final tufree statics&dynamics

FINAL EXAMTU - F R E E

STATICS&DYNAMICS

ตวฟร MECHANICS FINAL EXAM

1 MOTION OF PARTICLE VELOCITY AND ACCELERATION Velocity

Acceleration

with Constant Acceleration

RECTANGULAR COORDINATE(x-y)

v = ds

dt= !s

a = dv

dt= !v

ads = vdv

v = v 0 +at

s = s0 +v0t +12at 2

v 2 = v 20 + 2a(s − s0 )

!r = xi+ yi!v = !"r = vxi+vy j!a = !""r = axi+ ay j

vx =vy =

ax =ay =

v = θv =

a =

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are the same as for the differentiation of scalar quantities, except for thecase of the cross product where the order of the terms must be pre-served. These rules are covered in Art. C/7 of Appendix C and should bereviewed at this point.

Three different coordinate systems are commonly used for describingthe vector relationships for curvilinear motion of a particle in a plane: rec-tangular coordinates, normal and tangential coordinates, and polar coor-dinates. An important lesson to be learned from the study of thesecoordinate systems is the proper choice of a reference system for a givenproblem. This choice is usually revealed by the manner in which the mo-tion is generated or by the form in which the data are specified. Each ofthe three coordinate systems will now be developed and illustrated.

2/4 Rectangular Coordinates (x-y)This system of coordinates is particularly useful for describing mo-

tions where the x- and y-components of acceleration are independentlygenerated or determined. The resulting curvilinear motion is then ob-tained by a vector combination of the x- and y-components of the posi-tion vector, the velocity, and the acceleration.

Vector RepresentationThe particle path of Fig. 2/5 is shown again in Fig. 2/7 along with

x- and y-axes. The position vector r, the velocity v, and the accelerationa of the particle as developed in Art. 2/3 are represented in Fig. 2/7 to-gether with their x- and y-components. With the aid of the unit vectors i and j, we can write the vectors r, v, and a in terms of their x- and y-components. Thus,

(2/6)

As we differentiate with respect to time, we observe that the time deriv-atives of the unit vectors are zero because their magnitudes and direc-tions remain constant. The scalar values of the components of v and aare merely vx ! , vy ! and ax ! ! , ay ! . (As drawn inFig. 2/7, ax is in the negative x-direction, so that would be a negativenumber.)

As observed previously, the direction of the velocity is always tan-gent to the path, and from the figure it is clear that

If the angle ! is measured counterclockwise from the x-axis to v for theconfiguration of axes shown, then we can also observe that dy/dx !tan ! ! vy/vx.

a2 ! ax

2 " ay

2 a ! !ax

2 " ay

2

v2 ! vx

2 " vy

2 v ! !vx

2 " vy

2 tan ! ! vy

vx

xvy ! yxvxyx

a ! v ! r ! xi " yj

v ! r ! xi " yj

r ! xi " yj

Article 2/4 Rectangular Coordinates (x-y) 43

Figure 2/7

Path

j

ixi

yj r

A

x

y

Aθ

v avy

vx ax

ay

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NORMAL AND TANGENTIAL COORDINATE (n-t)

แกน tangential (t) คอ แกน normal (n) คอ

,

Note: 1. เมอวตถเคลอนทเปนเสนตรง

2. เมอวตถเคลอนทโคงดวยความเรวคงท

3. สามารถหาแกน tangential จาก

!v = vet

!a = anen +atet a = a2

n +a2t

ρ =1+ dy

dx( )2⎡⎣⎢

⎤⎦⎥

3/2

d 2ydx 2

an =

at =


12.7 CURVILINEAR MOTION: NORMAL AND TANGENTIAL COMPONENTS 53

1212.7 Curvilinear Motion: Normal andTangential Components

When the path along which a particle travels is known, then it is oftenconvenient to describe the motion using n and t coordinate axes whichact normal and tangent to the path, respectively, and at the instantconsidered have their origin located at the particle.

Planar Motion. Consider the particle shown in Fig. 12–24a, whichmoves in a plane along a fixed curve, such that at a given instant it is atposition s, measured from point O. We will now consider a coordinatesystem that has its origin at a fixed point on the curve, and at the instantconsidered this origin happens to coincide with the location of theparticle. The t axis is tangent to the curve at the point and is positive inthe direction of increasing s.We will designate this positive direction withthe unit vector A unique choice for the normal axis can be made bynoting that geometrically the curve is constructed from a series ofdifferential arc segments ds, Fig. 12–24b. Each segment ds is formed fromthe arc of an associated circle having a radius of curvature (rho) andcenter of curvature The normal axis n is perpendicular to the t axis withits positive sense directed toward the center of curvature Fig. 12–24a.This positive direction, which is always on the concave side of the curve,will be designated by the unit vector The plane which contains the nand t axes is referred to as the embracing or osculating plane, and in thiscase it is fixed in the plane of motion.*

Velocity. Since the particle moves, s is a function of time. Asindicated in Sec. 12.4, the particle’s velocity v has a direction that isalways tangent to the path, Fig. 12–24c, and a magnitude that isdetermined by taking the time derivative of the path function i.e., (Eq. 12–8). Hence

(12–15)

where

(12–16)v = s#

v = vut

v = ds>dts = s1t2,

un .

O¿,O¿.

r

ut .

*The osculating plane may also be defined as the plane which has the greatest contactwith the curve at a point. It is the limiting position of a plane contacting both the point andthe arc segment ds. As noted above, the osculating plane is always coincident with a planecurve; however, each point on a three-dimensional curve has a unique osculating plane.

s

O

O¿

n

un

ut

tPosition

(a)

O¿

ds

Radius of curvature

O¿

O¿ds

ds

rr

rr

rr

(b)

O¿

Velocity

r r

v

(c)

Fig. 12–24

12.7 CURVILINEAR MOTION: NORMAL AND TANGENTIAL COMPONENTS 53

1212.7 Curvilinear Motion: Normal andTangential Components

When the path along which a particle travels is known, then it is oftenconvenient to describe the motion using n and t coordinate axes whichact normal and tangent to the path, respectively, and at the instantconsidered have their origin located at the particle.

