faculty of engineering

(1)

FACULTY OF ENGINEERING CHULALONGKORN UNIVERSITY

2103213 ENG MECHANICS I

Year 2

^nd

, First Semester, Mid Term Examination. July 21, 2008. Time 13.00-15.00

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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สอบอื่นๆ 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 (………..……….)

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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

μs

. Also, take

2 / l b<

.

(3)

Figure 1: Free body diagram of problem 1

Solution: The approach to this problem is to determine the forceP 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. FAand FB denote the corresponding friction forces. If the beam is about to slip at endA,

FA=µs

mg 2 ,

assuming supportB has not yet reached the impending motion status. Take the moment about B along the z-axis,

[ΣMB^z = 0] −P b+µsmg

2 l = 0, P =µsmgl

2b .

Instead, had the beam is to slip at endB first, the friction atB would be the static friction;

FB =µs

mg 2 .

Chulalongkorn University Phongsaen PITAKWATCHARA

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Similarly, to determine the corresponding applied force P, we take the moment about A along the z-axis.

[ΣMA^z = 0] P(l−b)−µsmg

2 l= 0, P =µs mgl

2(l−b). Since the problem states b < l/2, we may conclude

b < l−b → 1 b > 1

l−b. µs

mgl 2b > µs

mgl 2(l−b).

Since the applied forceP for B to slip is less than that forA, B would slip first.

Therefore the maximum forceP which can be applied without causing the beam to slip is

Pmax =µs

mgl 2(l−b).

The corresponding friction atB is the static friction F_B =µ_s^mg₂ . The friction at A may be determined from the equilibrium condition along thex-axis as

[ΣFx = 0] FA+FB−P = 0

FA+µsmg

2 −µs mgl 2(l−b) = 0 FA =µsmg

2 b l−b

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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.

(6)

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 aboutO.

[ΣMO = 0] 40×1.125 + 40g×0.55×cos(30−θ)−2C×0.55 sinθ= 0, whereθ is the angle OAB that can be calculated by the cosine law.

0.175² = 0.55²+ 0.6²−2×0.55×0.6 cosθ, θ = 16.787^◦

Substitute the value into the above equation. The compressive hydraulic force may be determined.

C = 803 N.

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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.

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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

−6 3



×



 0 P 0



+





−2

−6 3



×





Rcos 30 Rsin 30

0



+





−1

−3 1.5



×



 0 0

−200g



= 0, whereP andRare the magnitude of reaction forces that the wallsC andDexert on the ball endB, 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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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.

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Solution 1: First, choose the originO and determine the equivalent force-couple resultant. Hence,

R= 100i+ 100j N

M_O = 0.4×100k−0.3×100k+ 0.4×100j−20j= 20j+ 10k Nm Next, project the couple M_O onto the direction parallel and perpendicular to nR, for which its value is

nR = R

|R| = 1

√2i+ 1

√2j Consequently, the components of M_O are

M_k = (MO·nR)nR = 10i+ 10j Nm M_⊥ =M_O−M_k =−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 pointP.

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 M_k, M_k = 10i+ 10j Nm

and the forceR,

R= 100i+ 100j N of which its line of action passes through the point P,

x= 0.1 m z= 0.1 m in thex-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 pointP in thex-zplane, where the wrench passes, has the coordinate (x,0, z). Consequently, the moment of the force system about P is

M_P = 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 atP must be equal to the couple of the wrench passing through P, which is parallel to the resultant force. That is, M_P k R. The constraint eqautions are obtained by comparing the ratio of their components:

100

100z = 100 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

M_P = 10i+ 10j Nm

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FACULTY OF ENGINEERING CHULALONGKORN UNIVERSITY

2103213 ENG MECHANICS I

Year 2

^nd

, First Semester, Final Examination. September 22, 2008. Time 13.00-15.00

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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สอบอื่นๆ 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 (………..……….)

(13)

2103213 Engineering Mechanics I

ID………..………….…Name………CR58…………

1. The robot arm is elevating and extending simultaneously. At a given instant,

θ =30^o

,

=

=10deg/s

θ^&

constant,

^l =⁰^.⁵^m

,

^l&=⁰^.²^m^/^s

, and

^l&&=−⁰^.³^m^/^s²

. 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.

