marlon f. sacedon dept. of mathematics, physics, statistics (dmps) college of arts and sciences,...
TRANSCRIPT
Marlon F. SacedonDept. of Mathematics, Physics, Statistics (DMPS)
College of Arts and Sciences, Visayas State UniversityVisca, Baybay City, Leyte, Philippines
GENERALIZED MATHEMATICAL FORMULA FOR DYNAMICS ON ONE-
DIMENSIONAL MOTION OF TWO-OBJECT SYSTEM
Formerly ViSCA
Baybay City, Leyte
What is the most important part in the solutions of solving problems in dynamics of motion? Free-body diagram or FBD
A vector diagram showing all external forces acting on a body
Baybay City, Leyte
What is the acceleration of the ball if air friction is negligible?
w = mg
ΣFy = ΣFΣF a
mg = ma
FBD
a = g = 9.8 m/s2
from FBD from Second Law
+y
Baybay City, Leyte
θ
μk L
What is the acceleration of the ball if air friction is negligible?
ΣFx = ΣF
+mgsinθ -f = ma
a = g(sinθ-μkcosθ)
from FBD from Second Law
mg
+y
+x
+N-f
+mgsinθ
-mgcosθ
+mgsinθ -μkN= ma+mgsinθ –μkmgcosθ = ma
aΣF
Baybay City, Leyte
What about this?
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mA=
mB=
+x
+y
+mAg
-T
+y
+x
+mBg
+mBgcosø+T
-mBg.sinθ
a ΣFA
aΣFB
-μmBg.cosθ
What about this?
Shows complicated FBD!
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Student’s response on FBD.
Wrong!Wrong!
Wrong!
Wrong!
Wrong!Correct!
Misconceptions on Free-Body diagram
Problem given to students…..
Baybay City, Leyte
This paper presents a generalized mathematical formula that can be used in solving problems of dynamics on one-dimensional motion of two-object system which was derived from the principles of Newton’s Laws of Motion.
The formula helps the difficulty of students in solving the problems because the solutions process escapes FBD.
Baybay City, Leyte
BA
AAB
mm
m
cossincossinBmga
Generalized formula for dynamics on two-object system
θ β
μAmA
mB
μBmBmA
Atwood-machine
mB
mA
β
mA
mB
mB
mA
mB
mA
mB
mA
mBmAmBmA
mB
mA
mB
mA
Baybay City, Leyte
BA
AAB
mm
m
cossincossinBmga
Generalized formula for dynamics on two-object system
θ β
μAmA
mB
μB
Assumptions:• maximum of one pulley and it’s a frictionless.• Air friction is negligible.• motion is due to gravity only.• each object moves on a straight line.• object B accelerates down the plane. If calculated a
is negative, then it moves up the plane.• Neglect the effects of rotation of pulley & masses.• Negligible weight of cord, and no elongations.
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Baybay City, Leyte
BA
AAB
mm
m
cossincossinBmga
Generalized formula for dynamics on two-object system
θ β
μAmA
mB
μB
Assumptions:• maximum of one pulley and it’s a frictionless.• Air friction is negligible.• motion is due to gravity only.• each object moves on a straight line.• object B accelerates down the plane. If calculated a
is negative, then it moves up the plane.• Neglect the effects on rotation of pulley & masses.• Negligible weight of cord, and no elongations.
