Determine the Reaction

1

Review 4.1

Rod AC is supported by a pin and bracket at A and rests against a peg atB. Neglecting the effect of friction, determine the reactions at A and Bwhen a = 20 cm.

Guide : Assuming that the rod AC can be treated as a rigid body. There are2 unknowns reactions at A and B. Use sine law.

A

B

C

a

.50 m

.25 m60 N

2

• Define the state of the problem

Give : a = 20 cmDetermine : reactions at A and B

Assuming that the rod AC can betreated as a rigid body.

A

B

C

a

.50 m

.25 m60 N

• Construct physical model

C

600 N

γ

α

.20 m

.30 m

B

A

2

1

G

D

F E

.25 m

A

B

tan α = α = 26.57o12

3

C

600 N

γ

α

.20 m

.30 m

B

A

2

1

G

D

F E

.25 m

A

B

tan α = α = 26.57o12

• Construct mathematical model and solve equations

Force Triangle:

600 N

300 N

300 N

γ = 26.57o

α = 26.57o

A

B

NNBA o 7.67057.26sin

300 ===

4

C

600 N

.20 m

.30 m

B

A

2

1

G

D

F E

.25 m

NBA 7.670==

• Conclusion

26.6o

671 N

26.6o

671 N

5

Review 4.2 (Problem 4.65)

Determine the reactions at B and D.

Guide : CD is a 2-force member line of action of reaction at D must passthrough C and D. Assuming that the member ABC can be treated as a rigidbody. There are 3 forces acting on it, reactions at B, C and 80 N at A. Useequilibrium equations ΣMB = 0 and ΣFy = 0.

80 N

B

D

C

250 mm

90 mm

60 mm

75 mm

A

6

• Define the state of the problem • Construct physical model

Determine : RD and RBCD is a 2-force member line of action ofreaction at D must pass through C and D.Assuming that the member ABC can betreated as a rigid body. There are 3 forcesacting on it, reactions at B, C and 80 N at A.

250 mm

80 N

BC 60 mm

90 mm160 mmA

160 mm

o35.41)250

16060(tan 1 =+= −β

• Construct mathematical model and solve equations

β

RB

RD45o

45o

+ ΣMA = 0: 80(250) + RD sin45(90) - RD cos 45 (60) = 0 -----(1)

ΣFy = 0: -80 - RD sin 45 + RB sin 41.35o = 0 -----(2)+

RD = -942.8 N RB = -888.0 N

80 N

B

D

C

250 mm

90 mm

60 mm

75 mm

A

7

• Conclusion

250 mm

80 N

BC 60 mm

90 mm160 mmA

41.4o

160 mm

RB = 888.0 N

RD = 943 N

45o

8

Review 4.3 (Problem 4.85)

A slender rod of length L = 200 mm is held in equilibrium as shown, withone end against a frictionless wall and the other end attached to a cord oflength S = 300 mm.Knowing that the weight of the rod is 10 N, determine(a) the distance h, (b) the tension in the cord, (c) the reaction at B.

L

S

A

B

C

h

Guide : Assuming that rod AB can be treated as a rigid body. There are 3forces acting on it,tension force by cable at A, reactions at B and its weight atmiddle of span . Use equilibrium equations ΣFy = 0 and ΣFx = 0.

9

GA

B

C

hD

h

E

• Define the state of the problem

L

S

A

B

C

h

∆ ACE: (2h)2 + (AE)2 = S2 (1)∆ ABE: h2 + (AE)2 = L2 (2)eq.(1)-eq.(2) : 3h2 = S2 - L2 (3)

• Construct physical model

θ

RBT

β

W = 10 N

3)( 22 LS

h−

=

• Construct mathematical model and solve equations

mmh 1.1293

)200300( 22

=−

=

8607.0300

)1.129(22cos: ===∆ShACE β

o609.30=β

ΣFy = 0: T cos β -10 = 0T = 11.62 N

+30.61o

ΣFx = 0: T sin β - RB = 0RB = 5.92 N

+11.62 N 30.61o

Assuming thatrod AB can betreated as a rigidbody. There are3 forces actingon it,tensionforce by cable atA, reactions at Band its weight atmiddle of span .

10

• Conclusion

θ

RB = 5.92 NT = 11.6 N

β

W = 10 N

GA

B

C

h = 129 mmD

E

11

Review 4.4 (Problem 4.88)

A uniform rod AB of length 2R rests inside a hemispherical bowl of radiusR as shown. Neglecting friction, determine the angle θ corresponding toequilibrium.

Guide : Assuming AB can be treated as a rigid body. There are 3 forceacting on it, W, RA and RB.RA and RB are intersection, RA passes through Oand RC is perpendicular to rod. Set : cos 2θ = 2 cos2θ - 1.

2 R

θA

B

C

12

θA

BORD

• Define the state of the problem

2 R

θA

B

• Construct physical model

2θ θ

RADetermine : θ for equilibriumAssuming AB can be treated as a rigid body.There are 3 force acting on it, W, RA and RB.The RA and RB are intersection on the circle,RA passes through O and RC is perpendicularto rod.

