𝐵 0,4,4𝐶 −2,0,0 𝐷 0,−5.6,0

6

442 kjiFF CBCB

4.10

46.9 kjFF BDBD


6

442 kjiFF ABAB

kW

1962

6

442 kjiFF CBCB

4.10

46.9 kjFF BDBD

00 WFFFF BDCBAB

Equations of Equilibrium

There are three unknowns.

03

1

3

10 CBABx FFF1

04.10

6.9

3

2

3

20 BDCBABy FFFF2

019624.10

4

3

2

3

20 BDCBABz FFFF3

kNF

kNF

kNF

BD

CB

AB

64.3

52.2

52.2

10. A small peg P rests on a spring that is contained inside the smooth pipe. The

spring exerts an upward force of 284 N on the peg. Determine the point of

attachment A (x, y, 0) of cord PA so that the tension in cords PB and PC equals

130 N and 84 N, respectively.

B

A

x

0.3 m

yx

z

y

0.6 m

C

P

0.2 m0.4 m


http://images.google.com/url?sa=i&rct=j&q=metal+spring&source=images&cd=&cad=rja&docid=dBpMB8OyVTmvlM&tbnid=uQFxmMPENo080M:&ved=0CAUQjRw&url=http://www.calypsoushaka.co.za/Newsletter&ei=8SVOUdnGLcSsPLfQgTg&bvm=bv.44158598,d.ZWU&psig=AFQjCNE0PZqBqMPFkWN4vFqLJHYuz0cSJw&ust=1364162413190132


Fspring = 284 N, TB = 130 N , TC = 84 N; determine coordinates of point A (x, y, 0).

B

Ax

0.3 m

y

x

z

y

0.6 m

C

P

0.2 m0.4 m

FBD of peg P

AT

BT

CT

springF

Coordinates

𝐴 𝑥, 𝑦, 0 𝐵 0,−0.4, 0

𝐶 −0.3,0.2,0 𝑃 0, 0, 0.6

a

AA

yx

kjyixTT

222 6.0

6.0

kjT

kjTT

B

BB

33.10822.72

72.0

6.04.0

130

kjiT

kjiTT

C

CC

722436

7.0

6.02.03.0

84

kFspring

284

From equilibrium of peg P 0F

360360 a

xT

a

xTF AAx

22.4802422.720 a

yT

a

yTF AAy

67.1036.0

07233.1086.0

2840 a

Ta

TF AAz

1

2

31

3mx

aT

a

xT

A

A

208.067.103

36

6.0

2

3my

aT

a

yT

A

A

279.067.103

22.48

6.0

TA = 119.845 N



11. Crate A weighing 580 N rests on the inclined surface by the cable AB and force P which is

parallel to the z-axis. Determine the tension in the cable AB and force P for equilibrium.

Since the crate is mounted on casters, the force exerted by the incline on the crate is

perpendicular to the incline.

P

A

1.5 m

4 m

3 m

2.2 m

x

y

zDO

C

B

E

3 m


𝑷𝑾

𝑻

𝑵

4 m

3 m

x

3 m

yFBD of Crate A

𝐴 (2.4; 1.2;−1.5)

𝐴 (3𝑐𝑜𝑠37; 3 − 3𝑠𝑖𝑛37;−1.5)

a

a

a 37o

A

𝐵 (0; 5.2; 0)jW

580

Forces acting on the crate

94

51442

94

5102125420

.

k.ji.TT

.

k.j..i.TT

kPP

Reaction (normal force) exerted by incline on crate

4 m

3 m

y

a

A

N

xa

jN.iN.N

isinNjcosNN

8060

3737

(Reaction force (normal force) must be perpendicular to the incline.)

SOLUTION:

jW

580

Forces acting on the crate

94

51442

.

k.ji.TT

kPP

jN.iN.N

8060

00 NTPWF

094

42600 T

.

.N.Fx

094

4805800 T

.N.Fy

094

510 T

.

.PFz

1

2

3

1 2 3From , and ; T=395 N, N=322 N, P=120 N


𝑷𝑾

𝑻

𝑵

155 mm 200 mm

460 mm

Coordinates: 𝐴 0,155,0 𝐵 200,0,460

𝑾

AxAz

𝑷

𝑻

𝑷

FBD of collar A

Reactions exerted by the vertical bar on collar A : kAiAN zxA

525

460155200 kjiTT

jP

351

jW

10

00 WTPNF A


12. Collars A and B are connected by wire AB and canslide freely on frictionless rods. If a force P=351 N isapplied to collar A, determine (a) the tension in thewire, (b) the magnitude of the force Q required tomaintain the equilibrium of the system and reactionson collar A by the vertical rod. WA=WB=10 N

Forces acting on collar A


SOLUTION:

kAiAN zxA

525

460155200 kjiTT

jP

351 jW

10

NT

TFy

1155

010351525

1550

1

NATAF xxx 4400525

2000 2

(correct sense)

3 NATAF zzz 10120525

4600

FBD of collar B

Ax

𝑻′

𝑸

𝑾

Bx

By

𝑾

AxAz

𝑷

𝑻

Reactions exerted by the horizontal on collar B : jBiBN yxB

525

4601552001155

kjiTT

jW

10 kQQ

NQQFz 101201155525

4600

NBBF xxx 44001155525

2000

NBBF yyy 33101155525

155100

13. If WA=WB=1400 N, determine the force P, TAB and the reactions between the

collars and bars.