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Baybay City, Leyte
θ β
μA
mA
μB
+y
+x
mBg
N -T
mBg.sinβ
aΣFB
-μBmBg.cosβ
+y
+x
+mAg
-μAmAgcosθ
+T
-mAg.sinθ
aΣFB N
-mBg.cosβ-mAg.cosθ
Derivation of Formula
mB
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+y
+x
mBg
N -T
mBg.sinβ
aΣFB
-μBmBg.cosβ
-mBg.cosβ
+y
+x
+mAg
-μAmAgcosθ
+T
-mAg.sinθ
aΣFA N
-mAg.cosθ
FBD of mA FBD of mB
ΣFx =
+T–mAgsinθ-μAmAgcosθ=mAa (eq.1)
ΣFx =
-T+mBgsinβ-μBmBgcosβ=mBa (eq.2)
Adding equations 1& 2
+mBgsinβ-μBmBgcosβ –mAgsinθ-μAmAgcosθ =mBa +mAa g[mB(sinβ-μBcosβ –mA(sinθ+μAcosθ )]=a(mB +mA)
BA
AAB
mm
m
cossincossinmg
a B
mAa mBaƩFy =0 ƩFy =0
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Common problems on Two-Object system
mBmA
Atwood-machine
mB
mA
mA
mB
BA
AAB
mm
m
cossincossinBmga
mB
mA
mB
mA
OR
mBmAmBmA
OR
Next
mB
mAmB
mA
OR
Assump
Click buttonsto solve
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BA
AAB
mm
m
cossincossinBmga
mBmA
Atwood-machine
θ = 90o β= 90o
μA = 0
Generalized Formula:
μB = 0
0 01 1
BA
A
mm
m
BmgaBack to menu
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mBmA
Atwood-machine
BA
A
mm
m
Bmga If mB >mA
If mB <mA
,then a>0
,then a<0
If mA = 0 ,then a=g
If mB = mA ,then a=0
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mB
mA
β
BA
AAB
mm
m
cossincossinBmga
θ = 90o
μA = 0
Generalized Formula:
μB
01
BA
AB
mm
m
cossinBmga
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BA
AAB
mm
m
cossincossinBmga
θ = 0o μB = 0
Generalized Formula:
0 0
mA
mB
μA
β= 90o
1
BA
AA
mm
m
Bmga
( 1 )
Recall: sin(-A) = -sinA cos(-A) = cosA
Baybay City, Leyte
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BA
AAB
mm
m
cossincossinBmga
θ = 0o
μB
Generalized Formula:
μA
β= 0o
0 0
BA
AAB
mm
m
Bm-ga
BA
AAB
mm
m
Bmg-a
mBmAmBmA
OR
Recall: sin(-A) = -sinA cos(-A) = cosA
1 1
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mBmA+y
+x
N
-R
-μAmAg
-mAg
FBD of mA
+y
+x
N
+R-μBmBg
-mBg
FBD of mB
ΣFx =
-R-μAmAg=mAa (eq.1)
ΣFx =+R-μBmBg=mBa (eq.2)
Adding equations 1& 2
-μBmBg -μAmAg =mBa +mAa
BA
AA
mm
m
BBmg-
a
mAa mBa
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mB
mA
BA
AAB
mm
m
cossincossinBmga
μB
Generalized Formula:
μA
β
BA
AAB
mm
m
cossincossinBmga
BA
AAB
mm
m
cossincossinBmga
-θ
BA
AAB
mm
m
cos)(sincossinmg
a B(- θ)
(- θ)
Recall: sin(-A) = -sinA cos(-A) = cosA
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mB
mA
mB
mA
OR
BA
AAB
mm
m
cossincossinBmga
Generalized Formula:
μB
μA -θ
BA
AAB
mm
m
cossincossinBmga(- θ)
(- θ)
BA
AAB
mm
m
cossincossinBmga
BA
AAB
mm
m
cossincossinBmga
β
BA
AAB
mm
m
cossincossinmg
a B
Recall: sin(-A) = -sinA cos(-A) = cosA
But : θ=β
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mB
mA
μB
μA
+y
+x
N
-R
mAg.sinβ
aΣFA
-μAmAg.cosβ
-mAg.cosβ
FBD of mA
+y
+x
N
mBg.sinβ
aΣFB
-μBmBg.cosβ
-mBg.cosβ+R
FBD of mB
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+y
+x
N
-R
mAg.sinβ
aΣFA
-μAmAg.cosβ
-mAg.cosβ
FBD of mA
+y
+x
N
mBg.sinβ
aΣFB
-μBmBg.cosβ
-mBg.cosβ+R
FBD of mB
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+y
+x
N
-R
mAg.sinβ
aΣFA
-μAmAg.cosβ
-mAg.cosβ
FBD of mA +y
+x
N
mBg.sinβ
aΣFB
-μBmBg.cosβ
-mBg.cosβR
FBD of mB
ΣFx =
-R+mAgsinβ-μAmAgcosβ=mAa (eq.1)
ΣFx =
+R+mBgsinβ-μBmBgcosβ=mBa (eq.2)
Adding equations 1& 2
+mBgsinβ-μBmBgcosβ +mAgsinβ-μAmAgcosβ =mBa +mAa
BA
AAB
mm
m
cossincossinmg
a B
mAa mBa
BA
AAB
mm
m
cossincossinmg
a B
But : θ=β
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mB
mA
mB
mA
mB
mA
Problems that cannot solve by the formula
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• The mathematical formula can solve many problems in dynamics on one-dimensional motion of two-object system.
• Solutions to the problems becomes shorter and simple.
• Escapes totally the constructions of FBD.
Conclusion
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