(14)

Solution: The most appropriate coordinate system for this problem is obviously ther-θcoodinates. From the given information, we can straightforwardly express the relevant parameters:

r = 1.25 m, r˙ = 0.2 m/s, r¨=−0.3 m/s², θ = 30^◦, θ˙= 10× π

180 rad/s, θ¨= 0 rad/s², 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

hv = ˙rer+rθe˙ θ

i v = 0.2er+ 1.25× 18^πe_θ = 0.2er+ 0.218eθ. h

a= (¨r−rθ˙²)er+ (rθ¨+ 2 ˙rθ)e˙ θ

i

a=

−0.3−1.25× ₁₈^π²²

e_r+ 2×0.2× ₁₈^π

e_θ =−0.338er+ 0.070eθ. 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 ac- knowledge the following transformation.

e_r = cos 30i+ sin 30j e_θ =−sin 30i+ cos 30j

Substitute e_r and e_θ 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.

(15)

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

^xA =0.4^m

, the velocity of A is

^vA =0.9^m/^s

to

the right. Determine the acceleration of each slider and the force in the bar at this instant.

(16)

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

x²_A+x²_B = 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;

x_Ax˙_A+x_B+ ˙x_B = 0.

Since v_A is given, we may solve for v_B:

0.4×0.9 + 0.3 ˙x_B= 0, x˙_B =−1.2.

That is, the sliderB 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:

xAx¨A+xBx¨B+ ˙x²_A+ ˙x²_B = 0.

For this particular instant, the following relation may be written

0.4¨xA+ 0.3¨xB+ 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

(17)

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,Aand B in turn, as follow;

[ΣFxA =mAx¨A] P −T ×^0.4_0.5 = 2¨xA. (2) [ΣFxB =mBx¨B] −T × ^0.30.5 = 3¨xB. (3) Solving (1), (2), and (3) simultaneously, the acceleration and the force in the bar are

¨

xA= 1.364 m/s²

¨

xB =−9.318 m/s² T = 46.6 N

(18)

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

45o

θ =

, determine the magnitude of the velocity of D and the angular velocity

ω

of ABD for this position.

(19)

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 ofDis 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 hencevB is directed perpendicular to OB.

Using the typical coordinate frame {xyz}, for this instant, vB =vA+ωABD ×rB/A

vB

cos 45 sin 45

= 4

0

+ωk×

0.4 cosα 0.4 sinα

, whereα =⁶ BAO that may be evaluated from the law of sine:

0.25

sinα = 0.4

sin 45, α= 26.23^◦.

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 =vA+ωABD×rD/A

v_D = 4

0

+ 7.47k×

0.6 cosα 0.6 sinα

=

2.02 4.02

m/s.

v_D = 4.5 m/s

(20)

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 =⁴^rad^/^s

during a short interval of motion. For the instant represented,

3 / 4 tan⁻¹

θ =

and AB is parallel to the x-axis. For this instant, determine the angular acceleration of both the

plate and link CB.

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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 linkOA, to the plate, and finally to link BC.

vA=ωOA×rA/O

vA=−4k×

0.1 sinα

−0.1 cosα

=

−0.24

−0.32

m/s.

where from the current mechanism posture,α= tan⁻¹ ⁴₃ . On the rigid plate, velocity of A and B are related by v_B =v_A+ω_AB ×r_B/A

v_B

−4/5 3/5

=

−0.24

−0.32

+ω_ABk×0.16i 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

v_B = 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 linkBC.

[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

[aA=rOA·ω²_OA] aA= 0.1×4²(−⁴5i+ ³₅j) =−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 pointB 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+ωAB×ωAB×rB/A+αAB×rB/A =aB =ωBC×ωBC×rB/C +αBC×rB/C.

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Substitute the position and velocity parameters into the vectorial equation, −1.28

0.96

+ 3.125k×3.125k×0.16i+αABk×0.16i

= 1.5k×1.5k×

0.2×3/5 0.2×4/5

+αBCk×

0.2×3/5 0.2×4/5

. Angular acceleration of the plate and the linkBC may then be solved:

αAB = 3.81 rad/s² CCW αBC = 16.08 rad/s² CCW

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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.

(24)

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 crankOAis transmitted to the slotCB through the pin A. Therefore, the following relative velocity equation is set up with the helpful velocity diagram shown in fig. 5:

v_A=v_P +v_A/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 v_A/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,

a_A=a_C +ω_CB×ω_CB×r_A/C+ ˙ω_CB×r_A/C + 2ω_CB×vrel +arel From the velocity analysis,

ω_CB×ω_CB×r_A/C

= 8.66 m/s²

2ω_CB×vrel

= 10 m/s² aA=v_A²/OA= 20 m/s²

Construct the acceleration diagram as depicted in fig.5and perform the geomet- rical analysis, the remaining acceleration can be determined.

arel = 20 cos 30−8.66 = 8.66 m/s² along the slot towards C

ω˙_CB×r_A/C

= 20 cos 60−10 = 0, ω˙CB = 0 rad/s²

(25)

Figure 5: Velocity and acceleration diagram