N ID DATE MTE 119 S P HOMEWORK 9 1 S 14mte119/s/MTE119 - Solutions Hw9.pdfstiffness of k =2lb/ ft and is originallyunstreched. Mechatronics Engineering NAME & ID DATE MTE 119 – STATICS

Mechatronics Engineering

NAME & ID DATE MTE 119 – STATICS HOMEWORK 9

SOLUTIONS

PAGE

PROBLEM 8-39

SOLUTION 8-39:

(a) (b)

Equations of Equilibrium: Since block A and B is either not moving or on the verge of moving, the spring force 0=spF ,

FBD (a):

∑ = 0xF 0sin10 =− θAF (1)

∑ = 0yF 0cos10 =− θAN (2)

From FBD (b):

∑ = 0xF 0sin6 =− θBF (3)

∑ = 0yF 0cos6 =− θBN (4)

Friction: Assuming block A is on the verge of slipping, then:

114

Two blocks A and B have a weight of 10 lb and 6 lb, respectively. They are resting on the incline for which the coefficients of static friction are

15.0=Aμ and 25.0=Bμ . Determine the angle θ which will cause motion of one of the blocks. What is the friction force under each of the blocks when this occurs? The spring has a stiffness of ftlbk /2= and is originally unstreched.



SOLUTIONS

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AAsAA NNF 15.0== μ

Solving Equations (1), (2), (3), (4) and (5) yields:

531.8=θ , lbFA 483.1= , lbN A 889.9= lbFB 8900.0= , lbN B 934.5=

Since BBsBB FlbNF >=== 483.1)934.5(25.0)( max μ , block B doesn’t slip. Therefore the above assumption is correct, thus:

531.8=θ , lbFA 483.1= , lbFB 8900.0=

PROBLEM 8-46

SOLUTION 8-46:

(a) (b)

Each of the cylinders has a mass of 50 kg. If the coefficient of static friction at the points of contact are

5.0=== CBA μμμ and 6.0=Dμ , determine the couple moment M needed to rotate cylinder E.

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SOLUTIONS

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FBD (a):

∑ = 0xF 0=− CD FN (1)

∑ = 0yF 05.490 =−+ DC FN (2)

∑ = 0OM 0)3.0()3.0( =−− DC FFM (3)

FBDD (b):

∑ = 0xF 0=−− DBA NFN (4)

∑ = 0yF 05.490 =−−+ DAB FFN (5)

∑ = 0PM 0)3.0()3.0()3.0( =−− DBA FFF (6)

Friction: Assuming cylinder E slips at points C and D and cylinder F does not move, then:

CCsCC NNF 5.0== μ and DDsDD NNF 6.0== μ . Substituting these values into Equations (1) and (2) and (3) and solving, we have:

NNC 31.377= NN D 65.188= mNM .6.90= Ans

If cylinder F is on the verge of slipping at point A, then AAsAA NNF 5.0== μ . Substitute this value into Equations (4), (5) and (6) and solving, we have:

NN A 92.150= NN B 15.679= NFB 73.37= Ans

Since BBsBB FNNF >=== 58.339)15.679(5.0)( max μ , cylinder F does not move. Therefore the above assumption is correct.

EXTRA PRACTICE PROBLEM 8- 63

Determine the largest weight of the wedge that can be placed between the 8-lb cylinder and the wall without upsetting equilibrium. The coefficient of static friction at A and C is

5.0=Sμ and at B, 6.0' =Sμ . Determine the couple moment M needed to rotate cylinder E.

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SOLUTIONS

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EXTRA PRACTICE SOLUTION 8-63:

(a) (b)

Equations of equilibrium:

FBD (a):

∑ = 0xF 060cos30cos =−− CBB NFN (1)

∑ = 0yF 060sin30sin =−++ WFFN CBB (2)

FBD (b):

∑ = 0xF 0860sin30sin =−−− BBA FNN (3)

∑ = 0yF 030cos60cos =−+ BBA NFF (4)

∑ = 0OM 0)5.0()5.0( =− BA FF (5)

Friction: Assume slipping occurs at points C and A, then CCsCC NNF 5.0== μ and AAsAA NNF 5.0== μ . Substituting these values into Equations (1),(2), (3),(4) and (5) and solving, we have:

lbW 64.66= Ans

lbN B 71.51= lbN A 71.59= lbNF CB 86.29==

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SOLUTIONS

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PROBLEM 8-70

SOLUTION 8-70:

(a)

If the beam AD is loaded as shown, determine the horizontal force P which must be applied to the wedge in order to remove it from under the beam. The coefficient of static friction at the wedge’s top and bottom surfaces are 25.0=CAμ and 35.0=CBμ respectively. If P=0, is the wedge self locking? Neglect the weight and size of the wedge and the thickness of the beam.

