FACULTY OF ENGINEERING
CHULALONGKORN UNIVERSITY
2103213 ENG MECHANICS I
Year 2nd, First Semester, Mid Term Examination. July 21, 2008. Time 13.00-15.00
---------------------------------------------------------------------------------------------------------------------------------------ชื่อ-นามสกุล.………………………….. เลขประจําตัว……………………. เลขที่ใน CR58.……
หมายเหตุ
1. ขอสอบมีทั้งหมด …4…….. ขอ ในกระดาษคําถาม ……4….. หนา แตละขอมีคะแนน 10 คะแนน
2. ไมอนุญาตใหนําตําราและเอกสารใดๆ เขาในหองสอบ
3. อนุญาตใหใชเครื่องคํานวณธรรมดาได
4. ใหเขียนชื่อ-เลขประจําตัวทุกแผน
5. ใหเขียนตอบลงในกระดาษคําตอบของขอเทานั้น
6. หามการหยิบยืมสิ่งใดๆ ทั้งสิ้น จากผูสอบอืน่ ๆ เวนแตผูคุมสอบจะหยิบยืมให
7. หามนําสวนใดสวนหนึ่งของขอสอบออกจากหองสอบ
8. ผูที่ประสงคจะออกจากหองสอบกอนหมดเวลาสอบ แตตองไมนอยกวา 45 นาที
9. เมื่อหมดเวลาสอบ ผูเขาสอบตองหยุดการเขียนใดๆ ทั้งสิ้น
10. ผูที่ปฏิบัติเขาขายทุจริตในการสอบ ตามประกาศคณะวิศวกรรมศาสตร
มีโทษ คือ ไดรับสัญลักษณ F ในรายวิชาที่ทุจริต และพักการศึกษาอยางนอย 1 ภาคการศึกษา
รับทราบ
ลงชื่อนิสิต (…………………..…………….)
2103213 Engineering Mechanics I
ID…………………..………….…Name……………………………………………………………CR58…………
1. The uniform I-beam of mass m is supported at its ends on two fixed horizontal rails as shown. Determine the
maximum horizontal force P which can be applied without causing the beam to slip, and find the corresponding
value of the friction force at A. The coefficient of static friction between the beam and the rails is μ s . Also, take
b <l /2.
Figure 1: Free body diagram of problem 1
Solution: The approach to this problem is to determine the force P that would
make the ends A and B start to slip. The one which requires less magnitude
would be the answer.
Free body diagram of the beam is drawn in fig. 1. By the symmetrical support
of the I-beam, the normal force developed at each end would be the same and
equal to half of the beam’s weight. FA and FB denote the corresponding friction
forces. If the beam is about to slip at end A,
FA = µs
mg
,
2
assuming support B has not yet reached the impending motion status. Take the
moment about B along the z-axis,
[ΣMBz = 0]
l = 0,
−P b + µs mg
2
.
P = µs mgl
2b
Instead, had the beam is to slip at end B first, the friction at B would be the
static friction;
mg
FB = µs
.
2
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Phongsaen PITAKWATCHARA
Similarly, to determine the corresponding applied force P , we take the moment
about A along the z-axis.
P (l − b) − µs mg
l = 0,
2
[ΣMAz = 0]
mgl
P = µs 2(l−b)
.
Since the problem states b < l/2, we may conclude
b<l−b
µs
→
1
1
>
.
b
l−b
mgl
mgl
> µs
.
2b
2(l − b)
Since the applied force P for B to slip is less than that for A, B would slip first.
Therefore the maximum force P which can be applied without causing the beam
to slip is
mgl
.
Pmax = µs
2(l − b)
. The friction at
The corresponding friction at B is the static friction FB = µs mg
2
A may be determined from the equilibrium condition along the x-axis as
[ΣFx = 0]
FA + FB − P = 0
mgl
FA + µs mg
− µs 2(l−b)
=0
2
FA = µs mg
2
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b
l−b
Phongsaen PITAKWATCHARA
2103213 Engineering Mechanics I
ID…………………..………….…Name……………………………………………………………CR58…………
2. It is desired that a person be able to begin closing the van hatch from the open position shown with a 40 N
vertical force P. As a design exercise, determine the necessary force in each of the two hydraulic struts AB. The
mass center of the 40 kg door is 37.5 mm directly below point A. Treat the problem as two dimensional.
