4/17/2012
EXAMPLES
1. The jet aircraft has a total mass of 22 Mg and a center of
mass at G. Initially at take-off the engines provide a thrust
of 2T = 4 kN and T ′ = 1.5 kN . Determine the acceleration of
the plane and the normal reactions on the nose wheel and
each of the two wing wheels located at B. Neglect the mass
of the wheels and, due to low velocity, neglect any lift
caused by the wings.
∑F
= m aGx
⇒ T ′ + 2T = m aGx
1,500 + 4,000 = 22,000 aGx ⇒ aGx = 0.25 m / s 2
1)
x
∑F
2)
= m aGy = 0
y
⇒ N A + 2 N B − 22,000 × 9.81 = 0
N A + 2 N B = 215,820
3) ∑ M G = 0 ⇒
− T ′(2.5 − 1.2) − 2T (2.3 − 1.2) − 2 N B × 3 + N A × 6 = 0
N A − N B = 1,058.33
Data:
m = 22 × 106 gr = 22 × 103 kg
2T = 4000 N
T ′ = 1500 N
aG = ?
NA = ?
NB = ?
N A = 72,646 N = 72.6 kN
1
2. The pipe has a length of 3 m and a mass of 500 kg . It is
attached to the back of the truck using a 0.6 m long chain
AB. If the coefficient of kinetic fiction at C is μ = 0.4,
determine the acceleration of the truck if the angle θ = 10D
with the road as shown.
Data:
LCA = 3m
LAB = 0.6m
m = 500kg
W = 4,905 N
μk = 0.4
θ = 10D
ax = aG = ?
ϕ c
G
a
•
3
b
SOLUTION
1) Geometry:
a
⇒ a = 0.52094 m
3
b
cos10D =
⇒ b = 2.9544 m
3
sin10D =
N C + 0.79843T = 4,905.0 (2)
10D
T sin 42.98D
G•
1.5sin10D
3) Moments:
∑M
C
= ∑ ( Mk
)
T sin 42.98D
W
C
G•
C•
T
52.98D − 10D
10D
T cos52.98D
1.5sin10D
maG
N B = 71,587 N = 71.6 kN
2
ϕ c
c = 1.0 − 0.52094 = 0.47906 m
sin ϕ =
b
0.47906
= 0.79843 ⇒ ϕ = 52.98D
0.6
2) Forces:
W
G•
Fμ
T
10D
T sin 42.98D
aG
G•
11.5sin10
5sin10D
NC
∑F
a
T
52.98D − 10D
10D
T cos52.98D
= maG ⇒ − μ N C + T cos52.98D = maG
x
−0.4 NC + 0.60209 T = 500 aG
∑F
y
=0
⇒
(1)
N C − W + T sin 52.98 = 0
4
3. The pipe has a mass of 460 kg and is held in place on the
truck bed using the two boards A and B. Determine the
greatest acceleration of the truck so that the pipe begins to
lose contact at A and the bed of the truck and starts to pivot
about B. Assume board B will not slip on the bed of the
truck, and the pipe is smooth. Also, what force does board
B exert on the pipe during the acceleration?
e = 1.5cos10D
Data:
−W (1.5cos10D ) + 3(T sin 42.98D ) = −maG (1.5sin10D )
2.0452 T − 72.457 = −130.23 aG
m = 460 kg
W = 4512.6 N
(3)
Solving the equations T = 3.39 kN , N C = 2.2 kN
aG = 2.33 m / s 2
5
SOLUTION
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1
4/17/2012
At the moment of interest
α = 0 and N = 0.
0.4
1) Equilibrium in the x - direction
∑F
= m ax ⇒ Px = m ax
x
d
P
Curvilinear Translation:
In a body subjected to a Curvilinear Translation all the points in
the body travel along parallel curvilinear paths.
Here it is convenient to write the equations of motion in normal
and tangential coordinates.
0.5
0.3
d = 0.1
∑F = ma
∑F = ma
∑M = 0
N=0
n
Gn
t
2) Equilibrium in the y - direction
∑F
y
= m a y ⇒ Py − W = 0 ⇒ Py = 4513 N
If the summation of moments about G is replaced by the
summation about another point B then we must account for the
kinetic moments
3) Equilibrium of moments about G
∑M
G
Gt
G
= 0 ⇒ Px × 0.4 − Py × 0.3 = 0 ⇒ Px = 3384 N
ax = 7.36 m / s 2
∑M
B
= ∑ ( Mk
)
B
EXAMPLES
7
1. The arm BDE of the industrial robot manufactured by Cincinnati
Milacron is activated by applying the torque of M = 50 N ⋅ m to
link CD. Determine the reactions at the pins B and D when the
links are in the position shown and have an angular velocity of
2 rad / s. The uniform arm BDE has a mass of 10 kg and a center
of mass at G1 . The container held in its grip at E has a mass of
12 kg and a center of mass at G2 . Neglect the mass of links AB
and CD
= e ⋅ (m aGt ) − h ⋅ (m aGn )
This is a curvilinear translation
aD = ω rD / C = 2 × 0.6
2
2
aD = 2.4 m / s 2 = aG
:
•
an
0.6
∑ M C = 0 ⇒ Dx × 0.6 − 50 = 0
Dx
M
:
Cy
Cx
2) In member BDE take ∑ M D = ∑ (Mk ) D , this eliminates Dy
Data:
mG1 = 10 kg
mG2 = 12 kg
M = 50 N ⋅ m
ω = 2 rad / s
By
Dy
aG W
1
aG
W2
W1 = 10 × 9.81 = 98.1 N
W2 = 12 × 9.81 = 117.72
− By × 0.22 − 98.1 × 0.365 − 117.72 × 1.1 = ( −10 × 2.4) × 0.365 − (12 × 2.4) × 1.1
By = −568 N
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10
2. The two 3 - lb rods EF and HI are fixed (welded) to the link
AC at E. Determine the normal force N E , the shear force VE ,
and moment M E , which the bar AC exerts on FE at E if at the
instant θ = 30D link AB has an angular velocity ω = 5 rad / s
and an angular acceleration α = 8 rad / s 2 as shown
3) Find Dy using equilibrium in the y-direction
By + Dy − W1 − W2 = − m1 aG1 − m2 aG2
Dy
aG W
1
Dy
1) Element CD: Rotation about fixed point C
Dx = 83.3 N
By
8
Data:
aG
W2
−567.54
567 54 + Dy − 98
98.11 − 117.72
117 72 = −24 − 28
28.8
8
Dy = 731 N
4) Find Bx using equilibrium in the x-direction. Because at
this instant ax = at = 0 we have
Bx + Dx = 0 ⇒ Bx = −83.3 N
11
WHI = WEF = 3 lb
mHI = mEF = 0.093168
0 093168 slug
ω = 5 rad / s
α = 8 rad / s 2
NE = ?