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SOLUTIONS

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(b)

Equations of equilibrium and friction:

If the wedge is on the verge of moving to the right, then slipping will have to occur at both contact surfaces. Thus, AAsAA NNF 25.0== μ and BBsBB NNF 35.0== μ .

FBD (a):

∑ = 0DM 0)5(16)2(6)7(10sin25.0)7(10cos =−−+ AA NN

kNN A 78.12=

FBD (b):

∑ = 0yF 010sin)78.12(25.080sin78.12 =−−BN

kNN B 14.13=

∑ = 0xF 0)14.13(35.010cos)78.12(25.080cos78.12 =−−+P

kNP 53.5= Ans

Since a force P(>0) is required to pull out the wedge, the wedge will be self-locking when

P = 0. Ans

EXTRA PRACTICE PROBLEM 9-13

The plate has a thickness of 0.25 ft and a specific weight of 3180 lb/ftγ = . Determine the location of the center of gravity. Also, find the tension in each of the cords used to support it.

614



SOLUTIONS

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Calculation of the area and the first moments:

Height of the differential element: 128 16y x x= − +

Area of the differential element: 128 16dA ydx x x dx

⎛ ⎞= = − +⎜ ⎟

⎝ ⎠

Centroid of differential element: 121and 8 16

2x x y x x

⎛ ⎞= = − +⎜ ⎟

⎝ ⎠

The total area is given by, 1616 1 32

22 2

0 0

168 16 16 42.67 ft2 3A

xA dA x x dx x x⎛ ⎞ ⎛ ⎞

= = − + = − + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

∫ ∫

Therefore: 1616 1 53

2 32 2

0 0

168 16 8 136.53 ft3 5A

xxdA x x x dx x x⎡ ⎤⎛ ⎞ ⎛ ⎞

= − + = − + =⎢ ⎥⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

∫ ∫

16 1 12 2

0

1 8 16 8 162A

ydA x x x x dx⎡ ⎤⎛ ⎞ ⎛ ⎞

= − + − +⎢ ⎥⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

∫ ∫

165 33 2 32 2

0

1 1 32 51248 256 136.53 ft2 3 5 3A

ydA x x x x x⎛ ⎞

= − + − + =⎜ ⎟⎝ ⎠

∫

The coordinates of the centroid are therefore given by:

(x,y)

( , )x y

dx 16

16

x

y

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SOLUTIONS

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136.53 3.2 ft ANS42.67

136.53 3.2 ft ANS42.67

A

A

A

A

xdxx

dA

ydxy

dA

= = =

= = =

∫

∫

∫

∫

Tension in the cables:

The weight of the plate is: W= 42.67(0.25)(180)=1920 lb

Equations of equilibrium:

( )0; 1920(3.20) (16) 0 384 lb ANS

0; 16 1920(3.2) 0 384 lb ANS

1152 lb ANS

x A A

y C C

B

M T T

M T T

T

= − = =

= − = =

=

∑∑

PROBLEM 9- 42

TA

TB TC

W

The hemisphere of radius r is made from a stack of very thin plates such that the density varies with height = kz, where k is a constant, Determine its mass and the

distance −

z to the center of mass G.

