mte119 - solutions hw5

Mechatronics Engineering

NAME & ID DATE MTE 119 STATICS HOMEWORK 5

SOLUTIONS

PAGE

Problem 1:

The toggle switch consists of a cocking lever that is pinned to a fixed frame at A and held in place by the spring which has an unstretched length of 200 mm. Determine the magnitude of the resultant force at A and the normal force on the peg at B when the lever is in the position shown.

Strategy:

Apply the rule of cosines to find the stretched length of the spring and find the spring force. Draw the free body diagram

and calculate the moment about point A to get the normal force on the peg. Apply the equilibrium equation to find the reaction forces at A.

2 20.3 0.4 2(0.3)(0.4)cos150 0.6766

sin sin150 12.8080.3 0.67664

5(0.67664 0.2) 2.3832s

L m

F k s N

= + =

= == = =

D

DD

0.4 m

0.3

m

L 150o

12.808o

0.1

0.3

Ax Ay

NB

0 (2.3832sin12.808 )(0.4) (0.1) 02.113

A B

B

M NN N

= + = = D

0 2.3832cos12.808 02.3239

x x

x

F AA N

= = = D

2 2

0 2.11327 2.3832sin12.808 0

1.585

2.3239 ( 1.585) 2.81

x y

x

A

F AA N

F N

= + = =

= + =

D

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SOLUTIONS

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Problem 2:

The Uniform rod has a length l and weight W. it is supported at one end A by a smooth wall and the other end by a cord of length s which is attached to the wall as shown. Show that for

equilibrium it is required that ( ) 1/ 22 2 / 3h s l =

Strategy:

Apply equations of equilibrium to obtain an expression for the tension in the cable and then apply the law of sine and cosine to find h.

Applying the law of Sine:

NA W T

L

h s

0 sin ( ) sin ( ) 02

sin2sin

0 cos( ) 0

sin cos( ) 02sin

sin cos( ) 2sin( ) 0

A

y

lM T l W

WT

F T WW W

= =

=

= = = =

180 sin sin sin sin

sin cos( ) 2 sin( ) 0

cos( ) 2 (1)

hh s s

hs

hs

= = =

=

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SOLUTIONS

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Using the cosine law: 2 2 2

2 2 2

2 cos( )

cos( ) (2)2

l h s hsh s l

hs

= + + =

Using equation (1) and (2) => 2 2 2 2 22

2 3h h s l s lhs hs

+ = =

Problem 3:

The uniform load has a mass of 600 kg and is lifted using a 30-kg strongback beam and four wire ropes as shown. Determine the tension in each segment of the rope and the force that must be applied to the sling at A.

Strategy:

Due to symmetry, all wires are subjected to the same tension. This condition satisfies moment equilibrium about x and y axis and force equilibrium along y axis.

Apply the force equilibrium along z axis to find the tension in the wires and the force at point A.

W

T 3

4

40 4 ( ) 600(9.81) 05

1839.37 1.84

zF T

T N kN

= = = =

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SOLUTIONS

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The force F applied to the sling A must support the weight

of the load and the strongback beam.

Problem 4:

The windlass is subjected to a load of 150 lb. Determine the horizontal force P needed to hold the handle in the position shown, and the components of the reaction at the ball-and-socket joint A and the smooth journal bearing B. The bearing at B is in proper alignment and exerts only force reactions on the windlass.

Strategy:

Draw the free body diagram and apply force and moment equilibrium for x, y and z axis and find the unknown forces.

F

W beam

W box

0 600(9.81) 30(9.81) 06180.3 6.18

zF FF N kN

= = = =

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SOLUTIONS

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Problem 5:

The pole is subjected to the two forces shown. Determine the components of reaction at A assuming it to be a ball-and-socket joint. Also, compute the tension in each of the guy wires, BC and ED.

Strategy:

Draw the free body diagram and then apply the equilibrium equation of force and moment and find the unknowns.