Planar Motion. Consider the particle shown in Fig. 12–24a, whichmoves in a plane along a fixed curve, such that at a given instant it is atposition s, measured from point O. We will now consider a coordinatesystem that has its origin at a fixed point on the curve, and at the instantconsidered this origin happens to coincide with the location of theparticle. The t axis is tangent to the curve at the point and is positive inthe direction of increasing s.We will designate this positive direction withthe unit vector A unique choice for the normal axis can be made bynoting that geometrically the curve is constructed from a series ofdifferential arc segments ds, Fig. 12–24b. Each segment ds is formed fromthe arc of an associated circle having a radius of curvature (rho) andcenter of curvature The normal axis n is perpendicular to the t axis withits positive sense directed toward the center of curvature Fig. 12–24a.This positive direction, which is always on the concave side of the curve,will be designated by the unit vector The plane which contains the nand t axes is referred to as the embracing or osculating plane, and in thiscase it is fixed in the plane of motion.*

Velocity. Since the particle moves, s is a function of time. Asindicated in Sec. 12.4, the particle’s velocity v has a direction that isalways tangent to the path, Fig. 12–24c, and a magnitude that isdetermined by taking the time derivative of the path function i.e., (Eq. 12–8). Hence

(12–15)

where

(12–16)v = s#

v = vut

v = ds>dts = s1t2,

un .

O¿,O¿.

r

ut .

*The osculating plane may also be defined as the plane which has the greatest contactwith the curve at a point. It is the limiting position of a plane contacting both the point andthe arc segment ds. As noted above, the osculating plane is always coincident with a planecurve; however, each point on a three-dimensional curve has a unique osculating plane.

s

O

O¿

n

un

ut

tPosition

(a)

O¿

ds

Radius of curvature

O¿

O¿ds

ds

rr

rr

rr

(b)

O¿

Velocity

r r

v

(c)

Fig. 12–24

54 CH A P T E R 12 KI N E M AT I C S O F A PA RT I C L E

12Acceleration. The acceleration of the particle is the time rate ofchange of the velocity. Thus,

(12–17)

In order to determine the time derivative note that as the particlemoves along the arc ds in time dt, preserves its magnitude of unity;however, its direction changes, and becomes Fig. 12–24d. As shown inFig. 12–24e, we require Here stretches between thearrowheads of and which lie on an infinitesimal arc of radius Hence, has a magnitude of and its direction is defined by

Consequently, and therefore the time derivative becomes Since Fig. 12–24d, then and therefore

Substituting into Eq. 12–17, a can be written as the sum of its twocomponents,

(12–18)

where

or (12–19)

and

(12–20)

These two mutually perpendicular components are shown in Fig. 12–24f.Therefore, the magnitude of acceleration is the positive value of

(12–21)a = 4at2 + an

2

an = v2

r

atds = v dvat = v#

a = atut + anun

u# t = u#un = s

#

run = v

run

u#= s

# >r,ds = rdu,u# t = u#un .

dut = duun ,un .dut = 112 du,dut

ut = 1.uœt ,ut

dutuœt = ut + dut .

uœt ,

ut

u# t ,

a = v# = v# ut + vu# t

O¿

(d)

rr

ut

u¿t

un

du

ds

ut

u¿t

un

dut

du

(e)

an

O¿

Acceleration

Pat

a

(f)

Fig. 12–24 (cont.)



1. รถยนตเคลอนทบนถนนดวยความหนวงคงท 0.6 m/s ตอวนาท ทจด A รถยนตมความเรว 16 m/s จงหาความเรงของรถยนตเมอเคลอนทเปนระยะทาง 120 m ไปถงจด B กำหนดรศมความโคงทจด B เปน 60 m




POLAR COORDINATE (r-ϴ) Velocity

โดยท

Acceleration

โดยท

Note: 1. การเคลอนทเปนวงกลม

2. เคลอนทเปนวงกลมดวยความเรวคงท

v = !r er + r !θ eθ

vr =vθ =

!a = (""r − r "θ 2)er + (r ""θ + 2 "r "θ)eθ

ar =aθ =




2. แกนไฮดรอลค OB หมนรอบจด O โดยความยาว l ทสวน AB สามารถยดหดได ถาแกนไฮดรอลคหมนดวยความเรวเชงมมคงท = 60 deg/s และความยาว l หดเขาดวยอตราเรวคงท 150 mm/s a) จงเขยนแสดงแกน r-θ เพอพจารณาการเคลอนทของจด B b) จงหาขนาดของความเรวลพธเปนหนวย m/s และความเรงลพธ ของจด B ขณะทความยาว l =125 mm. *

!θ

v = 545 mm/s, a = 632 mm/s2


2/139 An internal mechanism is used to maintain a con-stant angular rate about the z-axisof the spacecraft as the telescopic booms are ex-tended at a constant rate. The length l is variedfrom essentially zero to 3 m. The maximum accel-eration to which the sensitive experiment modulesP may be subjected is . Determine themaximum allowable boom extension rate .

Problem 2/139

2/140 The radial position of a fluid particle P in a certaincentrifugal pump with radial vanes is approxi-mated by cosh Kt, where t is time and is the constant angular rate at which the impellerturns. Determine the expression for the magnitudeof the total acceleration of the particle just prior toleaving the vane in terms of , R, and K.

Problem 2/140

Fixedreference

axis

R

r

r0

θ

P

r0

!K !r ! r0

l

l

z

P

P

1.2 m1.2 m

Ω

l0.011 m /s2

" ! 0.05 rad /s

72 Chapter 2 Kinematics of Particles

2/136 As the hydraulic cylinder rotates around O, the ex-posed length l of the piston rod P is controlled bythe action of oil pressure in the cylinder. If thecylinder rotates at the constant rate and l is decreasing at the constant rate of 150mm/s, calculate the magnitudes of the velocity vand acceleration a of end B when .

Problem 2/136

2/137 The drag racer P starts from rest at the start line Sand then accelerates along the track. When it hastraveled 100 m, its speed is 45 m/s. For that in-stant, determine the values of and relative toaxes fixed to an observer O in the grandstand G asshown.

Problem 2/137

2/138 In addition to the information supplied in the pre-vious problem, it is known that the drag racer is ac-celerating forward at when it has traveled100 m from the start line S. Determine the corre-sponding values of and .!r

10 m /s2

G

r

SP

35 m

Oθ

!r

375 mm

θO

l

B

l ! 125 mm

! 60 deg /s!

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3. [final 2556] อนภาค P เคลอนทไปดามเสนทางเดนรปกนหอยตามรป โดยท R=4+0.25t2 m และ θ=2πt rad เมอ t คอเวลาหนวยเปนวนาท จงหาขนาดของความเรวและความเรง เมอเวลาเทากบ 2 วนาท




RELATIVE MOTION (Translation Axes)

! โดยท !

! โดยท !