Figure 2: Free body diagram of problem 2
Solution: Free body diagram of the van hatch is shown in fig. 2. There are
four forces, namely the closing force P , the weight, the pin force at O, and two
hydraulic strut forces, each of it has the magnitude C. Because the pin force is
not of interest, we take the equilibrium moment condition about O.
[ΣMO = 0]
40 × 1.125 + 40g × 0.55 × cos(30 − θ) − 2C × 0.55 sin θ = 0,
where θ is the angle OAB that can be calculated by the cosine law.
0.1752 = 0.552 + 0.62 − 2 × 0.55 × 0.6 cos θ,
θ = 16.787◦
Substitute the value into the above equation. The compressive hydraulic force
may be determined.
C = 803 N.
Chulalongkorn University
Phongsaen PITAKWATCHARA
2103213 Engineering Mechanics I
ID…………………..………….…Name……………………………………………………………CR58…………
3. One of the vertical walls supporting end B of the 200 kg uniform shaft is turned through a 30 o angle as shown
here. End A is still supported by the ball and socket connection in the horizontal x-y plane. Calculate the
magnitudes of the forces P and R exerted on the ball end B of the shaft by the vertical walls C and D,
respectively.
Figure 3: Free body diagram of problem 3
Solution: From the free body diagram of the shaft in fig. 3, we see there are
5 unknowns due to the constraint types of both ends which cannot support the
moment. Number of the unknowns corresponds to the number of independent
equilibrium conditions that may be set up.
Since the reaction at A is not asked for, we may set up the equilibrium
moment condition around point A. Based upon the specified coordinate frame,
ΣM A = 0

 
 
 
 
 

−2
0
−2
R cos 30
−1
0
 −6  ×  P  +  −6  ×  R sin 30  +  −3  × 
 = 0,
0
3
0
3
0
1.5
−200g
where P and R are the magnitude of reaction forces that the walls C and D exert
on the ball end B, respectively.
It is trivial to verify that this three dimensional vector equation has only two
independent equations (consistent with the number of the unknowns). Solving
them to obtain the wall reactions, we have
R = 755.2 N, P = 1584.4 N
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Phongsaen PITAKWATCHARA
2103213 Engineering Mechanics I
ID…………………..………….…Name……………………………………………………………CR58…………
4. The resultant of the two forces and couple may be represented by a wrench. Determine the vector expression
for the moment M of the wrench and find the coordinates of the point P in the x-z plane through which the
resultant force of the wrench passes.
Solution 1: First, choose the origin O and determine the equivalent force-couple
resultant. Hence,
R = 100i + 100j N
MO = 0.4 × 100k − 0.3 × 100k + 0.4 × 100j − 20j = 20j + 10k Nm
Next, project the couple MO onto the direction parallel and perpendicular to
nR , for which its value is
nR =
1
R
1
= √ i+ √ j
|R|
2
2
Consequently, the components of MO are
Mk = (MO · nR ) nR = 10i + 10j Nm
M⊥ = MO − Mk = −10i + 10j + 10k Nm
Thirdly, transform the couple M⊥ into pair of forces R and −R. −R is at
the origin O. R is along the line through which it passes the point P in the x-z
plane. Let r be the position vector of that point P .
r = xi + zk m
Consistency in the transformation requires the induced moment of R about
O be equal to the couple M⊥ :
M⊥ = r × R
−10i + 10j + 10k = (xi + zk) × (100i + 100j)
Solving the above vector equation yields the coordinate values:
x = 0.1 m
z = 0.1 m
To conclude, the equivalent wrench system consists of the couple Mk ,
Mk = 10i + 10j Nm
and the force R,
R = 100i + 100j N
of which its line of action passes through the point P ,
x = 0.1 m
z = 0.1 m
in the x-z plane.