0 (150)(0.5) (1) 0

750 0

y

y y

M PP lbF A

= = =

= =

0 (150)(2) (4) 0750 75 150 0 75

x z

z

z z z

M BB lbF A A lb

= = =

= + = =

0 (4) 75(6) 0112.5 1120 112.5 75 0 37.5

z x

x

x x x

M BB lb lbF A A lb

= = = =

= + = =

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SOLUTIONS

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Force vector:

1

2

2 2 2

860{cos 45 sin 45 } 608.11 608.11

450{ cos 20 cos30 cos 20 sin 30 sin 20 } 366.21 211.43 153.91

( 6 0) ( 3 0) (0 6)( 6) ( 3) ( 6)

2 1 2 [3 3 3

ED ED

ED

F i k i k N

F i j k

i j k N

i j kF F

F i j k

= = = + = +

+ + = + + =

D D

D D D D D

GG

G

2 2 2

]

(6 0) ( 4.5 0) (0 4)(6) ( 4.5) ( 4)

12 9 8 [ ]17 17 17

BC BC

BC

i j kF F

F i j k

+ + = + + =

G

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SOLUTIONS

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Force Equilibrium:

Moment Equilibrium:

Solving Eqs. (4) and (5):

205.09 , 628.57BC EDF N F N= =

Substituting the results into Eqs. (1), (2) and (3):

32.4 , 107 , 1277.58 1.28x y zA N A N A N kN= = = =

0

2 120; 608.11 366.11 0 (1)3 17

1 90; 211.43 0 (2)3 17

2 80; 608.11 153.91 0 (3)3 17

x x ED BC

y y ED BC

z z ED BC

F

F A F F

F A F F

F A F F

=

= + + =

= + =

= =

G

1 2 0 4 8 ( ) 6 ( ) 0

360; 1691.45 2 0 (4)17

480; 1935.22 4 0 (5)17

A BC ED

x ED BC

y ED BC

M k F k F F k F

M F F

M F F

= + + + =

= + + =

= + =

G G G G G

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SOLUTIONS

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Problem 6:

Eight identical 20 30 -in rectangular plates, each weighing 50lb, are held in a vertical plane as shown. All connections consist of frictionless pins, rollers, or short links. For each case, determine whether (a) the plate is completely, partially or improperly constrained, (b) the reactions are statically determinate or indeterminate, (c) the equilibrium of the plate is maintained in the position shown. Also, wherever possible, compute the reactions.

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SOLUTIONS

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SOLUTIONS

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PROBLEM 7:

Determine the force in each member of the truss and state if the members are in tension or compression. Set P1=100 lb, P2=200 lb P3=300 lb.

SOLUTION:

FBD of the entire truss:

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SOLUTIONS

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0; 200(10) 300(20) cos30 (30) 0 307.9 lb

0; 100 200 300 307.9cos30 0 333.4 lb

0; 307sin 30 0 154.0 lb

oA D D

oy y y

ox x x

M R R

F A A

F A A

= + = =

= + = =

= = =

Joint A:

10; 333.4 -100 0 330 lb (C) ANS210; 154.0 (330) 0 79.4 lb (T) ANS2

y AB AB

x AF AF

F F F

F F F

= = =

= + = =

Ax

100 lb

200 lb 300 lb 30o

RD

Ay

A

B C

F E D

10 ft 10 ft 10 ft

A 154.0 lb

100 lb

333.4 lb

FAB

FAF 1 1

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SOLUTIONS

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Joint B:

10; (330) 0 233.3 lb (T) ANS2

10; (330) 0 233.3 lb (C) ANS2

y BF BF

x BC BC

F F F

F F F

= = =

= = =

Joint F:

10; - 200 233.3 0 47.14 lb (C) ANS210; (47.14) 79.37 0 113 lb (T) ANS2

y FC FC

x FE FE

F F F

F F F

= + = =

= + = =

F 79.37 lb

233.3 lb

200 lb

FFC

FFE 1 1

B

330 lb FBF

FBC

1 1

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SOLUTIONS

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Joint E:

0; 300 lb (T) ANS

0; 113 lb (T) ANSy EC

x ED

F F

F F

= == =

Joint C:

1 10; (47.14) 233.3 0 377.1 lb (C) ANS2 2

1 10; (47.14) 300 (377.1) 0 Check!2 2

x CD CD

y

F F F

F

= + = =

= + =

C 233.3 lb

FCD 1

1 47.14 lb

E 113 lb

FEC

300 lb

FED

300 lb

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SOLUTIONS

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PROBLEM 8:

Determine the force in each member of the truss and state if the members are in tension or compression. Hint the horizontal force component at A must be zero. Why?