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SOLUTIONS

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SOLUTION 9-42:

Mass and Moment Arm: The density of the material is kz=ρ . The mass of the thin disk

differential element is ))(( 222 dzzrkzdzydVdm −=== πρπρ and its Centroid zz =−

. Evaluating the integrals, we have:

))(( 2

0

2 dzzrkzdmmr

m

−== ∫∫ π

4

|)42

(4

0

222 krzzrk r ππ =−= Ans

]})([({ 2

0

2 dzzrkzzdmzr

m

−= ∫∫−

π

15

2|))53

(5

0

222 krzzrk r ππ =−=

Centroid: Applying Equation 9-4, we have:

rkrkr

dm

dmz

m

m

158

4/15/2

4

5

==∫

∫−

ππ

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SOLUTIONS

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SOLUTION 9-51:

The three members of the frame each have a weight per unite length of 4 lb/ft. Locate the position

),(−−

yx of the center of gravity. Neglect the size of the pins at the joints and the thickness of the members. Also, calculate the reactions at the fixed support A.

1014



SOLUTIONS

PAGE

lbW

ftlbWx

774.88724454)7(4

.073.14272)4(345)4(5.1.

=++=

=+=

∑∑

−

ftWWx

x 06.1774.88073.142.

===∑∑

−−

Ans

ftlbWy .241.62572)4(1045)4(7)7)(4(5.3. =++=∑−

ftWWy

y 04.7774.88241.625.

===∑∑

−−

Ans

∑ = 0xF 0=xA

∑ = 0yF lbAy 14960774.88 =+=

∑ = 0AM 0)06.1(774.88)6(60 =+−− AM

ftlbM A .502= Ans

PROBLEM 9- 67

Locate the Centroid −

y of the beam’s cross-section built up from a channel and a wide-flange beam.

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SOLUTIONS

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SOLUTION 9-67:

Centroid: The area of each segment and its respective centroid are tabulated below:

Segment )( 2inA )(iny−

)( inAy−

1 14(0.4) 16.20 90.72

2 3.4(1.3) 14.7 64.97

3 10.3(0.76) 15.62 122.27

4 14.48(0.56) 8.00 64.87

5 10.3(0.76) 0.38 2.97

∑ 33.78 345.81

Thus:

ininAAy

y 2.1024.1078.3381.345.

====∑∑

−−

Ans

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SOLUTIONS

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∑ = 0xF lbFRx 16128107525376 =+=

∑ = 0yF lbFRy 604840322016 =+=

kiplbFR 2.1717225)6048()16128( 22 ==+= Ans

56.20)161286048(tan 1 == −θ

The storage tank contains oil having a specific weight of 3

0 /56 ftlb=γ . If the tank is 6 ft wide, calculate the resultant force acting on the inclined side BC of the tank, caused by the oil, and specify its location along BC, measured from B. Also compute the total resultant force acting on the bottom of the tank.

ftlbhbWB /672)2)(56(60 === γ

ftlbhbWC /3360)10)(56(60 === γ

lbFh 5376)672(81 ==

lbFh 10752)8)(6723360(2/12 =−=

lbFV 2016)56)(6)(2)(3(1 ==

lbFV 4032)56)(6)(8)(3(2/12 ==

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SOLUTIONS

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∑ = 0BRM )2(4032)5.1(2016)4(5376)8)(3/2(1075217225 +++=d

ftd 22.5= Ans

At bottom:

kiplbFR 8.1818816)56)(6)(14(4 ===

Problem 9-119

SOLUTION 9-119:

]8)1(

6[10 ++

=x

P

dxx

PdAFR 3].8)1(

6[102

0∫ ∫ +

+==

lblbxxFR 67875.677|8)1ln(6(30 20 ==++= Ans

dxx

xPdAx 3].8)1(

6)[10(2

0∫ ∫ +

+=

−

lblbxxx 67825.642|4)1ln((6(30 20

2 ==++−=

ftdAP

dAPxx 948.0

75.67725.642

.

..=== ∫

−

−

, fty 50.1=−

(by symmetry) Ans

The pressure loading on the plate is described by the function:

2/]8)1/(6[10 ftlbxp ++=

Determine the magnitude of the resultant force and the

coordinates −−

),( yx of the point where the line of action of the force intersects the plate.

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Documents

N ID DATE MTE 119 S P HOMEWORK 9 1 S 14mte119/s/MTE119 - Solutions Hw9.pdfstiffness of k =2lb/ ft and is originallyunstreched. Mechatronics Engineering NAME & ID DATE MTE 119 – STATICS