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Solution 2: The resultant force is just simply
R = 100i + 100j N
Another way to determine the wrench is to assume the point where the wrench
passes. Let point P in the x-z plane, where the wrench passes, has the coordinate
(x, 0, z). Consequently, the moment of the force system about P is
MP = 100 × zi + 100 × (0.4 − x) k + 100 × (0.4 − z) j − 100 × 0.3k − 20j
= 100zi + (20 − 100z) j + (10 − 100x) k Nm
Note that this moment at P must be equal to the couple of the wrench passing
through P , which is parallel to the resultant force. That is, MP k R. The
constraint eqautions are obtained by comparing the ratio of their components:
100
100
=
100z
20 − 100z
and
10 − 100x = 0
As the result,
x = 0.1 m, z = 0.1 m
and substituting back into the moment equation, we have
MP = 10i + 10j Nm
Chulalongkorn University
Phongsaen PITAKWATCHARA
FACULTY OF ENGINEERING
CHULALONGKORN UNIVERSITY
2103213 ENG MECHANICS I
Year 2nd, First Semester, Final Examination. September 22, 2008. Time 13.00-15.00
---------------------------------------------------------------------------------------------------------------------------------------ชื่อ-นามสกุล.………………………….. เลขประจําตัว……………………. เลขที่ใน CR58.……
หมายเหตุ
1. ขอสอบมีทั้งหมด …5….. ขอ ในกระดาษคําถาม ……5….. หนา แตละขอมีคะแนน 10 คะแนน
2. ไมอนุญาตใหนําตําราและเอกสารใดๆ เขาในหองสอบ
3. อนุญาตใหใชเครื่องคํานวณธรรมดาได
4. ใหเขียนชื่อ-เลขประจําตัวทุกแผน
5. ใหเขียนตอบลงในกระดาษคําตอบของขอเทานั้น
6. หามการหยิบยืมสิ่งใดๆ ทั้งสิ้น จากผูสอบอืน่ ๆ เวนแตผูคุมสอบจะหยิบยืมให
7. หามนําสวนใดสวนหนึ่งของขอสอบออกจากหองสอบ
8. ผูที่ประสงคจะออกจากหองสอบกอนหมดเวลาสอบ แตตองไมนอยกวา 45 นาที
9. เมื่อหมดเวลาสอบ ผูเขาสอบตองหยุดการเขียนใดๆ ทั้งสิ้น
10. ผูที่ปฏิบัติเขาขายทุจริตในการสอบ ตามประกาศคณะวิศวกรรมศาสตร
มีโทษ คือ ไดรับสัญลักษณ F ในรายวิชาที่ทุจริต และพักการศึกษาอยางนอย 1 ภาคการศึกษา
รับทราบ
ลงชื่อนิสิต (…………………..…………….)
2103213 Engineering Mechanics I
ID…………………..………….…Name……………………………………………………………CR58…………
1. The robot arm is elevating and extending simultaneously. At a given instant, θ = 30 o ,
θ& = 10 deg/ s = constant, l = 0.5m , l& = 0.2m / s , and &l& = −0.3m / s 2 . Compute the magnitudes of the
velocity v and acceleration a of the gripped part P. In addition, express v and a in terms of the unit vectors i and
j.
Solution: The most appropriate coordinate system for this problem is obviously
the r-θ coodinates. From the given information, we can straightforwardly express
the relevant parameters:
r = 1.25 m, ṙ = 0.2 m/s, r̈ = −0.3 m/s2 ,
θ = 30◦ , θ̇ = 10 ×
π
rad/s, θ̈ = 0 rad/s2 ,
180
using the simple relation
r = 0.75 + l
Therefore the velocity and the acceleration of the gripped part P may be
readily determined by direct substitution of the above parameters as
h
i
π
eθ = 0.2er + 0.218eθ .
v = 0.2er + 1.25 × 18
v = ṙer + r θ̇eθ
i
h
2
a = (r̈ − r θ̇ )er + (r θ̈ + 2ṙ θ̇)eθ
π2
π
a = −0.3 − 1.25 × 18
eθ = −0.338er + 0.070eθ .
er + 2 × 0.2 × 18
2
Their magnitudes are then
√
v = 0.22 + 0.2182 = 0.296 m/s.
√
a = 0.3382 + 0.0702 = 0.345 m/s2 .
To express the velocity and the acceleration in x-y coordinate frame, we acknowledge the following transformation.
er = cos 30i + sin 30j
eθ = − sin 30i + cos 30j
Substitute er and eθ into the above expressions, the same vectors expressed in
x-y coordinate frame may be obtained.
v = 0.2 (cos 30i + sin 30j) + 0.218 (− sin 30i + cos 30j)
= 0.064i + 0.289j m/s.
a = −0.338 (cos 30i + sin 30j) + 0.070 (− sin 30i + cos 30j)
= −0.328i − 0.109j m/s2 .
