MTE119 - Solutions Homework 3mte119/s/MTE119 - Solutions Hw3.pdfMechatronics Engineering NAME & ID DATE MTE 119 – STATICS HOMEWORK 3 SOLUTIONS PAGE Problem 2: Textbook Exercise 3

Mechatronics Engineering

NAME & ID DATE MTE 119 – STATICS HOMEWORK 3

SOLUTIONS

PAGE

Problem 1: Textbook Exercise 3-42

Goal:

Determine the magnitudes of the forces F1, F2 and F3 for equilibrium of the particle.

Solution:

Strategy:

Write the forces in Cartesian system and then use the equilibrium equations for the particle in x, y and z axis to find the magnitudes.

Writing Forces in Cartesian form:

NkFiFkiFF }ˆ866.0ˆ5.0{}ˆ60sinˆ60{cos 1111 +=+=

NjFiFjiFF }ˆ8.0ˆ6.0{}ˆ54ˆ

53{ 2222 −=−=

NjFiFjiFF }ˆ5.0ˆ866.0{}ˆ30sinˆ30cos{ 3333 −−=−−=

Equilibrium Equations for particle:

For the x-axis:

0866.06.05.00 321 =−+⇒=∑ FFFFx

For the y-axis:

030sin8005.08.00 32 =+−−⇒=∑ FFFy

For the z-axis:

030cos800866.00 1 =−⇒=∑ FFz

Solving the above system of equations:

NFNFNF 564,147,800 321 ===

112

Mechatronics Engineering

NAME & ID DATE MTE 119 – STATICS HOMEWORK 3

SOLUTIONS

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Problem 2: Textbook Exercise 3-51

Cables AB and AC can sustain a maximum tension of 500 N, and the pole can support a maximum compression of 300 N.

Goal:

Determine the maximum weight of the lamp that can be supported in the position shown.

Assumptions:

The force in the pole acts along the axis of the pole.

Solution:

Strategy:

First write the forces and the weight of the lamp in Cartesian form, then write the equilibrium equations for point A and solve for the weight.

Writing forces in Cartesian form:

NkjiFF AOAO }ˆ5.6

6ˆ5.65.1ˆ

5.62{ +−=

NkjiFF ABAB }ˆ96ˆ

93ˆ

96{ −+−=

NkjiFF ACAC }ˆ76ˆ

73ˆ

72{ −+−=

NkWW }ˆ{−=

Equilibrium Equations:

For the x-axis

072

96

5.620 =−−⇒=∑ ACABAOx FFFF

For the y-axis

073

93

5.65.10 =++

−⇒=∑ ACABAOx FFFF

212

Mechatronics Engineering

NAME & ID DATE MTE 119 – STATICS HOMEWORK 3

SOLUTIONS

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For the z- axis

076

96

5.660 =−−−⇒=∑ WFFFF ACABAOx

Assuming that one of the members is already under maximum load, we can solve the above equations to calculate what the load is in the other members.

First Case: NFAB 500=

072)500(

96

5.62

=−− ACAO FF

073)500(

93

5.65.1

=++−

ACAO FF

076)500(

96

5.66

=−−− WFF ACAO

NNFNF AOAC 30046.1444,9.388 >==⇒ Not acceptable

Second case: NFAC 500=

0)500(72

96

5.62

=−− ABAO FF

0)500(73

93

5.65.1

=++−

ABAO FF

0)500(76

96

5.66

=−−− WFF ABAO

NNFNNF AOAB 30014.1857,50085.642 >=>=⇒ Not acceptable

Third case: NFAO 300=

072

96)300(

5.62

=−− ACAB FF

073

93)300(

5.65.1

=++−

ACAB FF

076

96)300(

5.66

=−−− WFF ACAB

NWNFNF ABAC 138,104,8.80 ===⇒ Answer

312

Mechatronics Engineering

NAME & ID DATE MTE 119 – STATICS HOMEWORK 3

SOLUTIONS

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Problem 3: Textbook Exercise 3-65:

Determine the tension developed in cables OD and OB and the strut OC, required to support the 50-kg crate. The spring OA has an un-stretched length of 0.8 m and a stiffness mkNkOA /2.1= .