W

G

• Construct mathematical model and solve equations(AE) cos 2θ = (AG) cos θ(2R)cos 2θ = (R) cos θset : cos 2θ = 2 cos2θ - 1 4 cos2θ - 2 = cos θ4 cos2θ - cos θ - 2 = 0

cos θ = 0.84307 θ = 32.5o

cos θ = -0.59307 θ = 126.4o (discard)

C

RC

E

13

• Conclusion

32oA

BORD

RA

C

RC

E

W

G

14

Review 4.6

The frame ACD is hinged at A and D and is supported by a cable whichpasses through a ring at B and is attached to hooks at G and H. Knowingthat the applied force P is 1 kN, determine the tension force in the cable.

z

xy

0.875 m

0.75 m

0.5 m0.5 m

0.35 m

0.925 m

A

B

O

C D

P

H

G

0.75 m

Guide : Assuming that the frameACD can be treated as a rigid body.Draw FBD of it , the forces acting onthe FBD are reactions involving 8unknowns, namely, the force TBGand TBH exerted by the cables, 3force components at A and D. Useequilibrium equations :MAD = uAD

. [( rAC x P) + ( rAB x TBH ) + ( rAB x TBG )] = 0

15

• Define the state of the problem

Give : P = 1 kNDetermine : TBG and TBH

Assuming that frame ACD can betreated as a rigid body.

z

xy

0.875 m

0.75 m

0.5 m0.5 m

0.35 m

0.925 m

A

B

O

C D

P

H

G

0.75 m

• Construct physical modelz

xy

0.875 m

0.75 m

0.5 m0.5 m

0.35 m

0.925 m

A

B

O

C D

1 kN k

H

G

0.75 m

TBG

TBH

• Construct mathematical model

MAD = uAD. [( rAC x P) + ( rAB x TBH )

+ ( rAB x TBG )] = 0

16

z

x

0.875 m

0.75 m

0.5 m0.5 m

0.35 m

0.925 m

A B

O

C D

P

H

G

y0.75 m

uAD = (_ .6 i + 8 j)rAB = (0.5 m ) jrAC = (1.0 m ) j

• Solve equations

TBH = (_ 0.75 i + 0.375 j + 0.75 k)

= _ (.667 T) i + ( .333 T) j + (.667 T) k

T1.125

mdBH 125.1)75.0()75.0()375.0( 222 =−++=

TBG

TBHTBG = (_ 0.4 i - 0.5 j + 0.925 k)

= _ (.356 T) i - (. 444 T) j + (.822 T) k

T1.125

mdBG 125.1)4.0()925.0()5.0( 222 =−++−=

MAD = uAD. [( rAC x P) + ( rAB x (TBH+TBG))] = 0

+ -.6 0 .8 0 0.5 0-1.023T -.111T 1.489T

= 0

-.6 0 .8 0 1 0 0 0 -1

T = 1.343 kN

17

z

x

0.5 m0.5 m

0.35 m

0.925 m

A

B

O

C D

P

H

G

y0.75 m

• Conclusion

TBH = (_ 0.75 i + 0.38 j + 0.75 k)(1.34 kN)1.12

TBG = (_ 0.4 i - 0.5 j + 0.92 k)(1.34 kN)1.12

18

m

0.2 m 0.125 m

0.125 m

0.2 m

E

θ

0.1 mx

y

z

A

B

C

D

x

Review 4.6

A 200x250-mm panel of mass 20 kg is supported by hinges along edgeAB. Cable CDE is attached to the panel at C, passes over a small pulleyat D, and supports a cylinder of mass m. Determine mass of cylinder ifθ = 30o.

Guide : Assuming that plate can betreated as a rigid body. Draw FBD of it ,There are 8 forces acting on it, theweight of plate W, the force T exertedby the cable, and reactions involving 6unknowns, namely, 3 forcescomponents at each hinge. To eliminatethe reactions at hinge from thecomputations, we express that the sumof the moments of the forces about ABis zero (ΣMAB = 0).The forces reactionsis zero because there has not r can findT next assuming cylinder can be treatedas a particle ΣF = 0 can find m.

19

0.2 m 0.125 m

0.125 m

0.2 m

θ

0.1 mx

y

z

A

B

C

D

x

• Define the state of the problem • Construct physical model

m

0.2 m 0.125 m

0.125 m

0.2 m

E

30o

0.1 mx

y

z

A

B

C

D

x

G

WT

B (0, 0, 0)C (0.2sin30o, - 0.2cos30o, 0.125)D (0.2, 0.1, 0)G (0.1sin30o, - 0.1cos30o, 0.125)

Assuming that plate can be treated as arigid body. Draw FBD of it , There are 8forces acting on it, W, T and reactionsinvolving 6 unknowns, namely, 3 forcescomponents at each hinge. To eliminatethe reactions at hinge.