4. รถไฟเคลอนทดวยความเรว 120 km/h วงตดผานถนนทรถยนต A ซงเคลอนท 90 km/h จงคำนวณหา a) ขนาดความเรวของรถไฟทคนขบรถ A เหน b) ทศทางความเรวของรถไฟทคนขบรถ A เหน

!vA = !vB + !vA/B

!vA/B

!aA =

!aB +

!aA/B

!aA/B


12.10 RELATIVE-MOTION OF TWO PARTICLES USING TRANSLATING AXES 87

1212.10 Relative-Motion of Two ParticlesUsing Translating Axes

Throughout this chapter the absolute motion of a particle has beendetermined using a single fixed reference frame. There are many cases,however, where the path of motion for a particle is complicated, so that itmay be easier to analyze the motion in parts by using two or more framesof reference. For example, the motion of a particle located at the tip of anairplane propeller, while the plane is in flight, is more easily described ifone observes first the motion of the airplane from a fixed reference andthen superimposes (vectorially) the circular motion of the particlemeasured from a reference attached to the airplane.

In this section translating frames of reference will be considered for theanalysis. Relative-motion analysis of particles using rotating frames ofreference will be treated in Secs. 16.8 and 20.4, since such an analysisdepends on prior knowledge of the kinematics of line segments.

Position. Consider particles A and B, which move along thearbitrary paths shown in Fig. 12–42. The absolute position of eachparticle, and is measured from the common origin O of the fixed x,y, z reference frame.The origin of a second frame of reference isattached to and moves with particle A. The axes of this frame are onlypermitted to translate relative to the fixed frame. The position of Bmeasured relative to A is denoted by the relative-position vectorUsing vector addition, the three vectors shown in Fig. 12–42 can berelated by the equation

(12–33)

Velocity. An equation that relates the velocities of the particles isdetermined by taking the time derivative of the above equation; i.e.,

(12–34)

Here and refer to absolute velocities, sincethey are observed from the fixed frame; whereas the relative velocity

is observed from the translating frame. It is importantto note that since the axes translate, the components ofwill not change direction and therefore the time derivative of thesecomponents will only have to account for the change in theirmagnitudes. Equation 12–34 therefore states that the velocity of B isequal to the velocity of A plus (vectorially) the velocity of “B withrespect to A,” as measured by the translating observer fixed in the

reference frame.z¿y¿,x¿,

rB>Az¿y¿,x¿,vB>A = drB>A>dt

vA = drA>dtvB = drB>dt

vB = vA + vB>A

rB = rA + rB>A

rB>A .

z¿y¿,x¿,rB ,rA

z¿

z

A

O

x¿

x

y

y¿

rB

rA

rB/A

Translatingobserver

Fixedobserver

B

Fig. 12–42



5. ลกบอล B ถกโยนขนดวยทตำแหนงสงจากพน 12 m ดวยความเรว 18 m/s ขณะเดยวกนแทนเหลก E อยตำแหนง 5 m สงจากพน ดวยความเรวคงท 2 m/s จงคำนวณหา ก) เวลา t และตำแหนง s ทลกบอกปะทะแทนเหลก ข) ความเรวของลกบอกเทยบกบแทนเหลก ทตำแหนงปะทะ

*

t = 3.65 s, s = 12.3 m, vB/E = -19.81 m/s




6. รถยนต A เคลอนทไปขางหนาดวยอตราเรว 18 km/h และมอตราเรงท 3 m/s2 ดงแสดงในรป ในขณะทกลองถกตดตงทตำแหนง B บนชงชาสวรรคซงหมนดวยอตราเรวเชงมมคงท Ω = 3 rev/min ในทศทางตามเขมนาฬกา (ตำแหนง B ทำมม 45° กบแกน x) จงคำนวณหาa) ความเรวของรถยนต A เทยบกบกลอง Bb) ความเรงของรถยนต A เทยบกบกลอง B

*

vA/B=3.00i + 1.999j m/s vA/B=3.63i + 0.628j m/s2




2 FORCE AND ACCELERATION FRICTION

EQUATION OF MOTION

!

!

F = µN

6/3 Dry FrictionThe remainder of this chapter describes the effects of dry friction

acting on the exterior surfaces of rigid bodies. We will now explain themechanism of dry friction with the aid of a very simple experiment.

Mechanism of Dry FrictionConsider a solid block of mass m resting on a horizontal surface, as

shown in Fig. 6/1a. We assume that the contacting surfaces have someroughness. The experiment involves the application of a horizontal forceP which continuously increases from zero to a value sufficient to movethe block and give it an appreciable velocity. The free-body diagram ofthe block for any value of P is shown in Fig. 6/1b, where the tangentialfriction force exerted by the plane on the block is labeled F. This frictionforce acting on the body will always be in a direction to oppose motionor the tendency toward motion of the body. There is also a normal forceN which in this case equals mg, and the total force R exerted by the sup-porting surface on the block is the resultant of N and F.

A magnified view of the irregularities of the mating surfaces, Fig.6/1c, helps us to visualize the mechanical action of friction. Support isnecessarily intermittent and exists at the mating humps. The directionof each of the reactions on the block, R1, R2, R3, etc. depends not only onthe geometric profile of the irregularities but also on the extent of localdeformation at each contact point. The total normal force N is the sum

Article 6/3 Dry Friction 337

Pm P

F

αRN

mg

(a)

(c)

t

n

F

F = P

Fmax = s Nµ Fk = k Nµ

R1 R3R2

(b)

P

Staticfriction

(no motion)

Kineticfriction(motion)

Impendingmotion

(d)

Figure 6/1

6/3 Dry FrictionThe remainder of this chapter describes the effects of dry friction

acting on the exterior surfaces of rigid bodies. We will now explain themechanism of dry friction with the aid of a very simple experiment.

Mechanism of Dry FrictionConsider a solid block of mass m resting on a horizontal surface, as

shown in Fig. 6/1a. We assume that the contacting surfaces have someroughness. The experiment involves the application of a horizontal forceP which continuously increases from zero to a value sufficient to movethe block and give it an appreciable velocity. The free-body diagram ofthe block for any value of P is shown in Fig. 6/1b, where the tangentialfriction force exerted by the plane on the block is labeled F. This frictionforce acting on the body will always be in a direction to oppose motionor the tendency toward motion of the body. There is also a normal forceN which in this case equals mg, and the total force R exerted by the sup-porting surface on the block is the resultant of N and F.

A magnified view of the irregularities of the mating surfaces, Fig.6/1c, helps us to visualize the mechanical action of friction. Support isnecessarily intermittent and exists at the mating humps. The directionof each of the reactions on the block, R1, R2, R3, etc. depends not only onthe geometric profile of the irregularities but also on the extent of localdeformation at each contact point. The total normal force N is the sum

Article 6/3 Dry Friction 337

Pm P

F

αRN

mg

(a)

(c)

t

n

F

F = P

Fmax = s Nµ Fk = k Nµ

R1 R3R2

(b)

P

Staticfriction

(no motion)

Kineticfriction(motion)

Impendingmotion

(d)

Figure 6/1

MEE224: Engineering Mechanics Lecture 6

caused by the relative velocity between the layers account for the fluid

friction.

