NAME ID DATE MTE 119 STATICS PAGE H 4 1 S 14 ...mte119/s/MTE119 - Solutions...Mechatronics...
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Mechatronics Engineering NAME & ID DATE MTE 119 – STATICS HOMEWORK 4 SOLUTIONS PAGE PROBLEM 1 Systems 1 and 2 each consist of a couple. If they are equivalent, what is F ? SOLUTION 1: For couples, the sum of the forces vanish for both systems. For system 1, the two forces are located at: i r ˆ 4 11 = , j r ˆ 5 12 + = . The forces are: j i j i F ˆ 100 ˆ 21 . 173 ) 30 sin ˆ 30 cos ˆ ( 200 1 + = + = k k j i F r r M ˆ 05 . 1266 0 100 21 . 173 0 5 4 ˆ ˆ ˆ ) ( 1 12 11 1 = − = × − = ) . ( m N For system 2, the positions of the forces are i r ˆ 2 21 = and j i r ˆ 4 ˆ 5 22 + = . The forces are: ) ˆ 3420 . 0 ˆ 9397 . 0 ( )) 20 sin( ˆ ) 20 cos( ˆ ( 2 j i F j i F F − = − + − = The moment of the couple in System 2 is: k F k j i F F r r M ˆ 7848 . 4 0 3420 . 0 9397 . 0 0 4 3 ˆ ˆ ˆ ) ( 2 22 21 2 = − − − = × − = from which, if the systems are to be equivalent, 6 . 264 7848 . 4 1266 = = F ) ( N Ans 1 14
Transcript of NAME ID DATE MTE 119 STATICS PAGE H 4 1 S 14 ...mte119/s/MTE119 - Solutions...Mechatronics...
Mechatronics Engineering
NAME & ID DATE MTE 119 – STATICS HOMEWORK 4
SOLUTIONS
PAGE
PROBLEM 1
Systems 1 and 2 each consist of a couple. If they are equivalent, what is F ?
SOLUTION 1:
For couples, the sum of the forces vanish for both systems. For system 1, the two forces are located at:
ir ˆ411 = , jr ˆ512 += .
The forces are:
jijiF ˆ100ˆ21.173)30sinˆ30cosˆ(2001 +=+=
kkji
FrrM ˆ05.1266010021.173054
ˆˆˆ
)( 112111 =−=×−= ).( mN
For system 2, the positions of the forces are ir ˆ221 = and jir ˆ4ˆ522 += . The forces are:
)ˆ3420.0ˆ9397.0())20sin(ˆ)20cos(ˆ(2 jiFjiFF −=−+−=
The moment of the couple in System 2 is:
kFkji
FFrrM ˆ7848.403420.09397.0043
ˆˆˆ
)( 222212 =−
−−=×−=
from which, if the systems are to be equivalent,
6.2647848.4
1266==F )(N Ans
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Mechatronics Engineering
NAME & ID DATE MTE 119 – STATICS HOMEWORK 4
SOLUTIONS
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PROBLEM 2 Exercise Problem 4-106/107 (Textbook - page 175) 4-106: Goal: Replace the force and couple system by an equivalent force and couple moment at point O. 4-107: Goal: Replace the force and couple system by an equivalent force and couple moment at point P.
