Comprehensive Solution for EAS 208 Homework Assignment #3 (Fall 2019)

This document presents a comprehensive solution to Homework Assignment #3 from the EAS 208 course, covering various mechanics problems. The solution includes detailed step-by-step calculations for problems involving ball velocity at exit, time to exit, acceleration of blocks connected by ...Read More

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Solution
a) Velocity at exit (V)
Calculating the force exerted by the spring from Hooke’s law
F=Kx
K=5lb/ ft∧x=10∈¿ 10
12 ft
F= 5∗10
12 =4.167 lb
Calculating the acceleration of the ball
From the equation
F−mgsin 30=ma
4.167−0.33 sin 30=( 0.33
32.1 )a
a= 4.002∗32.1
0.33 =389.285 ft /s2
The component of ball weight normal to the tube does not work
The equation motion

v2=u2 +2 a x
Where u=0, a=389.285 ft / s2 and x= 16
12 ft
Substituting the values
v2=(0)2 + 2∗389.285∗16
12 =1038.093 ( ft
s )
2
v= √1038.093=32.21 ft / s
The velocity at exit v= 32.21 ft / s
b) Time required to exit (t)
From the equation of motion
v=u+at
Making t the subject
t= v−u
a =32.21−0
389.285 =0.083 seconds
Time required t=0.083 seconds
Solution
Total length of the cable connecting block A and B
X A −2 Y B =0
Taking double differentiation to get the acceleration
V A −2V B=0
a A−2 aB=0
a A=2 aB∧aB =1
2 aA
FBD and kinetic diagram of Block A

At equilibrium of the inclined planes
T −W A sin 20−F=M A aB
T −200 sin 20−0.16∗200 cos 20= 200
32.2∗2 aB
T −38.334=12.42 aB Equation1
FBD and kinetic diagram of Block B
Equilibrium of inclined planes
W B sin7 0−2 T −F=MA aB
300 sin 70−2 T −0.25∗300 cos 70= 300
32.2 aB
128.12−T =4.658 aB Equation 2
Solving both equation 1 and 2
128.12−4.658 aB =38.334+12.42 aB
( 128.12−38.334 )=¿
aB=5.257 ft /s2
Acceleration of Block A
a A=2 aB
2∗5.257 ft
s2
a A=10.514 /s2
Part B
From equation 1

T −38.334=12.42 aB
T =38.334+12.42 aB
Where aB=5.257 ft /s2
¿ 38.334+(12. 42∗5.257 ft /s2)
T =103.626 lb
Solution
Sketch
First section
Applying the Newton second law and considering the horizontal forces and vertical forces
∑ F x=Ma

T AC sin 30+T BC sin 45=m ( v2
ρ ) Equation 1
∑ FY =0
T AC cos 30+T BC cos 45=m g Equation 2
Where m=mass , ρ=distance ¿ the centre ,T ¿AC∧T B C are te nsionon the wires AC∧BC
Dividing equation 1 by equation 2
T AC sin 30+T BC sin 45
T AC cos 30+T BC cos 45 =
m v2
ρ
mg
T AC sin 30+T BC sin 45
T AC cos 30+T BC cos 45 = v2
ρ g Equation3
Case1:
When wire AC is slack tension T AC=0 and Equation 3 reduce to
T BC sin 45
T BC cos 45 = v2
ρ g Equation4
tan 45= v2
ρ g
v= √ ρ gtan 45
Where ρ=1.6 m∧g=9.81m/s
v= √ 1.6∗9.81∗tan 45= 3.96 m
s
Case2:
When wire BC is slack tension T B C=0 and Equation 3 reduce to
T AC sin 30
T AC cos 30 = v2
ρ g Equation5
tan 30= v2
ρ g
v= √ ρ gtan30
Where ρ=1.6 m∧g=9.81m/s
v= √1.6∗9.81∗tan30=3.01 m/s
The range of speed when both AC and BC will be taut is 3.01 m/s ≤ v ≥3.96 m/ s
Second section
Subtracting equation 1 by equation 2
T AC cos 30+T BC cos 45−T AC sin 30−T BC sin 45=m g−m ( v2
ρ )
Sin45=cos45; substituting cos45 with sin45
T AC cos 30+T BC sin 45−T AC sin 30−T BC sin 45=m g−m ( v2
ρ )