3. ________________________________. It occurs in all solid materials

subjected to cyclic loadings. Internal friction is present during deformation

and causes a loss of energy.

In this course, we will only concentrate on ___________________________.

Dry Friction

As stated before, dry friction occurs at the contact between two rough surfaces during

a sliding motion or under a tendency to slide. The friction force is always tangent to

the surface and its direction always opposes the motion or the impending motion.

Mechanism of Dry Friction

Let us consider the block in figure 1 (redrawn in figure 2a). The block is vertical

equilibrium as the force N is equal to its weight mg. However, the block may or may

not slide horizontally depending on the magnitudes of the force P and the friction F.

Figure 2

The curve in figure 2b shows us two regions on either side of the vertical dotted line

labeled ‘______________________________________________’. These regions

correspond to the types of behavior of the block upon the application of force P.

Stationary Block – No motion is present

∑F = ma

P




7. [final 1/2554] รถบรรทกลงวงดวยความเรวคงท 72 km/h ดงรป จงหาสมประสทธความเสยดทานสถตตำสด (μs)min ระหวางลงกบพนรถทจะทำใหลงไมลนไถลในขณะรถบรรทกลดความเรวอยางสมำเสมอจนหยดทระยะ 100 m


162 Chapter 3 Kinetics of Particles

SAMPLE PROBLEM 3/11Calculate the velocity v of the 50-kg crate when it reaches the bottom of the

chute at B if it is given an initial velocity of 4 m/s down the chute at A. The coef-ficient of kinetic friction is 0.30.

Solution. The free-body diagram of the crate is drawn and includes the nor-mal force R and the kinetic friction force F calculated in the usual manner. Thework done by the weight is positive, whereas that done by the friction force isnegative. The total work done on the crate during the motion is

The work-energy equation gives

Ans.

Since the net work done is negative, we obtain a decrease in the kinetic energy.

SAMPLE PROBLEM 3/12The flatbed truck, which carries an 80-kg crate, starts from rest and attains

a speed of 72 km/h in a distance of 75 m on a level road with constant accelera-tion. Calculate the work done by the friction force acting on the crate during thisinterval if the static and kinetic coefficients of friction between the crate and thetruck bed are (a) 0.30 and 0.28, respectively, or (b) 0.25 and 0.20, respectively.

Solution. If the crate does not slip on the bed, its acceleration will be that ofthe truck, which is

Case (a). This acceleration requires a friction force on the block of

which is less than the maximum possible value of !sN ! 0.30(80)(9.81) = 235 N.Therefore, the crate does not slip and the work done by the actual static frictionforce of 213 N is

Ans.

Case (b). For !s ! 0.25, the maximum possible friction force is 0.25(80)(9.81) !196.2 N, which is slightly less than the value of 213 N required for no slipping.Therefore, we conclude that the crate slips, and the friction force is governed by thekinetic coefficient and is F ! 0.20(80)(9.81) = 157.0 N. The acceleration becomes

The distances traveled by the crate and the truck are in proportion to their ac-celerations. Thus, the crate has a displacement of (1.962/2.67)75 = 55.2 m, andthe work done by kinetic friction is

Ans.U1-2 ! 157.0(55.2) ! 8660 J or 8.66 kJ[U ! Fs]

a ! F/m ! 157.0/80 ! 1.962 m/s2[F ! ma]

U1-2 ! 213(75) ! 16 000 J or 16 kJ[U ! Fs]

F ! 80(2.67) ! 213 N[F ! ma]

a ! v2

2s !

(72/3.6)2

2(75) ! 2.67 m/s2[v2 ! 2as]

v2 ! 3.15 m/s

12 (50)(4)2 " 151.9 ! 12 (50)v2

2

12 mv1

2 # U1-2 ! 12 mv2

2[T1 # U1-2 ! T2]

U1-2 ! 50(9.81)(10 sin 15!) " 142.1(10) ! "151.9 J[U ! Fs]

15°

50 kg

B

A10 m

50(9.81) N

R = 474 N

kR = 142.1 Nµ

80(9.81) N

aF

80(9.81) N

Helpful Hint

! The work due to the weight dependsonly on the vertical distance traveled.

Helpful Hints

! We note that static friction forces dono work when the contacting sur-faces are both at rest. When they arein motion, however, as in this prob-lem, the static friction force actingon the crate does positive work andthat acting on the truck bed doesnegative work.

" This problem shows that a kineticfriction force can do positive workwhen the surface which supports theobject and generates the frictionforce is in motion. If the supportingsurface is at rest, then the kineticfriction force acting on the movingpart always does negative work.

!

!

"

c03.qxd 2/9/12 7:39 PM Page 162



8. กลองลงหนก 50 kg จงคำนวณหาความเรงของกลองเมอ ก) P=150 ข) P=0 N




CURVILINEAR MOTION

! !

! !

∑Fn = man

∑Ft = mat

an =

at =

∑Fr = mar

∑Fθ = maθ

ar =

aθ =


54 CH A P T E R 12 KI N E M AT I C S O F A PA RT I C L E

12Acceleration. The acceleration of the particle is the time rate ofchange of the velocity. Thus,

(12–17)

In order to determine the time derivative note that as the particlemoves along the arc ds in time dt, preserves its magnitude of unity;however, its direction changes, and becomes Fig. 12–24d. As shown inFig. 12–24e, we require Here stretches between thearrowheads of and which lie on an infinitesimal arc of radius Hence, has a magnitude of and its direction is defined by

Consequently, and therefore the time derivative becomes Since Fig. 12–24d, then and therefore

Substituting into Eq. 12–17, a can be written as the sum of its twocomponents,

(12–18)

where

or (12–19)

and

(12–20)

These two mutually perpendicular components are shown in Fig. 12–24f.Therefore, the magnitude of acceleration is the positive value of

(12–21)a = 4at2 + an

2

an = v2

r

atds = v dvat = v#

a = atut + anun

u# t = u#un = s

#

run = v

run

u#= s

# >r,ds = rdu,u# t = u#un .

dut = duun ,un .dut = 112 du,dut

ut = 1.uœt ,ut

dutuœt = ut + dut .

uœt ,

ut

u# t ,

a = v# = v# ut + vu# t

O¿

(d)

rr

ut

u¿t

un

du

ds

ut

u¿t

un

dut

du

(e)

an

O¿

Acceleration

Pat

a

(f)

Fig. 12–24 (cont.)

