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TECHNOLOGICAL INSTITUTE OF THE PHILIPPINES
1338 ARLEGUI ST. QUIAPO, MANILA
COLLEGE OF ENGINEERING
ASSIGNMENT NUMBER 1
Submitted by:
Cadoy, Joshua Peter E.
BSChE / MECH363 / ES32FA1
Date Submitted:
NOVEMBER 8, 2010
Submitted to
ENGR. JEFFREY CABIGTING
IX. Principles of Dynamics
9.1 GeneralDynamics
-is a branch of mechanics that considers the study of effect of forces to bodies that are in motion-the study of the effect of forces in motion to the bodies also in motion
9.2 Kinematics and KineticsKinematics
- branch is a branch of classical mechanics which is solely focused on the movement of various bodies, without addressing the forces which can influence movement which considers the relation between displacement, velocity and acceleration
Kinetics- branch of mechanics that explains how and why things move in the ways they do which considers the
relation between mass and acceleration
9.3 Motion of ParticleDisplacement s (m)- is the vector distance travelled by an object from an origin to the point being occupied on its path of travel. If the change in displacement ∆s is positive (to the right of the origin), the change in the path travelled is increased. If the change in displacement ∆s is negative (to the left of the origin), the change in the path travelled is decreased. If the change in displacement ∆s is zero, there is no change in position at all.
Velocity v (ms
)- is the time rate of change of displacement. That is vave=ΔsΔt
.
The instantaneous velocity will be found as ∆t approaches zero as a limit. That is v= lim∆t→0
Δ sΔt
=dsdt
.
Acceleration a (m
s2)- is the time rate of change in velocity. That is aave=
ΔvΔt
.
The instantaneous acceleration will be found as ∆t approaches zero as a limit. That is
a= lim∆t →0
Δ vΔt
=dvdt
=d ( ds
dt)
dt=d2 s
dt2.
If v=dsdt
and a=dvdt
; therefore vdv=ads
9.4 Newton’s Law of Motion of a ParticleLaw of Inertia
A particle acted upon a balanced force system has no acceleration.
Law of AccelerationA particle acted upon n unbalanced force system has an acceleration in line with and directly proportional to
the resultant of the force system.
Law of Action and ReactionAction and reaction forces between two particles are always equal and opposite in magnitude.
9.5 Fundamental Equation of Kinetics of Particles1st Assumption: If a body has weight W is acted upon different forces(say F1, F2, F3, …) has a resultant R. Applying the Law of Acceleration: R=ka2nd Assumption: If a body has a weight W is in the vacuum. Since R=W which is only influenced by the acceleration due to gravity. Applying the Law of Acceleration: W=kg
If R=ka and W=kg; therefore R=Wga. Since m(mass)=
Wg
; R=ma.
R=F x+F y+F z=max+ma y+maz
9.6 Effective Force on a Particle. Inertia ForceEffective force on a particle is the resultant force on the particle. That is equal to R or ma.
Inertia Force is an equal but oppositely directed reaction to the resultant force acting on a particle.
Dynamic Equilibrium is a state where the inertia force is considered to act on a particle together with the resultant force; the particle will be in a state of equilibrium.
9.7 D’Alembert’s PrincipleWhen different forces act on a system such that it is in motion with acceleration in a particular direction, the vector sum of all the forces acting on the system including the inertia force (‘ma’ taken in the opposite direction to the direction of the acceleration) is zero or will be in dynamic equilibrium.
9.8 Motion of Center of Gravity of Any Body1st Assumption: Consider the X component of each term, we have ΣX=m1a1x+m2a2x+m3a3x+…2nd Assumption: From statics, the X component of the center of gravity is given asm x=m1 x1+m2 x2+m3 x3+…If we get the 2nd derivative: max=m1a1x+m2a2x+m3a3x+…We can get max=ΣXIf applied to Y and Z axisR=ma
9.9 Applying the Principles of DynamicsStudents should study the principles of dynamics if they’re going to venture engineering of other allied physical science course. Those principles were useless if they won’t apply it to various problems that require it. Although dynamics is very essential, it also has its limitation of it is only applicable to rigid bodies. Knowing that there are no rigid bodies, some solids can resist deformation depending on the applied pressure, so the deformation is almost negligible. So application of dynamics is concentrated on translational (rectilinear and curvilinear), rotational, and plane motion of bodies.