The force in the strut acts along the axis of the strut.

Solution:

Strategy:

First draw the free body diagram and write the equilibrium equation for point O. Weight and spring force are known and the other can be found by solving equilibrium equations.

Free Body Diagram:

412

Mechatronics Engineering

NAME & ID DATE MTE 119 – STATICS HOMEWORK 3

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Spring Force:

NkNksFsp 24024.0)8.01(2.1 ==−==

Writing forces in Cartesian form:

kFjFiFkjiFF OBOBOBOBOBˆ

32ˆ

32ˆ

31)

4)4()2(

ˆ4ˆ4ˆ2(222

+−−=+−+−

+−−=

kFiFkiFF OCOCOCOCˆ

53ˆ

54)

3)4(

ˆ3ˆ4(22

+−=+−

+−=

kFjFiFkjiFF ODODODODODˆ

32ˆ

32ˆ

31)

442

ˆ4ˆ4ˆ2(222

++=++

++=

NjFsp }ˆ240{−=

NkNkW }ˆ5.490{}ˆ)81.9(50{ −=−=

Equilibrium equations:

00 =++++⇒=∑ WFFFFF spODOCOB

0ˆ)5.49032

53

32(

ˆ)24032

32(ˆ)

31

54

31(

=−++

+−+−++−−⇒

kFFF

jFFiFFF

ODOCOB

ODOBODOCOB

1 4 1( ) 03 5 32 2

( 240) 03 3

2 3 2( 490.5) 03 5 3

OB OC OD

OB OD

OB OC OD

F F F

F F

F F F

⎧⎪ − − + =⎪⎪⎪⎪⎪⎪ − + − =⎨⎪⎪⎪⎪ + + − =⎪⎪⎪⎩

Solving the above equations, the forces are:

NFNFNF ODOCOB 480,150,120 ===

512

Mechatronics Engineering

NAME & ID DATE MTE 119 – STATICS HOMEWORK 3

SOLUTIONS

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Problem 4: Textbook Exercise 4-14 and 4-15:

4-14 Take lbFlbF CB 50,40 ==

and determine the moment of each force about the bolt located at A.

4-15 If lbFB 30= and lbFC 45= , determine the resultant moment about the bolt located at A.

Solution:

Strategy:

Find the force component normal to the rod and multiply it by it’s distance from point O.

For problem 4-14

ftlbM

lbFF

B

BBNormal

.6.90)5.2(25.36

25.3625cos

==

==

ftlbM

lbFF

B

CCNormal

.141)25.3(3.43

3.4330cos

==

==

For problem 4-15

ftlbM A .195)25.3(30cos45)5.2(25cos30 =+=

612

Mechatronics Engineering

NAME & ID DATE MTE 119 – STATICS HOMEWORK 3

SOLUTIONS

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Problem 5: Textbook Exercise 4-21:

The tool at A is used to hold a power lawnmower blade stationary while the nut is being loosened with the wrench at B in the direction shown.

Goals:

• Determine the moment it creates about the nut at C.

• What is the magnitude of force F at A so that it creates the opposite moment about C?

Solution:

Strategy:

Use the fact that the summation of moments must be zero

The magnitude of the force normal to the rod exerted by the wrench at B:

sin 60 50 sin 60BF =

The magnitude of the moment of BF about C is:

sin 60 (0.3) 12.99 13N.mBC BM F= = =

The magnitude of the moment exerted by the tool at A about C is:

( )( )12

0.4 N.m13

AC AM F=

Since the two moments balance each other:

12: 12.99 ( )(0.4) 35.2

13B AC C A AM M F F N= = ⇒ =

712

Mechatronics Engineering

NAME & ID DATE MTE 119 – STATICS HOMEWORK 3

SOLUTIONS

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Problem 6: Textbook Exercise 4-26:

The towline exerts a force of P=4kN at the end of the 20m long crane boom. If 30=θ

Goal:

• Determine the placement x of the hook at A so that this force creates a maximum moment about point O.

• What is this moment?