20

0.2 m 0.125 m

0.125 m

0.2 m

θ

0.1 mx

y

z

A

B

C

D

xG

WT

C (0.2sin30o, - 0.2cos30o, 0.125)

G (0.1sin30o, - 0.1cos30o, 0.125)• Construct mathematical model and solve equations

)1(0)(ˆ)(ˆ:0

−−=ו+ו

=Σ

TBCkWBGk

M AB

kjiBG ˆ125.0ˆ0866.0ˆ05.0 +−=

kjiBC ˆ125.0ˆ1732.0ˆ1.0 +−=

jjW ˆ2.196ˆ)81.920( −=×−=

)ˆ2538.0ˆ1421.0ˆ0711.0(

)1759.0

ˆ125.0ˆ2732.0ˆ1.0(

)ˆ125.0ˆ1732.0ˆ1.0(

kjiT

kjiT

kjiTBC

−+−=

−+

×+−=×

)ˆ2.196()ˆ125.0ˆ0866.0ˆ05.0( jkjiWBG −×+−=×

)ˆ81.9ˆ53.24( ki −=

)1759.0

ˆ125.0ˆ2732.0ˆ1.0()( kjiTCDCDTT −+==

B (0, 0, 0)D (0.2, 0.1, 0)

0)ˆ2538.0ˆ1421.0ˆ0711.0(ˆ)ˆ81.9ˆ53.24(ˆ

=++−•+

−•

kjiTk

kik

-9.81 + 0.2538T = 0 T = 38.66 N

0)(ˆ)(ˆ:0 =ו+ו=Σ TBCkWBGkM AB

0.1 -0.1732

-0.08660.05

21

0)(ˆ)(ˆ =ו+ו TBCkWBGk

kjiBG ˆ125.0ˆ0866.0ˆ05.0 +−=

kjiBC ˆ125.0ˆ1732.0ˆ1.0 +−=

jjW ˆ2.196ˆ)81.920( −=×−=

)1759.0

ˆ125.0ˆ2732.0ˆ1.0( kjiTT −+=

-9.81 + 0.2538T = 0 T = 38.66 N

Or use determinant

01759.0

125.02732.01.0125.01732.01.0100

06.1920125.00866.005.0100

=−

−

+−

−

T

0)(ˆ)(ˆ:0 =ו+ו=Σ TBCkWBGkM AB

T

mg

ΣFy = 0:T - mg = 0

38.66 - m(9.81) = 0m = 3.94 kg

+

(3.94)(9.81)

0.2 m 0.125 m

0.125 m

0.2 m

E

30o

0.1 mx

y

z

A

B

C

D

xG

W

• Conclusion

22

Review 4.7 (Problem 4.133)

The 50-kg plate ABCD is supported by hinges along edge AB and by wireCE. Knowing that the plate is uniform, determine the tension in the wire.

200

480

200

240 240

400

400

B

A

C

D

y

x

z

E

Dimension in mm

Guide : Assuming that plate ABCDcan be treated as a rigid body. DrawFBD of it , There are 12 forces actingon it, the weight of plate W, the force Texerted by the cable, and reactionsinvolving 10 unknowns, namely, 3forces components and 2 momentcomponents at each hinge F and H. Toeliminate the reactions at F and H fromthe computations, we express that thesum of the moments of the forcesabout AB is zero (ΣMAB = 0).The forcesreactions is zero because there hasnot r and moment reaction is zerobecause dot product is zero.

FH

23

• Define the state of the problem • Construct physical model

200

480

200

240 240

400

400

B

A

C

D

y

x

z

E

FH

Given : m = 50 kgDetermine : T

Assuming that plate ABCD can be treated as a rigid body. There are 12 forces acting on it,the weight of plate W, the force T exerted by the cable, and reactions involving 10unknowns, namely, 3 forces components and 2 moment components at each hinge F and H.

200

480

200

240 240

400

400

B

A

C

D

y

x

z

E

TG

-WjH

F

24

• Construct mathematical model and solve equations

)1(0)(ˆ)(ˆ:0

−−=ו+ו

=Σ

WrTr

M

AGABAEAB

AB

λλ

W = -(mg)j = -(50x9.81)j = -490.5j

mmCEkjiCE 760ˆ400ˆ600ˆ240 =−+−=

)ˆ400ˆ600ˆ240(760

kjiTCECETT −+−==

)ˆ5ˆ12(131

520ˆ200ˆ480ˆ jiji

ABAB

AB −=−==λ

jir AEˆ400ˆ240 +=

kjir AGˆ200ˆ100ˆ240 +−=

Substitute in Eq.(1) : 0131

05.49002001002400512

7601340060024004002400512

=−−−

+×

−−+−

T

(-12x400x400 - 5x240x400)T/760 + 12x200x490.5 = 0 T = 372.78 N

200

480

200

240 240

400

400

B

A

C

D

y

x

z

E

TG

-Wj

F

F

25

200

480

200

240 240

400

400

B

A

C

D

y

x

z

E

G

-Wj

• Conclusion

760)ˆ400ˆ600ˆ240()373( kjiNT −+−=

F

F