12.8 CURVILINEAR MOTION: CYLINDRICAL COMPONENTS 69

12Acceleration. Taking the time derivatives of Eq. 12–24, usingEqs. 12–25, we obtain the particle’s instantaneous acceleration,

To evaluate it is necessary only to find the change in the direction of since its magnitude is always unity. During the time a change will notchange the direction of however, a change will cause to become

where Fig. 12–30d. The time change in is thus For small angles this vector has a magnitude and acts in the

direction; i.e., Thus,

(12–27)

Substituting this result and Eq. 12–23 into the above equation for a, wecan write the acceleration in component form as

(12–28)

where

(12–29)

The term is called the angular accelerationsince it measures the change made in the angular velocity during aninstant of time. Units for this measurement are

Since and are always perpendicular, the magnitude ofacceleration is simply the positive value of

(12–30)

The direction is determined from the vector addition of its twocomponents. In general, a will not be tangent to the path, Fig. 12–30e.

a = 4(r$ - ru

#2)2 + (r u

$+ 2r

#u#)2

auar ,rad>s2.

u$

= d2u>dt2 = d>dt1du>dt2ar = r

$ - ru#2

au =$

ru + 2#ru#

a = arur + auuu

u# u = -u#ur

u# u = lim¢t:0

¢uu¢t

= - a lim¢t:0

¢u¢tbur

¢uu = -¢uur .-ur ,¢uu L 11¢u2 ¢uu .uuuœ

u = uu + ¢uu ,uuœ ,uu¢uuu ,

¢r¢t,uuu# u ,

+ ru$uu + ru

#u# ua = v# = r

$ur + r#u# r + r

#u#uu

O

rar

au

a

Acceleration

(e)

u

(d)

ur

u¿uuu

!uu

!u



9. [final 2/2554] รถมอเตอรไซคตองขบขามสนนนบนถนนขณะผานไฟสญญาณจราจร โดยกำหนดใหสนนนมรศมความโคง 15 m จงเขยน Free Body Diagram และ Kinetics Diagram ความเรงของมอเตอรไซคขณะอยตำแหนงสงสดของสนนน และหากรถอเตอรไซคมความเรวคงท จงหาความเรวสงสดทสามารถขบขไดโดยไมลอยพนจากสนนน เมอนำหนกรวมของรถและคนขบเทากบ 2000 นวตน

*

v = 12.1 m/s




10. แทง ABO หมนในแนวราบ รอบแกนแนวตง O และชนงาน C หนก 1.5 kg กำลงเคลอนทเขาหาจด O ดวยความเรวคงท 50-mm/sec จากแรงดงในเสนเชอก ในรป ABO หมนดวยความเรว ω = 6 rad/sec และ ความหนวง 2 rad/sec2 และ r = 250 mm จงหา (ก) แรงตงเชอก T และ (ข)แรงปฏกรยา N ทแทง ABO กระทำกบชนงาน C และบอกดวยวา กระทำทฝง A หรอ B


3/77 Small steel balls, each with a mass of 65 g, enter thesemicircular trough in the vertical plane with a hori-zontal velocity of 4.1 m/s at A. Find the force R ex-erted by the trough on each ball in terms of andthe velocity vB of the balls at B. Friction is negligible.

Problem 3/77

3/78 The flat circular disk rotates about a vertical axisthrough O with a slowly increasing angular velocity

. Prior to rotation, each of the 0.5-kg sliding blockshas the position mm with no force in its at-tached spring. Each spring has a stiffness of 400N/m. Determine the value of x for each spring for asteady speed of 240 rev/min. Also calculate the nor-mal force N exerted by the side of the slot on theblock. Neglect any friction between the blocks andthe slots, and neglect the mass of the springs. (Hint:Sum forces along and normal to the slot.)

Problem 3/78

80mm

80mm

x O

A

A

ω

x

x ! 25!

vA = 4.1 m/s

vB

320 mm

A

B

θ

"

3/79 The spring-mounted 0.8-kg collar A oscillates alongthe horizontal rod, which is rotating at the constantangular rate rad/s. At a certain instant, r isincreasing at the rate of 800 mm/s. If the coefficientof kinetic friction between the collar and the rod is0.40, calculate the friction force F exerted by the rodon the collar at this instant.

Problem 3/79

3/80 The slotted arm revolves in the horizontal planeabout the fixed vertical axis through point O. The 3-lb slider C is drawn toward O at the constant rateof 2 in./sec by pulling the cord S. At the instant forwhich in., the arm has a counterclockwise an-gular velocity rad/sec and is slowing downat the rate of 2 rad/sec2. For this instant, determinethe tension T in the cord and the magnitude N ofthe force exerted on the slider by the sides of thesmooth radial slot. Indicate which side, A or B, ofthe slot contacts the slider.

Problem 3/80

! ! 6r ! 9

Vertical

A

θr.

" ! 6

Article 3/5 Problems 149

c03.qxd 2/9/12 7:39 PM Page 149



3 WORK AND ENERGYWORK

Definition of Work ! Work of a Constant Force

! Work of Friction

! !

Note :

dU = F ⋅ds

U1−2 = PL cosθ

(1) Work Associated with a Constant External Force. Considerthe constant force P applied to the body as it moves from position 1 toposition 2, Fig. 3/4. With the force P and the differential displacementdr written as vectors, the work done on the body by the force is

(3/9)

As previously discussed, this work expression may be interpreted as theforce component P cos ! times the distance L traveled. Should ! be be-tween 90! and 270!, the work would be negative. The force component P sin ! normal to the displacement does no work.

(2) Work Associated with a Spring Force. We consider here thecommon linear spring of stiffness k where the force required to stretchor compress the spring is proportional to the deformation x, as shown inFig. 3/5a. We wish to determine the work done on the body by the springforce as the body undergoes an arbitrary displacement from an initialposition x1 to a final position x2. The force exerted by the spring on thebody is F ! "kxi, as shown in Fig. 3/5b. From the definition of work, wehave

(3/10)

If the initial position is the position of zero spring deformation sothat x1 ! 0, then the work is negative for any final position x2 ! 0. Thisis verified by recognizing that if the body begins at the undeformedspring position and then moves to the right, the spring force is to theleft; if the body begins at x1 ! 0 and moves to the left, the spring force isto the right. On the other hand, if we move from an arbitrary initial po-sition x1 ! 0 to the undeformed final position x2 ! 0, we see that thework is positive. In any movement toward the undeformed spring posi-tion, the spring force and the displacement are in the same direction.