SimplifyingT AC (cos 30−sin30)=m g−m ( v2
ρ )Equation 7
Multiplying Equation 1 by cos30
T AC sin 30 cos 30+T BC sin 45 cos 30=m ( v2
ρ ) cos 30 Equation8
Multiplying Equation 2 by sin30
T AC cos 30 sin 30+T BC cos 45 sin 30=m g sin 30 Equation9
Subtracting Equation 9 from Equation 8
T AC sin 30 cos 30+T BC sin 45 cos 30− ( T AC cos 30 sin 30+T BC cos 45 sin 30 ) =m ( v2
ρ ) cos 30−m g sin 30 Equation 10
T BC ( sin 45 cos 30−cos 45 sin 30 ) =m v2
ρ c o s 30−m g sin 30
Simplifying equation 10 further
T BC ¿
Tension in both wire AC and BC should not exceed 60N
Case1:
When T A C=60 N
Substituting the values of T A C=60 N , m=5 kg , ρ=1.6 m∧g= 9.81 m
s into Equation 7
60 ( cos 30−sin 30 )=5∗9.81−5 ( v2
1.6 )
21.96=49.05−3.125 v2
3.125 v2=27.09
v= √ 27.09
3.125 =2.94 m/s
When T A C ≤60 N the value of speed is 2.94 m/s
Case2:
When T B C=60 N
Substituting the values of T A C=60 N , m=5 kg , ρ=1.6 m∧g= 9.81 m
s into Equation 11
T BC ¿
60 ¿
15.53=2.71 v2−24.52
v= √ 15.53+24.52
2.71 = √ 40.05
2.71 =3.84 m/ s
When T B C ≤60 N the value of speed is 3.8 4 m/ s
The range of speed for second section is 2.94 m/ s ≤ v ≥ 3.84 m/ s
Combining the results from the first section and second section thus the range of speed that
satisfy all conditions are:
3.01 m/s ≤ v ≥3.84 m/s

Solution
Data
Speed of the car 72km/hr (72 km/hr∗10 m/ s)/36 k m/hr=20 m/s
Coefficient of friction μk =0.2
ρA =30 m∧ρB=40 m
Part A
Sketch
Summation of normal forces
+↑∑ FN =M aN
F−W =m v2
ρA
But W=mg
F=m ( g+ v2
ρA )
Frictional forceFr
Fr=μk∗F

¿ 0.2 m (g+ v2
ρA )
Considering
+→∑ FT =M aT
−Fr=M aT
μk M (g+ v2
ρA )=M aT
aT=μk ( g+ v2
ρA ) =−0.2 ( 9.81+ 202
30 ) =−4.63 m/s2
The deceleration before point A is 4.63 m/ s2
Part B
Summation of forces
+↑∑ FN =0
F−W =0
F=mg
Frictional forceFr
Fr=μk∗F
Considering
+→∑ Fx=M ax
−Fr=M ax
−0.2 mg=M ax
ax=−0.2 g
¿ ( −0.2∗9.81 ) =−1.96 m/ s2
The deceleration between point A and B is1.96 m/s2
Part C
Sketch
Summation of normal forces
+↓∑ FN =M aN
−F+ W= m v2
ρA
But W=mg

F=m (g− v2
ρB )
Frictional force Fr
Fr=μk∗F
¿ 0.2 m ( g− v2
ρA )
Considering
+→∑ FT =M aT
−Fr=M aT
−μk M ( g− v2
ρA )=M aT
aT=−μk (g− v2
ρA )=−0.2 (9.81− 202
30 )=−0.184 m/s2
The deceleration after point B is 0.184 m/ s2
Solution