SOLUTION 2
Solution 4-106
The total forces acting on the x- axis are:
∑∑ =→+
xRx FF 30769.060cos4)135(6 =−=RxF )(kN
The total forces acting on the y-axis are:
∑∑ =↑+
yRy FF 0744.260sin4)1312(6 =−=RyF )(kN
10.2)0744.2()30769.0( 22 =+=RF )(kN -Ans
The angle of direction of total force (with respect to x-axis) is:
6.8130769.00744.2tan 1 =⎟
⎠⎞
⎜⎝⎛−= −θ Ans
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Mechatronics Engineering
NAME & ID DATE MTE 119 – STATICS HOMEWORK 4
SOLUTIONS
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The total moment acting at point O is:
∑=↓+
OO MM 62.10)4(60cos4)5)(135(6)4)(
1312(68 −=−+−=OM
62.10−=OM ).( mkN Ans
Solution 4-107
Comparing to the previous problem, there would be no change in total force, but the Changes will present in the moment which act at point P. The total forces acting on the x- axis are:
∑∑ =→+
xRx FF 30769.060cos4)135(6 =−=RxF )(kN
The total forces acting on the y-axis are:
∑∑ =↑+
yRy FF 0744.260sin4)1312(6 =−=RyF )(kN
10.2)0744.2()30769.0( 22 =+=RF )(kN -Ans
The angle of direction of total force (with respect to x-axis) is:
6.8130769.00744.2tan 1 =⎟
⎠⎞
⎜⎝⎛−= −θ Ans
The total moment acting at point P is:
∑=↓+
PP MM )3(60sin4)4(60cos4)5)(135(6)7)(
1312(68 +−+−=PM
8.16−=PM ).( mkN Ans
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Mechatronics Engineering
NAME & ID DATE MTE 119 – STATICS HOMEWORK 4
SOLUTIONS
PAGE
PROBLEM 3 Exercise Problem 4-113 (Textbook – page 175) Goal: Replace the three forces acting on the shaft by a single resultant force. Specify where the force acts, measured from end B.
SOLUTION 3
The total forces acting on the x- axis are:
∑∑ =→+
xRx FF )(300)135(260)
54(500 lbFRx −=+−=
The total forces acting on the y-axis are:
∑∑ =↑+
yRy FF )(740)1312(260200)
53(500 lbFRy −=−−−=
)(798)740()300( 22 lbFR =−+−= -Ans
The angle of direction of total force (with respect to x-axis) is:
9.67300740tan 1 =⎟
⎠⎞
⎜⎝⎛−= −θ Ans
The total moment acting at point B is:
∑=↓+
BRB MM )4)(1312(260)6(200)9)(
53(500)(740 ++=x
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Mechatronics Engineering
NAME & ID DATE MTE 119 – STATICS HOMEWORK 4
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)(8.16 ftx −= Ans
PROBLEM 4 The tension in cable AB is 400N, and the tension in cable CD is 600N
a) If you represent the forces exerted on the left post by the cables by a force F
acting at the origin O and a coupleM , what are F andM ?
b) If you represent the forces exerted on the post by the cables by the force F alone, where does its line of action intersect the y-axis?
SOLUTION 4a
From the right triangle, the angle between the positive x axis and the cable AB is:
6.26800400tan 1 −=⎟
⎠⎞
⎜⎝⎛−= −θ
The tension in AB is:
jijiTABˆ89.178ˆ77.357))6.26sin(ˆ)6.26cos(ˆ(400 −=−+−= )(N
The angle between the positive x axis and the cable CD is:
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NAME & ID DATE MTE 119 – STATICS HOMEWORK 4
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6.20800300tan 1 −=⎟
⎠⎞
⎜⎝⎛−= −α
The tension in CD is:
jijiTCDˆ67.210ˆ8.561))6.20sin(ˆ)6.20cos(ˆ(600 −=−+−= )(N
The equivalent force acting at the origin O is the sum of the forces acting on the left post:
jiF ˆ)67.21089.178(ˆ)8.56177.357( −−++=
jiF ˆ6.389ˆ61.919 −= Ans
The sum of moments acting on the left post is the product of the moment arm and the x-component of the tensions:
∑ −=−−= kkkM ˆ419ˆ)8.561(3.0ˆ)77.357(7.0 ).( mN
Check: The position vectors at the point of application are jrABˆ7.0= , and jrCD
ˆ3.0= .