Solution:

Strategy:

The maximum moment occurs when the force is normal to the crane boom.

Free-Body Diagram:

By geometry:

mLLx

mLLL

mL

mL

DCAC

ACAC

BC

BC

DC

2495.23

63.660tan

5.1130sin205.1

32.1730cos20

==+=

=⇒=

=+=

==

The maximum moment of P about O is then:

4000(20) 80kN.mMAXO OBM P L= ⋅ = =

B

A O

C

20 m

1.5 m

x

30o

60o

D

812

Mechatronics Engineering

NAME & ID DATE MTE 119 – STATICS HOMEWORK 3

SOLUTIONS

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Problem 7: Textbook Exercise 4-65:

If a torque or moment of 80lb.in is required to looser the bolt at A, determine the force P that must be applied perpendicular to the handle of the flex-headed ratchet wrench.

Solution:

Strategy:

Since the force is perpendicular to the handle the resultant moment is the force magnitude multiplied by the distance from the bolt at A.

lbP

PM A

5.841.9

8080)60sin1075.0(

==⇒

=+=

912

Mechatronics Engineering

NAME & ID DATE MTE 119 – STATICS HOMEWORK 3

SOLUTIONS

PAGE

Problem 8: Textbook Exercise 4-74:

The resultant couple moment created by the two couples acting on the disk is

ˆ{10 }kipRM k= .in.

Goal:

Determine the magnitude of force T.

Solution:

Let us do:

1 2in.L =

( )2 4 2 3 in. 10in.L = + + =

The resultant couple moment is: 2

1

1 210kip (2) (9) 11

R ii

R

M M

M TL TL T T T=

=

= = + = + =

∑

The magnitude of the force T:

0.909kipT⇒ =

L2

L1

1012

Mechatronics Engineering

NAME & ID DATE MTE 119 – STATICS HOMEWORK 3

SOLUTIONS

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Problem 9: Textbook Exercise 4-85:

Two couples act on the frame. If d=4 ft, determine the resultant couple moment. Compute the result by resolving each force into x and y components and

a) Finding the moment of each couple (Eq. 4-13) b) Summing the moments of all the force components about point B.

Solution:

a) Finding the moments of each couple:

1

1

2

2

ˆ4

4 3ˆ ˆ80( ) 80( )5 5

ˆ3

ˆ ˆ50 sin 30 50cos 30

R j

F i j

R i

F i j

=

= − −

=

= − −

2

1

( )C i ii

M R F=

= ×∑

ˆ3 0 0 0 4 0 {126 }lb.ft

64 48 050 sin 30 50 cos 30 0

C

i j k i j k

M k= + =

− −− −

b) Find the component of the forces that creates a moment about B and multiply by the distance from B.

ftlbM C .126)5)(80(54)1)(80(

54)5(30cos50)2(30cos50 =+−−=

B R2

R1

F2

F1

2 ft

1 ft

4 ft

3 ft

1112

Mechatronics Engineering

NAME & ID DATE MTE 119 – STATICS HOMEWORK 3

SOLUTIONS

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Problem 10: Textbook Exercise 4-93:

The gear reducer is subject to the couple moments shown. Determine the resultant couple moment and specify its magnitude and coordinate direction angles.

Solution:

Strategy:

Write the couples in Cartesian system and calculate the sum of them to obtain the resultant couple moment. Find the couple unit vector to find the coordinate direction angles.

The moments in Cartesian form:

ftlbkjikjiM

ftlbiM

.}ˆ40ˆ99.48ˆ99.48{)ˆ30sinˆ45cos30cosˆ45sin30cos(80

.}ˆ60{

2

1

−−−=−−−=

=

Resultant Moment:

ftlbkjiMMM RR .}ˆ40ˆ49ˆ01.11{ −−=⇒=∑

Magnitude of the resultant moment:

ftlbM R .2.64)40()99.48(01.11 222 =−+−+=

Coordinate direction angles: , , 120α β γ ≤

129)2.64

40(cos

140)2.6499.48(cos

1.80)2.64

01.11(cos

1

1

1

=−

=

=−

=

==

−

−

−

γ

β

α

1212