14. The 100-kg collar A rests on the smooth straight fixed bar CD by the cable AB.

Determine the tension in the cable and the reaction between the collar and bar CD.

C


C

FBD of collar A

xz

y

C

D

Length of fixed bar CD: mCD 9447 222

Coordinates of point A: m.zx AA 6729

64 m.yA 332

9

37 𝐴 (2.67; 2.33; 2.67)

𝐵 (0; 7; 4)

jj.W

981819100

Forces acting on the collar

kjiTkji

TT

24.084.048.054.5

33.167.467.2

kNjNiNN zyx

(Reaction exerted by bar on the collar)

15. The 100-kg collar A rests on the smooth straight fixed bar CD by the cable AB. Determine the tension in

the cable and the reaction between the collar and bar CD.

𝑻

𝑵

𝑾

C

jW

981Forces acting on the collar

kjiTT

24.084.048.0 kNjNiNN zyx

(There are four unknowns: T, Nx, Ny, Nz)

(But, there are there equations of equilibrium.)

SOLUTION METHOD:Reaction force between the collar and bar CD is

perpendicular to the axis of the bar. ( 𝑪𝑫𝑵 or

𝒏𝑪𝑫𝑵 ). So, scalar product of these vectors must be equal to zero.

0 CDnN

Additionally, since 0F

0 CDnF

00

0

CDCDCDABCDAB nWnNnTnWNT

𝐶 (4; 0; 4)

𝐷 (0; 7; 0)

9

474 kjinCD

076311066021009

474981240840480

T.T.T.

kjijk.j.i.T

k.j.i.TNT AB

0824428854324881017

k.j.i.NWTNF AB

0824472126324880

3.4 m

r =3.4 m

4.5 m

158

16. Collar A weighing 170 N, which can slide freely

on the quarter circle, is held in equilibrium by cable

AB.

a) Determine the tension in the cable.

b) Determine the magnitude and the components of

the contact force acting on the collar from the circle.


xz

y

FBD of collar A

W

ABT

N

Coordinates of points A and B:

0

17

843

17

1543 ;.;.A 𝐴 (3; 1.6; 0) 𝐵 (0; 3.4; 4.5)

3.4 m

r =3.4 m

4.5 m

158

x

y

z

y

W

ABT

N

Forces acting on the collar

75

54813

.

k.j.iTTAB

jW

170

xyN

kNjNiNN zxyxy

17

8

17

15

xy plane yz plane

zN

(Reaction exerted by bar on the collar)

00 NTWF AB

017

15

75

30 xyx NT

.F

017

8

75

811700 xyy NT

.

.F

075

540 T

.

.NF zz

1

2

3

Equation of Equilibrium

1 2 3From , and ; T=285 N, Nxy=170 N, Nz=225 N

17. Smooth collars A, B and C, each weighing 360 N, are connected by the

wires AB and BC and may slide freely on the smooth rod having the shape

shown. Determine the magnitude of the horizontal force P which must be

applied to the collar A to maintain equilibrium. DEFG portion of the rod is

parallel to xy-plane.


P

NAy

NAx

TAB

W

xz

y

xz

y

W

TBA

TBC

NBz

NBxy

xz

y

TCB

W

NCxNCz

FBD of collar A

FBD of collar BFBD of collar C

Four unknowns

Four unknowns

Three unknowns

xz

y

TCB

W

NCxNCz

FBD of collar C

A (0;9;3)

B (2;5;7)

C (6;1;0)

E

B

F

6 m

1.5 m4.5 m xB

65.4

5.1 Bx NT

TF

kNiNN

jW

kjiTT

NWTF

BC

BCy

CCC

BCBC

CBC

zx

810

03609

40

360

9

744

00

6.0cos

6.0sin

87.36

q

q

q o

NT

TNF

TNF

kNjNiNN

jW

kjiTT

kjiTT

NWTTF

BA

BAxyBy

BAxyBx

BxyBxyBB

BABA

CBBC

BABBC

z

1080

03606

43608.00

06

23606.00

8.06.0

360

6

442

630360360

00

FBD of collar B

xz

y

W

TBA

TBC

NBzNBxy

F

E

E

B

F

6 m

4.5 mNBxy qq

NP

PF

kPP

jNiNN

jW

kjiTT

PNWTF

z

yAxAA

BAAB

AAB

720

07200

360

720720360

00

P

NAy

NAx

TAB

W

xz

y

FBD of collar A

18. Smooth collars A and B are connected by the spring. Spring has a

constant of 120 N/cm and its unstretched length is 30 cm. Determine the

magnitude of the force P which must be applied to the collar A to maintain

equilibrium and the reaction between the collar and bar. Neglect the weight of

the collars. Take A (40;0;40) and B (0;20;80).

yx

z

30 cm

40 cm40 cm

20 cm80 cm

80 cm

P

Q

A

B


FBD of collar A

P

NAxz

NAy

Fspring

yx

z

yx

z

30 cm

40 cm

40 cm

20 cm

80 cm80 cm

P

Q

A

B

NPNN

NPF

NNNF

NPF

jNkNiNNkPiPP

cmABkjiABkjiF

kjiABkji

F

Axz

Axzz

AyAyy

Axzx

AyAxzAxzA

spring

spring

87.3483387

05

3

5

424000

1200012000

05

4

5

324000

5

3

5

4

5

4

5

3

60402040240012002400

)4080()020()400(60

402040)3060(120

40 cm

x

A

z

30 cm

NAxz

q

q

00 Aspring NPFF

Correct sense

problems (equilibrium of particles) 1. -...

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