Calculating the equilibrium forces in the radial Fr and Tangential F θ direction
∑ Fr :mar=Ncosθ−mgsinθ Equation1
∑ Fθ :m aθ=F +Nsinθ−mgcosθ Equation 2Where m=mass of the ball
N=Normal reaction on the ball
g= Gravitational acceleration
F=Normal force acting on arm OA
ar=Radial acceleratio n
aθ=Tangential acceleratio nEvaluating ar∧aθ
ar= ¨r −r θ2 Equation 3
aθ=r ˙θ−2 ˙r ˙θ Equation 4Where r=radial position related to the angle θ
˙r =radial velocity
¨r =radial acceleration
˙θ=Angular acceleration(ω)
Evaluating the value of r , ˙r∧ ¨r ∈terms of radial∧angular dislacement(John J. Uicker, 2017)
r =2 Rcos θ Equation5
˙r = dr
dt =−2 R ˙θ sin θ Equation6
¨r = d2 r
d t2 = d
dt ˙r=−2 R( ¨θ sinθ + ˙θ2 cosθ)
But angular velocity is constant thus ¨θ=0
¨r =−2 R ˙θ2 cosθ Equation 7
Substituting equation 3 into equation 1

m ( ¨r−r θ2 ) =Ncosθ−mgsinθ Equation8
Substituting the Equations 5 and 7 into equation 8
m (−2 R ˙ω2 cosθ−2 R ω2 cos θ )=Ncosθ−mgsinθ Equation 9
When the ball is at the exiting the contact Normal reaction (N=0) is zero. Thus equation 9
reduces to:
m ( −2 R ˙ω2 cosθ−2 R ω2 cos θ )=−mgsi nθ
gsinθ=4 R ˙ω2 cosθ
θ=tan−¿ ( 4 R ˙ω2
g )Equation 10¿
The values of R=1m, w=0.87rad/s and g=9.81m/s2
θ=tan−¿ ( 4∗1∗ ˙0.872
9.81 )=17.1515¿
θ=17.150
Angle at which theball leaves the circular surface ∅ =2 θ
∅ = ( 2∗17.15 ) =34.30
Solution
Data
B=3kg slide
along AA’
B constant
Outward speed
˙r =0.5 m/s
D=rotates
about point O
at speed=
¨
θ=0.75 t ( rad
s )
Kinematics

collar speed
dr
dt =˙r =0.5 m/s
Expressing the distance r as a function of time
At initial values of r =0∧t=0
∫
0
r
dr=∫
0
r
0.5 dt=0.5 t (m)
r =0.5 t( m)
collar Accelarion
d2 r
d t2 = ˙dr
dt = ¨r =0
Drum speed∧ Accelarion
Speed= ˙θ=0.75 t rad /s
Accelarion= ¨θ=0.75 rad / s2
Calculating the Accelarion during sliding∧rotation
S liding ar= ¨r −r ¨θ2
ar=0− ( 0.5 t )∗( 0.75 t )2=−9
32 t3 m/ s2
R otation aθ=r ¨θ+2 ˙r ¨θ
aθ= ( 0.5 t ) ( 0.75 ) +2 ( 0.5 ) ( 0.75 t ) = 9
8 t m
s
Kinetics
Summation vertical forces when the collar is sliding Fr and horizontal forces when the drum is
rotating Fθ (Beer, Jr, & Mazurek, 2015)

collar sliding Fr ( Normal force)
+↑∑ Fr =m ar
−Fr= ( 3 kg )∗
( −9
32 t3 m
s2 ) =−27
32 t3 N
Fr= 27
32 t3 N
Drum rotating Fθ ( Tensional force)
+←∑ Fθ =m aθ
Fθ= ( 3 kg )∗
( 9
8 t m
s2 )= 27
8 t N
Fθ= 27
8 t N
Equating both forces
Fr=Fθ
27
32 t3= 27
8 t

t2=( 27
8 ∗32
27 )=4 s2
t= √ 4 s2=2 seconds
Reference
Beer, F. P., Jr, E. R., & Mazurek, D. F. (2015). Vector Mechanics for Engineers: Statics. McGraw-Hill
Education.
John J. Uicker, J. G. (2017). Theory of Machines and Mechanisms. Oxford University Press.