The sum of the moments is:
067.2108.56103.00
ˆˆˆ
089.17877.35707.00
ˆˆˆ
)()(−
+−
=×+×=∑kjikji
TrTrM CDCDABAB
kkkM ˆ419ˆ)8.561(3.0ˆ)77.357(7.0 −=−−=∑ Ans
SOLUTION 4b
Check: The equivalent single force retains the same scalar components, but must act at appoint that duplicates the sum of the moments. The distance on the y axis is the ratio of the sum of the moments to the x-component of the equivalent force. Thus:
456.06.919
419==D )(m
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Mechatronics Engineering
NAME & ID DATE MTE 119 – STATICS HOMEWORK 4
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Check: The moment is:
kkDDkji
FrM Fˆ419ˆ6.919
06.3896.91900
ˆˆˆ
2 −=−=−
=×=
from which
456.06.919
419==D )(m , check.
PROBLEM 5 Exercise Problem 4-131 (Textbook – page 178) Goal: Replace the system of forces acting on slings by an equivalent force and couple moment at point O.
SOLUTION 5
Force vectors:
)}(ˆ00.6{1 kNkF =
)ˆ45sinˆ30cos45cosˆ30sin45cos(52 kjiF ++−=
}ˆ536.3ˆ062.3ˆ768.1{2 kjiF ++−= )(kN
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)ˆ45cosˆ60cosˆ60(cos42 kjiF ++=
}ˆ828.2ˆ00.2ˆ00.2{2 kjiF ++= )(kN
Equivalent force and couple moment at point O:
kjiFFFFRˆ)828.2536.36(ˆ)00.206.3(ˆ)00.2768.1(321 ++++++−=++=
}ˆ4.12ˆ06.5ˆ232.0{ kjiFR ++= )(kN Ans
The position vectors are )}(62{1 mjir += and )}(4{1 mir =
)()( 22210 FrFrMM Ro ×+×== ∑
536.3062.3768.1004
ˆˆˆ
00.600062
ˆˆˆ
−+=
kjikji
}ˆ2.12ˆ1.26ˆ36{ kjiM Ro +−= ).( mkN Ans
PROBLEM 6 Exercise Problem 4-134 (Textbook – page 178)
Goal: Determine the equivalent resultant force and specify its location (x, y) on the slab.
kNFkNF 50,20 21 ==
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Mechatronics Engineering
NAME & ID DATE MTE 119 – STATICS HOMEWORK 4
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SOLUTION 6
∑∑ =↓+
zR FF 50205020 +++=RF )(140 kNFR = Ans
∑= yRoy MM )10(50)10(20)4)(50()(140 ++=x )(43.6 mx = Ans
∑= xRox MM )13(50)11(20)3)(50()(140 −−−=− y )(29.7 my = Ans
PROBLEM 7 Exercise Problem 4-135 (Textbook – page 179)
SOLUTION 7
Force and moment vectors:
)}(ˆ300{1 NkF = , )}(ˆ100{3 NjF =
)ˆ45sinˆ45(cos52 kiF −=
}ˆ42.141ˆ42.41.1{2 kiF −−= )(N
).}(ˆ100{1 mNkM =
)ˆ45sinˆ45(cos1802 kiM −=
Goal: Replace the two wrenches and forces, acting on the pipe assembly, by an equivalent resultant force and couple moment at point O.
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}ˆ28.127ˆ28.127{2 kiM −−= ).( mN
Equivalent force and couple moment at point O:
kjiFFFFRˆ)42.141300(ˆ100ˆ42.141321 −++=++=
kjiFRˆ159ˆ100ˆ141 ++= )(N Ans
The position vectors are: )}(5.0{1 mjr = and )}(1.1{2 mjr =
2122210 )()( MMFrFrMM Ro ++×+×== ∑
kikkjikji
ˆ28.127ˆ28.127ˆ10042.141042.141
01.10
ˆˆˆ
3000005.00
ˆˆˆ
−++−
+=
}ˆ183ˆ122{ kiM Ro −= ).( mN Ans
PROBLEM 8 Exercise Problem 4-137 (Textbook – page 179)
Goal: Replace the three forces, acting on the plate by a wrench. Specify the magnitude of the force and couple moment for the wrench and the point P(x,y) where its line of action intersect the plate.