TECHNOLOGICAL INSTITUTE OF THE PHILIPPINES
1338 ARLEGUI ST. QUIAPO, MANILA
COLLEGE OF ENGINEERING
ASSIGNMENT NUMBER 2
Submitted by:
Cadoy, Joshua Peter E.
BSChE / MECH363 / ES32FA1
Date Submitted:
NOVEMBER 8, 2010
Submitted to
ENGR. JEFFREY CABIGTING
X. Rectilinear Translation
10.1Definition and Characteristics of TranslationTranslation motion of a rigid body such that a line connecting any two points of the body is shifted parallel to itself. During translational motion, all points of the body describe identical trajectories, that is, trajectories coincident when superposed, and have at every instant velocities and accelerations that are the same in magnitude and direction. The translational motion of a body is therefore treated in much the same way as the kinematics of a particle. Translational motion is the movement of an object through space without rotation.Translation may either be Rectilinear or Curvilinear An object has a rectilinear motion when it moves along a straight line. Meaning all point on a rigid body move in a straight line.An object has a curvilinear motion when its translation is along a curve path.
10.2 Rectilinear Motion with Constant Acceleration1st Formula:
dv=adt ∫vo
v
dv=a∫0
t
dt
v=at+vo2nd Formula:
ds=vdt ds=(at+vo)dt ∫0
s
ds=∫0
t
(at+vo)dt
s=vo t+a t 2
23rd Formula:
vdv=ads ∫vo
v
vdv=a∫0
s
ds
v2−vo2
2=as
v2=vo2+2as
10.3 Freely Falling Bodies, Air Resistance NeglectedIn the absence of air resistance we find that all bodies, regardless of their size, weight or composition, fall with the same acceleration at the same point on the earth’s surface, and if the distance covered are not too great. The acceleration remains constant throughout the fall. This ideal motion, in which air resistance and the small change in acceleration with altitude is neglected, is called “free fall. The acceleration of a freely falling body is called acceleration due to gravity and is denoted by g (a=g). Near the earth’s surface, its magnitude is approximately:
g=9.8ms2
=980 cms2
=32 fts2
and is directed down toward the center of the earth. Formulas of freely falling body
is the same as the formulas of motion with constant acceleration but take note to change a to g. In some cases, there were problems where the initial direction of motion is not always downward, with that; it will tell whether the displacement, velocity, and acceleration are positive or negative.
10.4Rectilinear Motion with Variable AccelerationIn this case, the acceleration is varying. We must work with the given data then apply the principles of motion using differential equations since only s, v, a, and t are interrelated to each other.The 1st case is where displacement is given with respect to time i.e. s=f(t). In this case, v and a can be solved by successive differentiation.The 2nd case is where the acceleration is expressed in terms of time i.e. a=f(t). In this case, v and s can be solved using definite integration.The 3rd case is where the given is velocity in terms of time i.e. v=f(t). In this case, s and a can be solved by differentiation and definite integration.
The 4th case is where one of the principal variables is given in terms of the adjacent variable i.e. a=f(v) or v=f(s). One can solve this type by manipulating and simplifying different differential equations involving the parameters.The 5th case is where the given variables are not adjacent i.e. a=f(s). Same method is applied as case 4 in solving.
10.5 Motion CurvesIn the analyzation of the performance of machines in motion, motion curves are the best way to represent them.In the displacement-time curve, the displacement at any instant together with time is plotted. The slope of the said curve is the same as the velocity.In the velocity-time curve, the velocity at any instant together with time is plotted. The slope of the said curve is also the acceleration.Lastly, the acceleration-time curve is a curve where the acceleration at any instant with the corresponding time is plotted.In the reverse process, getting the area under the curve of acceleration-time and velocity-time using integration will give ∆v and ∆s respectively.