SOLUTION:

Joint C:

0; 800cos 60 0 =400 lb (C) ANS

0; 800sin 60 0 =693 lb (C) ANS

ox CB CB

oy CD CD

F F F

F F F

= == =

Joint B

B

FBA

x

y

4 3 FBD

600 lb

400 lb

C FCB

FCD

800 lb

x

y

60o

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SOLUTIONS

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30; 400 0 =666.7 lb (T) ANS 5

40; (666.7) 600 0 =1133 lb (C) ANS 5

x BD BD

y BA BA

F F F

F F F

= =

= =

PROBLEM 9:

Determine the force developed in member GB and GF of the bridge truss and state if these members are in tension or compression.

FBD:

Ax

600 lb 800 lb Ay

A B C

D

10 ft

4 ft

10 ft

10 ft

F G 4 ft E

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SOLUTIONS

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0; - 600(10) -800(18) (28) 0 =728.571 lb

0; 0

0; 600 800 728.571 0 671.429 lb

A y y

x x

y y y

M D D

F A

F A A

= + =

= =

= + = =

Consider Section a-a

Take moment about the joint B:

0; - 671.429(10) (10) 0 F =671.429 lb (C) ANS

0; 671.429 =0 =671.429 lb (T) ANS

B GF GF

y GB GB

M F

F F F

= + =

=

Ax

Ay

A B

10 ft

10 ft

G

a

FGB

FAB

FGF a

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SOLUTIONS

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PROBLEM 10:

For the truss shown determine:

a. The force in members DG and FH and indicate if the members are in tension or compression. Hint: Use section a-a

Ans. 3.60DGF kN= C, 3.60FHF kN= T b. The force in members IL, GJ, and HK and indicate if the members are in tension or

compression. Hint: Begin with pins I and J and use section b-b

Ans. 3.60ILF kN= C, 3.60HKF kN= T, 0.833GJF kN= C

SOLUTION:

Free-Body Diagram of the Truss:

Equilibrium Equations:

0xF+ = : 0xA =

0yF+ = : 6.4yA N kN+ = (1)

0AM+ =2 : ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( )1 1.6 3.2 1.2 2.4 8 1.2 6.4 8 1 9.6 0N + =

Therefore, 3.2N kN= Using (1): 6.4 3.2y yA N A kN= =

Ay

Ax

N

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SOLUTIONS

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Or, noticing the symmetry of the structure: yA N=

Using (1): 6.4 3.22y y

A A N kN= = =

a. Free-Body Diagram of Section a-a:

0FM+ =2 : ( )( ) ( )( ) ( )( )2.4 3.2 3.2 1.6 1 0DGF + =

3.60DGF kN= C Ans. 0DM+ =2 :

( )( ) ( )( ) ( )( )2.4 1.6 1 3.2 3.2 0FHF + = 3.60FHF kN= T Ans.

b. Joint I:

By inspection GI ILF F= and 1.2IJF kN= C Joint J:

By inspection ( ) ( )GJ HJF C F T= (2) Free- Body Diagram of Section b-b:

0xF+ = : ( )4 0

5IL HK GJ HJF F F F + =

0IL HKF F = (3)

Ay

DGFDEF

EFFFHF

N

ILFGJF

HJF

HKF

1.2kN1.2m

1.2m

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SOLUTIONS

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0JM+ =2 : ( )( ) ( )( ) ( )( )3.2 3.2 1.2 1.6 1 0IL HKF F + = so, 7.2IL HKF F+ = (4) (3) in (4):

3.60ILF kN= C Ans. 3.60HKF kN= T Ans.

Finally, 0yF+ = : ( )33.2 1 1.2 0

5 GJ HJF F + =

53GJ HJ

F F+ = Using (2):

5 0.8336GJ HJ GJ

F F F kN= = = C Ans.

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mte119 - solutions hw5

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