problem 1-exercise 6-73 - university of waterloomte119/s/mte119 - solutions hw6.pdf · problem...

Mechatronics Engineering

NAME & ID DATE MTE 119 – STATICS

HOMEWORK 6

SOLUTIONS

PAGE

PROBLEM 1-EXERCISE 6-73

The compound beam is pin-supported at C and supported by a roller A and B. There is a

hinge (pin) at D. Determine the reactions at the supports. Neglect the thickness of the beam.

SOLUTION 1-EXERCISE 6-73

6 ft 4 ft 2 ft

4 kip

8 kip

Dx

DyAy

30

(a)

Equation of Equilibrium: from FBD of (a)

∑ = 0DM 0)6()2(8)12(30cos4 =−+ yA�

kipAy 595.9=

∑ = 0yF 0830cos4595.9 =−−+�

yD kipDy 869.1=

∑ = 0xF 030sin4595.9 =−+�

xD kipDx 00.2=

1 22

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HOMEWORK 6

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From FBD of (b):

∑ = 0CM 0)16()8.0(1215)24(869.1 =−++ yB kipBy 541.8=

∑ = 0yF 0)8.0(12869.1541.8 =−−+yC kipC y 93.2=

∑ = 0xF 0)6.0(1200.2 =−−xC kipCx 2.9=


Determine the horizontal

and vertical components

of force which the pin at

A, B and C exert on

member ABC of the

frame.

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HOMEWORK 6

SOLUTIONS

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FBD:

0; - (3.5) 400(2) 300(3.5) 300(1.5) 0

657 N ANS

E y

y

M A

A

= + + + =

=

∑

FBD of member CD:

0; (3.5) 400(2) 0 229 N ANSD y yM C C= − + = =∑

Cy Dy

Dx Cx

400 N

1.5 m 2 m

Ay

Ey

Ex

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HOMEWORK 6

SOLUTIONS

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FBD of member ABC:

0; 0

0; F = F

50; 657.1 229 2 0

74

368.7 N

0 .

5(368.7)(2) 429 N Ans.

74

B x

x BD BE

y BD

BD BE

x

y

M C

F

F F

F F

B Ans

B

= =

=

= − − =

= =

= ←

= = ←

∑

∑

∑

PROBLEM 1-EXTRA PRACTICE 6-76

229 N

FBD

Cx

2.5 m

2.5 m

FBE

5

5

7 7

4

22

The three-hinged arch

support the loads

kNF 81 =

and kNF 52 = . Determine

the horizontal and vertical

components at the pin

supports A and B.

Take mh 2= .



HOMEWORK 6

SOLUTIONS

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SOLUTION 1-EXTRA PRACTICE 6-76

Member AC:

∑ = 0AM 0)7()8()4(8 =++− xy CC

∑ = 0xF 08 =−− xx CA

∑ = 0yF 0=+− yy CA yy AC =

Member BC:

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HOMEWORK 6

SOLUTIONS

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∑ = 0BM 0)9()6()2(5 =−+ xy CC

∑ = 0xF 0=− xx BC

∑ = 0yF 05 =−+− yy BC

Solving the above equations we have:

kNAx 61.56141.5 ==

kNAy 91.19122.1 ==

kNC x 39.23859.2 ==

kNC y 91.19122.1 ==

kNBx 39.23859.2 ==

kNBy 91.69122.6 ==


Determine the horizontal

and vertical components

of force at each pin. The

suspended cylinder has a

weight of 80 lb.

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HOMEWORK 6

SOLUTIONS

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3 ft 3 ft 1 ft

80 lb

80 lb

By

Bx

Ax

Fcd3

2

∑ = 0BM 0)4(80)3)(13

2( =−CDF lbFCD 3.192=

lbDC xx 160)3.192(13

3===

lbDC yy 107)3.192(13

2===

∑ = 0yF 080)3.192(13

2=−+− yB lbBy 7.26=

Ey

Ex

Bx

80 lb

26.7 lb

3 ft

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HOMEWORK 6

SOLUTIONS

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∑ = 0EM 0)3(7.26)3(80)4( =++− xB lbBx 0.80=

∑ = 0xF 08080 =−+xE 0=xE

∑ = 0yF 07.26 =+− yE lbE y 7.26=

∑ = 0xF 080)3.192(13

380 =−++− xA lbAx 160=



FBD of dismembered machine:

FBC Ax

Ay

10 lb

Ax

F DY

Dx

Ay

If a force P=6 lb is applied

perpendicular to the

handle of the mechanism,

determine the magnitude

of force F for equilibrium.