10.6 Kinetics of Rectilinear Translation. Analysis as a ParticleIn the lesson 9.8, we got the equation R=ma. Since in rectilinear, only one direction has a value. So if ma=ΣX , ΣY and ΣZ is zero.Here are some steps that can help in solving some problems involving this:a. Draw the free body diagram and indicate the given data with proper signs.b. Determine the direction of each motion and indicate it by dashed arrowc. Determine the kinematic relationship between the bodies involvedd. Use ΣY=0 and ma=ΣX to each bodye. Solve the unknowns using additional derived equations in relation with s, v, and t.
10.7 Dynamic Equilibrium in Translation. Analysis as a Rigid BodyIf we already know how to calculate the magnitude of the resultant of the translating body, here we will predict the direction or location of the resultant.Using the D’Alembert’s Principles where dynamic equilibrium is associated, we can predict where an accelerated body will reach its state of equilibrium.
Problems:1002.) On a certain sketch of track, trains run at 60 mph. How far back of a stopped train should a warning torpedo be placed to signal an oncoming train? Assume that the brakes are applied at once and retard the train at the uniform rate of 2ft per sec².
Given: vo=60mpha=−2 fts2
Req’d: sSol’n: V f
2−V o2=2as ; vo=60 x5280 /3600=88 ft /sec ;v f=0
0−(88 )2=−2 (2)sS = 1936 ft
1003.) A stone is thrown vertically upward and returns to earth in 10sec. What was its initial velocity and how high did it go?Given: t t=10 sReq’d: vo∧sSol’n: v=at+vo g=−9.8m /s
At the top; v f=0 and t=tt2=102
=5 s
vo=−¿vo=−(−9.8)5=49m / s
s=vo t+g t2
2
s=49(10)+−9.81(5)2
2s=122.5m
1004.) A ball is dropped from the top of a tower 80 ft. high at the same instant that a second ball is thrown upward from the ground with an initial velocity of 40 ft per sec. When and where do they pass and with what relative velocity?Given:
vos a
Dropped 0 s 32.2 ft/s2
Thrown up 40 ft/s 80-s -32.2 ft/s2
Req’d: t , s, vrel
Sol’n: s=vo t+g t2
2@dropped:
s= (0 ) t+ 32.2t2
2=16.1t 2
@thrown up:
80−s=(40 ) t+−32.2 t2
280−16.1t 2=(40 ) t−16.1t 2
t=2 s s=16.1(2)2=64.4 ft ¿ t h e top
V f2=2a s+V o
2 →v f=2√2as+V o
2
@dropped: v f=2√2(32.2)(64.4 )+0=64.4 ft / s
@thrown up: v f=2√2(−32.2)(64.4 )+402=2 4.4 ft /s
vrel=v f (dropped )−v f (thrownup )=64.4−24.4vrel=40 ft /s
1005.) A stone is dropped down a well and 5 sec later the sound of the splash is heard. If the velocity of sound is 1120 ft per sec, what is the depth of the well?Given:
vot a
Stone 0 t 32.2 ft/s2
sound 1120 ft/s 5-t
Req’d: sSol’n:
s=vo t+g t2
2
@stone: s= (0 ) t+ 32.2t2
2=16.1t 2
t=√ s16.1
@sound v= st 1120= s
5−t
1120= s
5−√ s16.1
s=353.31 ft
1006.) Repeat Prob. 1005 if the sound of the splash is heard after 4sec.Given:
vot a
Stone 0 t 32.2 ft/s2
sound 1120 ft/s 4-t
Req’d: sSol’n:
s=vo t+g t2
2
@stone: s= (0 ) t+ 32.2t2
2=16.1t 2
t=√ s16.1
@sound v= st 1120= s
5−t
1120= s
4−√ s16.1
s=231.65 ft
1007.) A stone is dropped from a captive balloon at an elevation of 1000ft. Two seconds later another stone is projected vertically upward from the ground with a velocity of 248 ft per sec. How far above the ground will the stones be at the same level?Given:
vot a s
Stone1 0 t 32.2 ft/s2 1000-sStone 2 248 ft/s t-2 -32.2 ft/s2 s
Req’d: s
Sol’n: s=vo t+g t2
2@stone1: 1000−s=0+½(32 .2)(t )²
@stone2:s=248(t−2)+−32.2( t−2)2
21000−16 t ²=248 t – 496−16 t ²+64 t−64t=5 s
s=248(5−2)+−32.2(5−2)2
2s=600 ft
1008.) A stone is thrown vertically upward from the ground with a velocity of 48.3 ft per sec. one second later another stone is thrown vertically upward with a velocity of 96.6 ft per sec. how far above the ground will the stones be at the same level?Given:
vot a
Stone1 48.3 ft/s t -32.2 ft/s2
Stone 2 96.6 ft/s t-1 -32.2 ft/s2
Req’d: s
Sol’n: s=vo t+g t2
2@stone1: s=48.3 t+½(−32 .2)( t) ²
@stone2:s=96.6(t−1)+−32.2(t−1)2
2s=128.8t – 16.1 t ²−112.748.3 t−16.1 t ²=128.8 t –16.1 t ²−112.7t=1.4 sec .