Chulalongkorn University
Phongsaen PITAKWATCHARA
2103213 Engineering Mechanics I
ID…………………..………….…Name……………………………………………………………CR58…………
2. The sliders A and B are connected by a light rigid bar of length l = 0.5m and move with negligible
friction in the horizontal slots shown. For the position where x A = 0.4m , the velocity of A is v A = 0.9m / s to
the right. Determine the acceleration of each slider and the force in the bar at this instant.
Figure 4: Free body diagram of problem 6
Solution: The underlying kinematic constraint for the motion of two sliders is
the fact that OAB forms the triangle. Let xA and xB are the displacement of
slider A and B measured from O positively outward. The constraint may then
be expressed as
x2A + x2B = 0.52 .
With the current value of xA be 0.4 m, the corresponding xB will be 0.3 m.
Differentiating the equation to obtain the velocity constraint;
xA ẋA + xB + ẋB = 0.
Since vA is given, we may solve for vB :
0.4 × 0.9 + 0.3ẋB = 0, ẋB = −1.2.
That is, the slider B is traveling downward with the velocity of 1.2 m/s.
Differentiating the equation again, we obtain the acceleration constraint which
contains the acceleration terms of both sliders:
xA ẍA + xB ẍB + ẋ2A + ẋ2B = 0.
For this particular instant, the following relation may be written
0.4ẍA + 0.3ẍB + 2.25 = 0.
(1)
It is seen that we cannot yet solve for the individual acceleration. We must
search for additional equation(s). This is achieved by considering the kinetics of
the problem. Free body diagrams of both sliders are depicted in fig. 4. T is the
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developing tension force in the massless connecting bar. Weighting forces do not
show up because the apparatus is oriented in the horizontal plane.
From this, we may apply Newton’s second law of motion to each slider, A and
B in turn, as follow;
[ΣFxA = mA ẍA ]
P −T ×
[ΣFxB = mB ẍB ]
−T ×
0.4
0.5
0.3
0.5
= 2ẍA .
= 3ẍB .
(2)
(3)
Solving (1), (2), and (3) simultaneously, the acceleration and the force in the bar
are
ẍA = 1.364 m/s2
ẍB = −9.318 m/s2
T = 46.6 N
Chulalongkorn University
Phongsaen PITAKWATCHARA
2103213 Engineering Mechanics I
ID…………………..………….…Name……………………………………………………………CR58…………
3. The hydraulic cylinder produces a limited horizontal motion of point A. If v A = 4m / s when
θ = 45 o , determine the magnitude of the velocity of D and the angular velocity ω of ABD for this position.
Solution: If one rush to determine the velocity of D at the far end, he will
encounter the problem of having more unknowns than the available equations.
Motion of D is arbitrary. Instead, it is suggested that we should try to determine
the velocity of B first. B is constrained to move along the circular path and
hence v B is directed perpendicular to OB.
Using the typical coordinate frame {xyz}, for this instant,
0.4 cos α
4
cos 45
,
+ ωk ×
vB = v A + ω ABD × r B/A
=
vB
0.4 sin α
0
sin 45
where α = 6 BAO that may be evaluated from the law of sine:
0.4
0.25
=
, α = 26.23◦ .
sin α
sin 45
Therefore, the magnitude of the velocity of point B and the angular velocity of
the member ABD may be determined.
vB = 3.79 m/s
ωABD = 7.47 rad/s CCW
The magnitude of the velocity of point D may now be calculated directly
from the relative velocity equation as
vD = v A + ω ABD × rD/A
2.02
0.6 cos α
4
m/s.
=
+ 7.47k ×
vD =
4.02
0.6 sin α
0
vD = 4.5 m/s
Chulalongkorn University
Phongsaen PITAKWATCHARA
2103213 Engineering Mechanics I
ID…………………..………….…Name……………………………………………………………CR58…………
4. In the design of this linkage, motion of the square plate is controlled by the two pivoted links. Link
OA has a constant angular velocity ω = 4rad / s during a short interval of motion. For the instant represented,
θ = tan −1 4 / 3 and AB is parallel to the x-axis. For this instant, determine the angular acceleration of both the
plate and link CB.
Solution: Even the parameters of interest are the acceleration, it is a common
paradigm that the kinematics analysis cannot be jumped directly to the acceleration level. The velocity analysis must be completed before the acceleration one
can be carried out.