In the general case, of course, neither x1 nor x2 is zero. The magni-tude of the work is equal to the shaded trapezoidal area of Fig. 3/5a. Incalculating the work done on a body by a spring force, care must be

U1-2 ! !2

1 F " dr ! !2

1("kxi) " dx i ! "!x2

x1

kx dx ! 12

k(x1

2 " x2

2)

! !x2

x1

P cos ! dx ! P cos !(x2 " x1) ! PL cos !

U1-2 ! !2

1 F " dr ! !2

1 [(P cos !)i # (P sin !)j] " dx i


P

dr

x

y

α

L1 2

Figure 3/4

c03.qxd 2/9/12 7:39 PM Page 156

U1−2 = PL cosθ − µkNL


14.1 THE WORK OF A FORCE 171

14

Work of a Constant Force Moving Along a Straight Line.If the force has a constant magnitude and acts at a constant angle from its straight-line path, Fig. 14–3a, then the component of in thedirection of displacement is always The work done by whenthe particle is displaced from to is determined from Eq. 14–1, inwhich case

or

(14–2)

Here the work of represents the area of the rectangle in Fig. 14–3b.

Work of a Weight. Consider a particle of weight W, which movesup along the path s shown in Fig. 14–4 from position to position Atan intermediate point, the displacement Since

applying Eq. 14–1 we have

or

(14–3)

Thus, the work is independent of the path and is equal to the magnitudeof the particle’s weight times its vertical displacement. In the case shownin Fig. 14–4 the work is negative, since W is downward and is upward.Note, however, that if the particle is displaced downward thework of the weight is positive. Why?

1-¢y2,¢y

U1-2 = -W ¢y

= Ly2

y1

-W dy = -W1y2 - y12U1-2 = LF # dr = L

r2

r1

1-Wj2 # 1dxi + dyj + dzk2W = -Wj,

dr = dxi + dyj + dzk.s2 .s1

Fc

U1-2 = Fc cos u1s2 - s12U1-2 = Fc cos uL

s2

s1

ds

s2s1

FcFc cos u.Fc

uFc

Fc

Fc cos u s2s1s

(a)

u

s

(b)

F cos u

Fc cos u

s2s1

Fig. 14–3

dr

s r1r2

y

Ws2

s1

z

xy1

y2

Fig. 14–4



WORK AND ENERGY EQUATION # Kinetic Energy

#Potential Energy

#

#

#

No Force acting on the Particle

#

POWER

# (# )

EFFICIENCY

#

T1 +V1 +U1−2 =T2 +V2

T = 1

2mv 2

V =Ve +Vg

Ve = 1

2kx 2

Vg = mgh

T1 +V1 =T2 +V2

P = dUdt

= F ⋅v 1 W = 1 J/s

e =PoutputPinput


202 CH A P T E R 14 KI N E T I C S O F A PA RT I C L E : WO R K A N D EN E R G Y

14

Elastic Potential Energy. When an elastic spring is elongated orcompressed a distance s from its unstretched position, elastic potentialenergy can be stored in the spring. This energy is

(14–14)

Here is always positive since, in the deformed position, the force ofthe spring has the capacity or “potential” for always doing positive workon the particle when the spring is returned to its unstretched position,Fig. 14–18.

Ve

Ve = +12 ks2

Ve

Ve ! 0

Unstretchedposition, s ! 0

Elastic potential energy

"s

#s

k

k

k

Ve ! # ks212

Ve ! # ks212

Fig. 14–18

The weight of the sacks resting on thisplatform causes potential energy to bestored in the supporting springs.As eachsack is removed, the platform will riseslightly since some of the potential energywithin the springs will be transformedinto an increase in gravitational potentialenergy of the remaining sacks. Such adevice is useful for removing the sackswithout having to bend over to pick themup as they are unloaded.

3/7 Potential EnergyIn the previous article on work and kinetic energy, we isolated a

particle or a combination of joined particles and determined the workdone by gravity forces, spring forces, and other externally applied forcesacting on the particle or system. We did this to evaluate U in the work-energy equation. In the present article we will introduce the concept ofpotential energy to treat the work done by gravity forces and by springforces. This concept will simplify the analysis of many problems.

Gravitational Potential EnergyWe consider first the motion of a particle of mass m in close proxim-

ity to the surface of the earth, where the gravitational attraction(weight) mg is essentially constant, Fig. 3/8a. The gravitational poten-tial energy Vg of the particle is defined as the work mgh done against thegravitational field to elevate the particle a distance h above some arbi-trary reference plane (called a datum), where Vg is taken to be zero.Thus, we write the potential energy as

(3/18)

This work is called potential energy because it may be convertedinto energy if the particle is allowed to do work on a supporting bodywhile it returns to its lower original datum plane. In going from onelevel at h ! h1 to a higher level at h ! h2, the change in potential energybecomes

The corresponding work done by the gravitational force on the particleis "mg#h. Thus, the work done by the gravitational force is the nega-tive of the change in potential energy.

When large changes in altitude in the field of the earth are encoun-tered, Fig. 3/8b, the gravitational force Gmme/r2 ! mgR2/r2 is no longerconstant. The work done against this force to change the radial positionof the particle from r1 to r2 is the change (Vg)2 " (Vg)1 in gravitationalpotential energy, which is

It is customary to take (Vg)2 ! 0 when r2 ! !, so that with thisdatum we have

(3/19)

In going from r1 to r2, the corresponding change in potential energy is

#Vg ! mgR2 ! 1r1

" 1r2"

Vg ! "mgR2

r

# r2

r1

mgR2 drr2 ! mgR2 ! 1

r1 " 1

r2" ! (Vg)2 " (Vg)1

#Vg ! mg(h2 " h1) ! mg#h

Vg ! mgh

Earthme

r

m

R

(a)

(b)

Vg = 0

Vg = mgh

mgh

Vg = –mgR2—–—

r

mgR2—–—

r2

Figure 3/8

Article 3/7 Potential Energy 175

c03.qxd 2/9/12 7:39 PM Page 175



11. กลอง A มวล 50 kg เรมจากหยดนงท A ซงเปนระยะทสปรงยดออกมา x1=0.233 m (สปรงม k=80 N/m) จงหาความเรวทจด B เมอมแรงดงทเชอก 300 N (ไมคดแรงเสยดทานระหวางกลองกบรองเลอน)‑ *

v = 2.00 m/s




12. เครองรอนนำหนก 8 ตน ยดตดกบเชอกดงรป มม θ = 60° เมอเชอก AC ถกตดออก เครองรอนเรมเคลอนทจากหยดนง จงหา ก. ความเรวของเครองรอนกอนจะถงพนทตำแหนง B (θ=15°) ข. แรงตกเชอกทตำแหนง B *