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SOLUTION 8
kjiFRˆ800ˆ300ˆ500 ++= )(N
)(990)800()300()500( 222 NFR =++−=
kjiuFRˆ808.0ˆ3030.0ˆ5051.0 ++=
∑= '' xRxMM ; )4(800' yM
Rx−=
∑= '' yRyMM ; xM
Ry800' =
∑= '' zRzMM ; )6(300500' xyM
Rz−+=
Since RM also acts in the direction of FRu ,
)4(800)5051.0( yM R −=
xM R 800)3030.0( =
)6(300500)8081.0( xyM R −+=
).(07.3 mkNM R = Ans
)(16.1 mx = Ans
)(06.2 my = Ans
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NAME & ID DATE MTE 119 – STATICS HOMEWORK 4
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Extra Practice Problems: For those of you interested in a little bit more of practice.
PROBLEM 9 Exercise Problem 4-121 (Textbook - page 177)
SOLUTION 9
The total forces acting on the x- axis are:
∑∑ =→+
xRx FF )(450)60(cos500)54(250 NFRx −=−−=
The total forces acting on the y-axis are:
∑∑ =↑+
yRy FF )(0127.883)60(sin500)53(250300 NFRy −=−−−=
)(991)0127.883()450( 22 NFR =−+−= -Ans
The angle of direction of total force (with respect to x-axis) is:
0.63450
0127.883tan 1 =⎟⎠⎞
⎜⎝⎛−= −θ Ans
The total moment acting at point B is:
Goal: Replace the loading on the frame by a single resultant force. Specify where its line of action intersects member CD, measured from end C.
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∑= CRA MM
)1(60sin500)2(60cos500)6)(53(250)3(300400)(0127.883 ++++−=x
)(64.2 mx = Ans
PROBLEM 10
The handpiece of a miniature industrial grinder weighs 2.4N, and its center of gravity
is located on the y-axis. The head of the handpiece is offset in the x-z plane in such
a way that line BC forms an angle of 25 with the x-direction. Show that the weight
of the handpiece and the two couples 1M and 2M can be replaced with a single
equivalent force. Further assuming that 1 0.068 N mM = ⋅ and 2 0.065 N mM = ⋅ ,
determine:
(a) the magnitude and the direction of the equivalent force
(b) the point where its line of action intersects the x-z plane
SOLUTION 10a
First we find the total forces and moments acting on the present system. We know that the only force acting on the system is the weight of the handpiece of grinder, so the total equivalent force should have exactly the same amount of force, with the same direction, but acting on a point in the space. We represent both the moments in vector form:
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kijiM
kMˆ0275.0ˆ0589.0)ˆ25sinˆ25cos(065.0
ˆ068.0
2
1
−−=−−=
=
kikkiMM ˆ0405.0ˆ0589.0ˆ)0275.0ˆ068.0(ˆ0589.021 +−=−+−=+ (1)
We can see that the moment does not have any component in “y” direction. We should expect that the line of action of the resultant force intersect the x-z plane at point D with the following position vector:
jrirr yxˆˆ +=
and since the only force in the system is W, the resultant force should have the following magnitude and direction:
)ˆ(ˆˆˆ jWkFjFiFF RzRyRxR −=++=
)(ˆ4.2 NjFR −= Ans
SOLUTION 10b
We find the total moment with respect to point O, using )ˆ( jWFR −= and jrirr yxˆˆ += :
krirkWriWrrr
Wkji
FrM xzxz
zx
RRˆ)4.2(ˆ)4.2(ˆ)(ˆ)(
000
ˆˆˆ
−=−==×= (2)
By equating the above resultant moment (Equation (2)) with the present moment in the system (Equation (1)), the components of the position vector can be found:
kikrir xzˆ0405.0ˆ0589.0ˆ)4.2(ˆ)4.2( +−=−
)(54.24)(02454.04.2
0589.0 mmmrz −=−=−
=
)(88.16)(01688.04.2
0405.0 mmmrx −=−=−
=
)(ˆ9.16ˆ5.24 mmjir −−= Ans
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