The members are pin-

connected at A, B, C and

D.

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HOMEWORK 6

SOLUTIONS

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D

0 : (4) 6(25) 0 37.5

0 : 6 0 6 lb

0 : 37.5 0 =37.5b lb

0 : 5(6) 37.5(9) (39)( ) 0 9.42 .

A BC BC

x x x

y y y

M F F lb

F A A

F A A

M F F lb Ans

= − = → =

= − + = → =

= − + = →

= − − + = → = ←

∑

∑

∑

∑



FBD

Ax

Ay

Dy

Dx

20 lb

Determine the required

force P that must be

applied at the blade of the

pruning shears so that the

blade exerts a normal force

of 20 lb on the twig at E.

922



HOMEWORK 6

SOLUTIONS

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0; (5.5) (0.5)+20(1)=0

5.5 0.5 0

0; = A

0; 20 0

D x

x

x x x

y y y

M P A

P A

F D

F D A

= − −

+ =

=

= − − =

∑

∑

∑

0; (0.75)+ (0.5) 4.75 0

30; 0

3

20; + - =0

13

B y x

x x CB

y y CB

M A A P

F A F

F A P F

= − =

= − =

=

∑

∑

∑

Solving:

13.3 lb, 6.46 lb

13.3 lb, 28.9 lb

2.42 lb .

16.0 lb

x y

x y

CB

A A

D D

P Ans

F

= =

= =

= ←

=

2

3

Ay

Ax

FCB

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HOMEWORK 6

SOLUTIONS

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A) FBD

Bar:

0; 2( / 2) 2(87.5) 0 175 lb ANSyF F F= − = =∑

Man:

0; 175 2(87.5) 0 350 lb ANSy c cF N N= − − = =∑

F/2 F/2

87.51 lb 87.51 lb

Nc

87.51 lb 87.51 lb 175 lb

A man having a weight of

175 lb attempts to lift

himself using one of the

two methods shown.

Determine the total force

he must exerts on bar AB

in each case and the

normal reaction he exerts

on the platform at C.

Neglect the weight of the

platform.

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HOMEWORK 6

SOLUTIONS

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B) FBD

Bar:

0; 2(43.75) 2( ) 0 87.5 lb ANSyF F F= − = =∑

Man:

0; 175 2(43.75) 0 87.5 lb ANSy c cF N N= − − = =∑


F/2 F/2

43.75 lb 43.75 lb

Nc

43.75 lb 43.75 lb 175 lb

1222

The tractor boom supports

the uniform mass of 500

kg in the bucket which

has a center of mass at G.

Determine the force in

each hydraulic cylinder

AB and CD and the

resultant force at pins E

and F. The load is

supported equally on each

side of the tractor by

similar mechanism.



HOMEWORK 6

SOLUTIONS

PAGE


From The bucket FBD:

∑ = 0EM 0)25.0()1.0(5.2452 =− ABF NFAB 981=

∑ = 0xF 0981 =+− xE NEx 981=

∑ = 0yF 05.2452 =−yE NE y 5.2452=

kNFE 64.2)5.2452()981( 22=+=

(a) The Bucket FBD (b) The Tractor Boom FBD

From the tractor boom FBD:

∑ = 0FM 0)25.1)(2.12(sin)7.0)(2.12(cos)8.2(5.2452 =+−��

CDCD FF

kNNFCD 3.1616349 ==

∑ = 0xF 02.12sin16349 =−�

xF NFx 3455=

∑ = 0yF 02.12cos163495.2452 =+−−�

yF NFy 13527=

kNFF 0.14)13527()3455( 22=+=

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HOMEWORK 6

SOLUTIONS

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PROBLEM 3:

SOLUTION 3

Free-body diagram of the entire beam:

0yF =∑ :

750 0B C+ − =

. 0pt BM =∑ :

( )( )0.375 0.25 750 0C − =

Solving: 250B N= , 500C N=

The distributed load is:

( )7507500

0.1

N N

m mω = =

From the equilibrium equations:

0xF =∑ : 0A

P =

0yF =∑ :

( )250 0.05 7500 0AV− − =

0right endM−

=∑ :

Model the ladder rung as a simply

supported (pin supported) beam and

assume that the 750-N load exerted

by the person’s shoe is uniformly

distributed. Determine the internal

forces and moment at A.