13.5 m/s, 149 kN




13. [final 2/2554] ลกตมมวล 0.75 กก. ถกยดอยดวยเชอก ถกยงออกจากจดอยนงทตำแหนง A โดยการปลอยสปรงทถกกดอย 50 มลลเมตร ณ.ตำแหนงน จงหาคา k (คานจสปรง หรอ คาความแขงแกรงของสปรง) ในกรณท ก) ความเรวของลกตมเปนศนยทตำแหนง B ข) ความตงในเสนเชอกเปนศนยทตำแหนง C




4 IMPULSE AND MOMENTUM

โดยท เรยกวา

Note :

∑F = ma = m dv

dt

∑F dt

t1

t2

∫ = m dvv1

v2

∫

mv1 + ∑F dt

t1

t2

∫ = mv2

∑F dt

t1

t2

∫

mvx1 + ∑F dtt1

t2

∫ =mvx2

mvy1 + ∑F dtt1

t2

∫ =mvy2

∑F dt

t1

t2

∫


momentum of the particle over a finite period of time simply by inte-grating Eq. 3/25 with respect to the time t. Multiplying the equation bydt gives ΣF dt ! dG, which we integrate from time t1 to time t2 to obtain

(3/27)

Here the linear momentum at time t2 is G2 ! mv2 and the linear momen-tum at time t1 is G1 ! mv1. The product of force and time is defined asthe linear impulse of the force, and Eq. 3/27 states that the total linearimpulse on m equals the corresponding change in linear momentum of m.

Alternatively, we may write Eq. 3/27 as

(3/27a)

which says that the initial linear momentum of the body plus the linearimpulse applied to it equals its final linear momentum.

The impulse integral is a vector which, in general, may involvechanges in both magnitude and direction during the time interval.Under these conditions, it will be necessary to express ΣF and G in com-ponent form and then combine the integrated components. The compo-nents of Eq. 3/27a are the scalar equations

(3/27b)

These three scalar impulse-momentum equations are completelyindependent.

Whereas Eq. 3/27 clearly stresses that the external linear impulsecauses a change in the linear momentum, the order of the terms in Eqs.3/27a and 3/27b corresponds to the natural sequence of events. Whilethe form of Eq. 3/27 may be best for the experienced dynamicist, theform of Eqs. 3/27a and 3/27b is very effective for the beginner.

We now introduce the concept of the impulse-momentum diagram.Once the body to be analyzed has been clearly identified and isolated, weconstruct three drawings of the body as shown in Fig. 3/12. In the firstdrawing, we show the initial momentum mv1, or components thereof. In

m(v1)z " ! t2

t1

ΣFz dt ! m(v2)z

m(v1)y " ! t2

t1

ΣFy dt ! m(v2)y

m(v1)x " ! t2

t1

ΣFx dt ! m(v2)x

G1 " ! t2

t1

ΣF dt ! G2

! t2

t1

ΣF dt ! G2 # G1 ! $G


ΣF dtt1

t2

G1 = mv1

G2 = mv2

+ =

Figure 3/12

c03.qxd 2/9/12 7:39 PM Page 192



14. ตเกบเครองมอหนก 10 kg รบแรง F=(12+8t) N กระทำขนานกบพนเอยง เมอ t มหนวยเปนวนาท ในตอนตนตกำลงเคลอนทลงตามพนเอยง 20° ดวยความเรว 2 m/s จงหาเวลาซงแรงนทำใหตหยดเคลอนท

15. กลองมมวล 10 kg กำลงเคลอนทไปทางขวาดวยความเรว 0.6 m/s มแรง P มากระทำกบกลองดงกราฟ จงหาความเรวของกลองเมอเวลาผานไป 0.4 วนาท *

1.5 m/s


10 kg



CONSERVATION OF MOMENTUM #

Impact #

# สมประสทธการชน (Coefficient of Impact)

การชนแบบ 2 มต

∑mv1 = ∑mv2

mAvA +mBvB =mA ′vA +mB ′vB

e = ′vB − ′vA

vA −vB




16. รถลาก A มวล 80 Mg. วงเขาหารถพวง B มวล 50 Mg ดวยความเรว 0.65 m/s หลงปะทะ Coupling ไดตอตดกนเรยบรอยในเวลา 0.5 วนาท จงหา ก) ความเรว v2 ของรถ A กบ B ทตดไปดวยกน ข) แรงเฉลย F ท A ทำตอ B ขณะตอกน




17. [final 2/2554] ลกบอลกระทบพนดวยความเรว v0 = 24 m/s โดยทำมม 60° กบพนดงรป ถาคา coefficient of restitution e = 0.8 จงหาความเรว v และทศทาง θ ทลกบอลกระดอนออกจากพน




18. [final 1/2555] ลกบอลเหลกกระทบแผนเหลกทมมวลเทากน ดวยความเรว v0 = 24 m/s โดยทำมม 60˚ กบแนวราบ ถาแผนเหลกวางอยบนสปรงซงเคลอนไดในแนวดงเทานน ดงรปโดยมคา coefficient of restitution e = 0.8 จงหาความเรว v และทศทาง θ ของลกบอล ตลอดจนความเรวของแผนเหลก *

v1’ = 12.20 m/s, θ = -9.83˚, v2’ = 18.71 m/s down




ANGULAR IMPULSE AND MOMENTUM

โดยท r คอ

โดยท

Conservation of Angular Momentum

# #

HO = r ×mv

HO1+ ∑MO dt∫ =HO2

∑MO dt∫ =

∑HO1= ∑HO2

rAmAvA + rBmBvB = rAmA ′vA + rBmB ′vB


262 CH A P T E R 15 KI N E T I C S O F A PA RT I C L E : IM P U L S E A N D MO M E N T U M

15

15.5 Angular Momentum

The angular momentum of a particle about point O is defined as the“moment” of the particle’s linear momentum about O. Since this conceptis analogous to finding the moment of a force about a point, the angularmomentum, is sometimes referred to as the moment of momentum.

Scalar Formulation. If a particle moves along a curve lying in thex–y plane, Fig. 15–19, the angular momentum at any instant can bedetermined about point O (actually the z axis) by using a scalarformulation. The magnitude of is

(15–12)

Here d is the moment arm or perpendicular distance from O to the lineof action of mv. Common units for are or Thedirection of is defined by the right-hand rule. As shown, the curl ofthe fingers of the right hand indicates the sense of rotation of mv aboutO, so that in this case the thumb (or ) is directed perpendicular to thex–y plane along the axis.