s=48.3 (1.4 )+½ (−32.2 ) (1.4 )2
s=36.06 ft
1009.) A ball is shot vertically into the air at a velocity of 193.2 ft per sec. After 4 sec, another ball is shot vertically into the air. What initial velocity must the second ball have in order to meet the first ball 386.4 ft from the ground?Given: s = 386.4 ft
vot a
Stone1 193.2 ft/s t -32.2 ft/s2
Stone 2 vot-4 -32.2 ft/s2
Req’d: vo
Sol’n: s=vo t+g t2
2@stone1: s=193.2t+½(−32.2)(t) ²386.4=193.2 t−16.1t ²t=9.46 s
@stone2:s=vo(t−4)+−32.2(t−4 )2
2
386.4=vo(9.46−4)+−32.2(9.46−4)2
2.
vo=158.69 ft /s
1010.) A stone is thrown vertically up from the ground with a velocity of 300ft per sec. How long must one wait before dropping a second stone from the top of a 600-ft tower if the two stones are to pass each other 200ft from the top of the tower?Given: s = 400 ft
vot a
Stone1 300 ft/s t -32.2 ft/s2
Stone 2 0 t-x 32.2 ft/s2
Req’d: x
Sol’n: s=vo t+g t2
2@stone1: 400=300 t+½(−32 .2)( t) ²t=17.19 s
@stone2:600−400=0+32.2(17.19−x )2
2200=16.1(17.19−x )2.x=13.67 s
1011. A ship being launched slides down the ways with a constant acceleration. She takes 8 sec to slide the first foot. How long will she take to slide down the ways if their length is 625 ft?Given: @t=8 s ; s=1 ftst=625 ftReq’d: t tSol’n: vo=0
s=vo t+g t2
2
1=0+ a82
2
a= 132
ft /s ²
@condition:
625=½( 132 ) (t 2 )
t t=200 s
1012. A train moving with constant acceleration travels 24ft during the 10th sec of its motion and 18ft during the 12th sec of its motion. Find the initial velocity and its constant acceleration.Given: @condition1: t=10 s ; s=24 ft@condition2 : t=12 s ; s=18 ftReq’d: a , vo
Sol’n:s=vo t+a t 2
2
@9th second:s1=9vo+a92
2
@10th second:s2=10 vo+a102
2∆ s1=s2−s1=24
(10 vo+a102
2 )−(9vo+a92
2)=24
24=vo−9.5a
@11th second:s3=11 vo+a112
2
@12th second:s4=12vo+a122
2∆ s2=s4−s3=18
(12vo+a122
2¿−(11 vo+
a112
2 )=1818=vo−11.5a
Combining the two equations, a=−3 ms2
∧vo=52.5 ft /s
1013. An automobile starting from rest speeds up to 40ft per sec with a constant acceleration of 4 ft per sec², runs at this speed for a time, and finally comes to rest with deceleration of 5 ft per sec². If the total distance traveled is 1000ft, find the total time required.