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HOMEWORK 6

SOLUTIONS

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( ) ( )250 0.25AM − +

( )( ) ( )0.05 7500 0.025 0+ = Therefore, the internal forces at A are:

0A

P = , 125A

V N= − , 53.1A

M = N-m


(a) Point C, which is just to the right of the bearing at A.

(b) Point D, which is just to the left of the 3000-lb force.


∑ = 0BM 0)2(3000)8(900)20(2500)14( =+++− yA

lbAy 4514=

∑ = 0xF 0=xB

∑ = 0yF 0300090025004514 =−−−+yB lbBy 1886=

The shaft is supported

by a journal bearing

at A and a thrust

bearing at B.

Determine the normal

force, shear force, and

moment at a section

passing through

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HOMEWORK 6

SOLUTIONS

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∑ = 0CM 0)6(2500 =+ CM ftlbM C .15000−=

∑ = 0xF 0=CN

∑ = 0yF 045142500 =+−− CV lbVC 2014=

∑ = 0DM 0)2(1886 =− DM ftlbM D .3771=

∑ = 0xF 0=DN

∑ = 0yF 018863000 =+−− DV lbVD 1114=


Determine the normal

force, shear force, ad

moment at a section

passing through point

C. Take kNP 8=

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HOMEWORK 6

SOLUTIONS

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∑ = 0AM 0)25.2(8)6.0( =+− T kNT 30=

∑ = 0xF kNAx 30=

∑ = 0yF kNAy 8=

∑ = 0CM 0)75.0(8 =− CM mkNM C .6=

∑ = 0xF kNNC 30−=

∑ = 0yF 08 =+CV kNVC 8−=


Determine the internal

normal force, shear force,

and moment acting at point

C, and point D, which is

acted just to the right of the

roller support at B.

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HOMEWORK 6

SOLUTIONS

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Support Reactions: From FBD (a):

FBD (a)

∑ = 0AM 0)10(800)4(2400)2(800)8( =−−+yB lbBy 2000=

Internal Forces: Applying the equations of equilibrium to segment ED (FBD (b)), we have:

FBD (b)

∑ = 0DM 0)2(800 =−− DM ftlbM D .1600−=

∑ = 0xF 0=DN

∑ = 0yF 0800 =−DV lbVD 800=

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HOMEWORK 6

SOLUTIONS

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FBD (c)

Applying the equations of equilibrium to segment EC (FBD(c)), we have:

∑ = 0CM 0)6(800)2(1200)4(2000 =−−+− CM ftlbM C .800=

∑ = 0xF 0=CN

∑ = 0yF 080012002000 =−−+CV 0=CV



Support Reactions:

From FBD (a):

∑ = 0AM 0)45cos63(12)45cos66( =+−+��

yB kNBy 485.8=

∑ = 0xF 0=xA

∑ = 0yF 00.12485.8 =−+yA kNAy 515.3=

Determine the internal

force, shear force, and the

moment at point C and D.

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HOMEWORK 6

SOLUTIONS

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FBD (a) FBD (b)

From FBD (b):

∑ = 0CM 0)2(45cos515.3 =−�

CM mkNM C .97.4=

∑ = 0xF 045sin515.3 =− CN�

kNNC 49.2=

∑ = 0yF 0)2(45cos515.3 =− CV�

kNVC 49.2=

FBD (c)

∑ = 0DM 0)5.1(6)3(485.8 =−− DM mkNM D .5.16=

∑ = 0xF 0=DN

∑ = 0yF 06485.8 =−+DV kNVD 49.2−=

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HOMEWORK 6

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Free Body Diagram: The support reactions need not be computed.

Determine the x, y, z

components of force and

moment at point C in the

pipe assembly. Neglect the

weight of the pipe. Take

lbjiF }ˆ400ˆ350{1 −=

and

lbkjF }ˆ150ˆ300{2 +−=

2122



HOMEWORK 6

SOLUTIONS

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Internal Forces: Applying the equations of equilibrium to segment BC. We have:

;0=∑ xF 0350 =+CN lbNC 350−=

;0=∑ yF 0300400)( =−−yCV lbV yC 700)( =

;0=∑ zF 0150)( =+zCV lbV zC 150)( −=

;0=∑ xM 0)3(400)( =+xCM ftkipftlbM xC .2.1.1200)( −=−=

;0=∑ yM 0)2(150)3(350)( =−+yCM ftlbM yC .750)( −=

;0=∑ zM 0)2(400)2(300)( =−−zCM ftkipftlbM zC .4.1.1400)( ==

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problem 1-exercise 6-73 - university of waterloomte119/s/mte119 - solutions hw6.pdf · problem...

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