Vector Formulation. If the particle moves along a space curve,Fig. 15–20, the vector cross product can be used to determine the angularmomentum about O. In this case

(15–13)

Here r denotes a position vector drawn from point O to the particle. Asshown in the figure, is perpendicular to the shaded plane containingr and mv.

In order to evaluate the cross product, r and mv should be expressed interms of their Cartesian components, so that the angular momentum canbe determined by evaluating the determinant:

(15–14)HO = 3 i j krx ry rz

mvx mvy mvz

3HO

HO = r * mv

+zHO

HO

slug # ft2>s.kg # m2>s1HO2z1HO2z = 1d21mv2

HO

HO ,

x

yOd

mv

z

HO

Fig. 15–19

x

y

z

O

mv

HO ! r " mv

r

Fig. 15–20

274 CH A P T E R 15 KI N E T I C S O F A PA RT I C L E : IM P U L S E A N D MO M E N T U M

15

•15–97. The two spheres each have a mass of 3 kg and areattached to the rod of negligible mass. If a torque

, where t is in seconds, is applied to therod as shown, determine the speed of each of the spheres in2 s, starting from rest.

15–98. The two spheres each have a mass of 3 kg and areattached to the rod of negligible mass. Determine the time thetorque , where t is in seconds, must beapplied to the rod so that each sphere attains a speed of 3 starting from rest.

m>sM = (8t) N # m

M = (6e0.2t) N # m

*15–96. The ball B has a mass of 10 kg and is attached tothe end of a rod whose mass can be neglected. If the shaft issubjected to a torque , where t is inseconds, determine the speed of the ball when . Theball has a speed when .t = 0v = 2 m>s t = 2 s

M = (2t2 + 4) N # m

15–95. The 3-lb ball located at A is released from rest andtravels down the curved path. If the ball exerts a normalforce of 5 lb on the path when it reaches point B, determinethe angular momentum of the ball about the center ofcurvature, point O. Hint: Neglect the size of the ball. Theradius of curvature at point B must first be determined.

15–99. An amusement park ride consists of a car which isattached to the cable OA. The car rotates in a horizontalcircular path and is brought to a speed when

. The cable is then pulled in at the constant rate of0.5 . Determine the speed of the car in 3 s.ft>sr = 12 ft

v1 = 4 ft>s

M

0.4 m

0.4 m

Probs. 15–97/98

A

O

r

Prob. 15–99

10 ft

A

B

O

r

Prob. 15–95

M0.5 m

vB

Prob. 15–96



19. ทรงกลมสองลกมวล 3 kg อยบนโครงเหลมไมคดมวล จงหาเวลาททำใหทรงกลมทงสองมความเรวจากหยดนงเปน 3 m/s เมอมโมเมนต M = (8t) N·m โดยท t มหนวยเปน วนาท *

20. [final 1/2549] เมอเรมตนแกน OA ซงมมวล m ตดไวทปลาย หมนรอบจด O ในแนวราบดวยความเรวเชงมม ω ตอมาแกน OA ทำมม 30° กบแนวระดบดงรป จงหา ก) หลงจากการตกลงมาแลวความเรวเชงมมในการหมนรอบจด O เปนกเทาของ ω ข) พลงงานจลนหลงจากทแกนตกลงมาแลวลดลงเปนกเทาของพลงงานจลนเรมตน

36. t = 1.34 s



!แบบประเมนความพงพอใจ

วชา ......................................................................................... เรยนวน ................................................................................................... ชอตวเตอร ....................................................................................................................................................................................................

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เปดโอกาสใหนกเรยนซกถามขอสงสย

เอกสาร

เนอหาครบถวน

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จำนวนแบบฝกหด

เอกสารอานเขาใจงาย

เวลาและสภานท

ความเหมาะสมของสถานทเรยน

ความเหมาะสมของระยะเวลาเรยน

ใหเกรดตวเตอร(A/B/C/D/F)

แบบสอบถามการเรยนพเศษของนกศกษาขอมลสวนบคคล

• เพศ [ ] ชาย [ ] หญง

• คณะ [ ] วศวกรรมศาสตร ภาควชา.................................... [ ] วทยาศาสตร ภาควชา.................................... [ ] ครศาสตรอตสาหกรรมและเทคโนโลย ภาควชา.................................... [ ] อนๆ ระบ................................................

• ชนป [ ] ป 1 [ ] ป 2 [ ] ป 3 [ ] ป 4 [ ] ป.โท-เอก

ตอนท 1 พฤตกรรมการเรยนพเศษ

• ขณะเปนนกศกษาเคยเรยนพเศษหรอไม

[ ] เคย [ ] ไมเคย (ขามไปทำตอน 2)• เรยนวชาอะไร (ตอบไดมากกวา 1 ขอ)

[ ] กลมวชาวศวกรรมศาสตร ระบ ...................................................... [ ] กลมวชาคณต-วทย ระบ ............................................................ [ ] อน ระบ ...................................................................

• ชอบเรยนพเศษรปแบบใด

[ ] เรยนตวตอตว [ ] เรยนกลมยอย (4-6 คน) [ ] เรยนกลมใหญ (8-10 คน) [ ] เรยนกลมใหญมาก (มากกวา10 คน)

• เรยนทไหน

[ ] สถาบนกวดวชา ชอ......................... ตวเตอรอสระ ชอ............................• จำนวนวชาตอเทอม

[ ] 1 วชา [ ] 2 วชา [ ] 3 วชา [ ] 4 วชา [ ] มากกวา 4 วชา

• คาเรยนพเศษทงหมดตอเทอม(midterm+final) [ ] ไมเกน 2,000.- [ ] 2,000-4,000.- [ ] 4,000-6,000.- [ ] มากกวา 6,000.

• วชาทอยากใหเปดสอนในเทอมหนา ..............................................................................

ตอนท 2 เหตผลททำใหตองเรยนพเศษ

• คณคดวา ทำไมตองมการเรยนเพมเตมขณะเปนนกศกษา

[ ] อยากเพมความมนใจ [ ] เรยนในหองไมรเรอง [ ] เพอนเรยนเลยเรยนดวย

• (เรยงลำดบจากมากไปนอย , 1 มากทสด) สงทคณพจารณาในการเลอกเรยนกบตวเตอร [__] ราคา [__] การเดนทาง [__] คณภาพตวเตอร [__] เพอนชวน

• รจกสถาบนกวดวชาจากอะไร

[ ] เหนสถานทเรยน [ ] ใบปลว-ปายประกาศ [ ] เพอนแนะนำ

[ ] Facebook [ ] อนๆ ระบ.....................................

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