Given: @condition1: vo=0 ;a=4m
s2
@condition2 :v=40 fts
@condition3 :v f=0 ; a=−5 ms2;vo=40
fts
st=1000 ftReq’d: t
Sol’n: v f=at+vo ;s=vo t+a t 2
2 ; v f
2=vo2+2as
@condition1: 40=4 t+0t 1=10 s
s1=0+4(10)2
2=200 f t
@condition2 :v=s2/t 2s2=40 t2@condition3 :0=40−5t 3t 3=8 s
s3=40(8)+−5 (8 )2
2=160 ft
st=s1+s2+s3=200+40 t 2+160=1000
t 2=16 st=t 1+ t2+t 3=10+16+8t=34 s
1014. A train travels between two stations ½ mile a part in a minimum time of 41 sec. if the train accelerates and decelerates at 8 ft per sec², starting from rest at the first station and coming to a stop at the second station, what is its maximum speed in mph? How long does it travel at this top speed?Given: s=0.5mile t=41s@condition1: a=8 ft / s2
@condition2 :a=−8 ft / s2Req’d: v , tSol’n:Vf – Vo = αt₁ →Vf = 8t₁Vf = S²/t² →-Vf = -8tssз = Vftз – ½ (8)tз²s₁ + s₂ + sз = 0.3 milet1 = t3 : 2640 = 4t₁² + Vft₂ + Vft₁ - 4t₁²2640 = Vf (t1 + t2)T1 + t2 + t3 = 41T1+ t2 = 41 – t32640 = Vf (41 – t3)2640 = Vf (41 – Vf/8)2640 = 41 Vf -Vf²/8By Quadratic Formula you got Vf = 240 ft/s & Vf = 88 ft/sUse or get 88ft/s88ft/s x 3600/h x mile/5280ft = 60mph
1015. Two cars A and B have a velocity of 60mph in the same direction. A is 250ft behind B when brakes are applies to car B, causing it to decelerate at the constant rate of 10ft/s2. In what time will A overtake B, and how far each car traveled?Given
voS a
Car A 60mph=88ft/s s+250Car B 60mph=88ft/s s -10 ft/s2
Req’d: sa , sb , t
Sol’n: v= st
@car A: sa=s+250=88t
@car B:sb=s=vot+a t 2
2
s=88 t+−10 t2
2=88 t−5 t 2
Combining the eq’ns88 t−5 t2+250=88 tt=7.07 s
sa=88 (7.07 )=62.22 ft
sb=88 (7.07 )−5 (7.07)2=26.87 ft
1016. An automobile moving at a constant velocity of 45 ft per sec passes a gasoline station. Two seconds later, another automobile leaves the gasoline station and accelerates at the constant rate of 6ft per sec². How soon will the second automobile overtake the first?Given:
vot a
auto1 45 ft/s tauto2 0 t-2 6 ft/s2
Req’d: tSol’n:@auto1: v = s/t 45t = s@auto2:s = ½ α (t-2)² Combining the eq’ns45t = ½ (6)(t²-4t +4)15t = t² + 4t = 18.79 st = 18.79 – 2 = 16.79 s
1017. A balloon rises from the ground with a constant acceleration of 4ft/s2. Five seconds later, a stone is thrown vertically up from the launching site. What must be the minimum initial velocity of the stone for it to just touch the balloon? Note that the balloon and stone have same velocity at contact.Given:
vot a v f
Balloon 0 t 4 ft/s2 v f
stone vot-5 -32 ft/s2 v f
Req’d: vo
Sol’n:v2=v1+atv2=0+2 (5 )=10 ft /s
s=v1t+a t 2
2=0+
2(5)2
2=25 ft
s1=v2t 2+a t2
2
2=10 t2+
2t 22
2=10t 2+t 2
2
s2=v1t 2+g t 2
2
2=v1t 2+
−32.2 t22
2=v1 t 2−16.1t 2
2
s2=s+s1=25+s1=25+10 t2+t 22
v1t 2−16.1t 22=25+10 t 2+ t2
2
v f=¿ v2+a t2¿
v f=¿10+2 t2¿
v f=¿ v1+g t 2¿
v f=¿ v1−32.2 t 2¿
v1−32.2t 2=10+2 t 2v1=10+34.2 t225+20 t 2+ t2
2=v1t 2−16.1 t22
Substituting v125+20 t 2+ t2
2=(10+34.2 t2)t 2−16.1 t22
25=17.1t 22
t 2=1.21 s
v1=10+34.2 t2v1=10+34.2(1.21)v1=51.38 s