Motion of the link OA is given. From this, with the help of the relative
motion equations, the analysis may be propagated to the required members.
According to the given x-y-z coordinate system, we first perform the velocity
analysis starting from link OA, to the plate, and finally to link BC.
−0.24
0.1 sin α
m/s.
=
vA = ω OA × r A/O
vA = −4k ×
−0.32
−0.1 cos α where from the current mechanism posture, α = tan−1 34 .
On the rigid plate, velocity of A and B are related by
−4/5
−0.24
vB = v A + ω AB × rB/A
vB
=
+ ωAB k × 0.16i
3/5
−0.32
recognizing that point B is constrained to move along the circular path
and assuming its velocity is pointing northwest perpendicular to link BC. Also
we assume the plate is rotating in the counter-clockwise direction. Solving two
scalar equations, we have
vB = 0.3 m/s
ωAB = 3.125 rad/s CCW.
Finally, velocity of point B may be used to solve for the angular velocity of
link BC.
[vB = ωBC · rBC ]
0.3 = ωBC × 0.2, ωBC = 1.5 rad/s CCW.
Now the acceleration analysis could be embarked. Starting at link OA, with
the zero angular acceleration at the moment, we may determine the acceleration
of A simply by
2
[aA = rOA · ωOA
]
aA = 0.1 × 42 (− 54 i + 35 j) = −1.28i + 0.96j m/s2 .
Propagating the acceleration to point B requires the angular acceleration of
the plate, which is not known yet. Alternatively, the acceleration of point B may
be determined from the motion of link BC that constrain it to move along the
circular path. These two relations may then be equated and use to determine
the angular acceleration of the members without solving the acceleration of B.
Mathematically,
aA + ω AB × ω AB × r B/A + αAB × r B/A = aB = ω BC × ω BC × r B/C + αBC × r B/C .
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Substitute the position and velocity parameters into the vectorial equation,
−1.28
+ 3.125k × 3.125k × 0.16i + αAB k × 0.16i
0.96
0.2 × 3/5
0.2 × 3/5
.
+ αBC k ×
= 1.5k × 1.5k ×
0.2 × 4/5
0.2 × 4/5
Angular acceleration of the plate and the link BC may then be solved:
αAB = 3.81 rad/s2 CCW
αBC = 16.08 rad/s2 CCW
Chulalongkorn University
Phongsaen PITAKWATCHARA
2103213 Engineering Mechanics I
ID…………………..………….…Name……………………………………………………………CR58…………
5. The crank OA revolves clockwise with a constant angular velocity of 10rad / s within a limited arc
of its motion. For the position θ = 30 o determine the angular velocity of the slotted link CB and the
acceleration of A as measured relative to the slot in CB.
Solution:
Typically, the velocity information must be evaluated before calculating the
acceleration because of the appearance of the velocity terms in the acceleration
equation. Motion of the crank OA is transmitted to the slot CB through the pin
A. Therefore, the following relative velocity equation is set up with the helpful
velocity diagram shown in fig. 5:
vA = vP + vA/P
From the given data, vA = 0.2 × 10 = 2 m/s. Completing the velocity
diagram in fig. 5, the pertinent velocities can be determined as
vP = 2 cos 30 = 2 × 0.2 cos 30 × ωCB , ωCB = 5 rad/s CW
vA/P = vrel = 2 sin 30 = 1 m/s
Similarly, apply the relative acceleration equation between the point A on
the crank and the fixed point C on the slot linkage. If the observer is at C and
is rotating along with the slot, he would see A to be moving along the straight
slot. Therefore,
aA = aC + ω CB × ω CB × rA/C + ω̇ CB × rA/C + 2ω CB × vrel + arel
From the velocity analysis,
ω CB × ω CB × rA/C = 8.66 m/s2
2ω CB × v = 10 m/s2
rel
aA = vA2 /OA = 20 m/s2
Construct the acceleration diagram as depicted in fig. 5 and perform the geometrical analysis, the remaining acceleration can be determined.
arel = 20 cos 30 − 8.66 = 8.66 m/s2 along the slot towards C
ω̇ CB × rA/C = 20 cos 60 − 10 = 0, ω̇CB = 0 rad/s2
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Figure 5: Velocity and acceleration diagram
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