VE = ?
SOLUTION
ME = ?
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2
4/17/2012
5) We must have aG = a A and we can find a A :
The bars EFHI are undergoing a curvilinear translation, therefore
their angular velocity and acceleration must be zero.
1) Find the position of G:
Gy = 0
3 lb × (−1 ft ) + 3 lb × (−2 ft )
Gx =
= −1.5 ft
6 lb
G
•
2) Equilibrium in the x-direction:
a A = α × rA / B − ω 2rA / B rA / B = − 3cos30D i + 3sin 30D
a A = (8k ) × (−2.5981i + 1.5 j) − 52 × (−2.5981i + 1.5 j)
a A = 53 i − 58.3 j
∑ Fx = N E = 2 m aGx
VE M
E
3) Equilibrium in the y-direction:
∑F
y
= −2W − VE = 2 m aGy
N E = 9.87 lb
VE = −4.86 lb
M E = −7.29 lb ⋅ ft
NE
4) Equilibrium of moments about G:
∑M
G
= M E − VE × 1.5 = 0
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The moment equation can be replaced by a summation about
any point P lying inside or outside the body. In that case we
must take into account the moments ∑ ( Mk ) P due to I gα ,
m aGn and m aGt .
ROTATION ABOUT A FIXED AXIS
aG = α rG aGt + ω 2 rG aGn
∑ F = m a = mω r
∑ F = m a = mα r
∑M = I α
2
n
Gn
t
Gt
G
G
In many problems it is convenient to choose moments about O.
This eliminates the unknown rection FO , and the kinetic moments
become
∑ M O = ∑ (Mk )O = m rG aGt + IGα
G
G
Free Body Diagram
Note that the component m aGn does not appear, because its
line of action goes through the point O.
Kinetic Diagram
Furthermore, using aGn = rG α we have ∑ M O = rG2 m α + I Gα
=
or ∑ M O = ( rG2 m + I G )α and from the parallel axis theorem
15
EXAMPLES
SOLUTION
1) Moment of inertia:
IO =
I O = I G + m rG2 . Therefore we also have ∑ M O = I Oα
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2) Equilibrium of moments:
1. The 10 lb bar is pinned at its center O and connected to a
torsional spring. The spring has a stiffness k = 5 lb ⋅ ft / rad ,
so that the torque developed is M = (5θ ) lb ⋅ ft , where θ is
in radians. If the bar is released from rest when it is vertical
at θ = 90D , determine its angular velocity when θ = 45D.
There are no forces involved
so only the moments
equation applies:
∑ MO = IO α
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Data:
W = 10 lb
m = 0.31056 slugg
k = 5 lb ⋅ ft / rad
M = 5θ lb ⋅ ft
θ0 = π / 2
θ1 = π / 4
ω (π / 4) = ?
1
1
m A 2 = × 0.31056 × 22 = 0.10352 slug ⋅ ft 2
12
12
17
∑M
O
= −5θ = 0.10356α ⇒ α = −48.281θ
3) Kinematics:
−∫
π /4
π /2
α dθ = ω d ω
ω
48.281θ dθ = ∫ ω dω ⇒
0
π /4
1 2
ω = −24.141θ 2
2
π /2
ω = 9.45 rad / s
2. The lightweight turbine consists of a rotor which is powered
from a torque applied at its center.
at the instant the rotor is horizontal
it has an angular velocity 0f 15 rad/s and a
clockwise angular acceleration of 8 rad/s2 .
Determine the internal normal force, shear force
and moment at a section through A. Assume the
rotor is a 50 m long slender rod, having a mass
18
of 3 kg/m.
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4/17/2012
Data:
2) Equilibrium in the Tangential direction: r = 17.5 m
MA
15m
•A
ω = 15 rad / s
α = 8 rad / s 2
γ = 3 kg / m
SOLUTION
∑F
T
MA
7.5m
•A
NA
VA
•G
W
m aGN
1) Equilibrium in the Normal direction:
m aGN
•G
m aGT
W
VA
VA = 5858.6 N = 5.86 kN
3) Find the Moment of Inertia:
Kinetic Diagram
m = 45 kg r = 17.5 m
W = 441.45 N
•A
15m
IG α
•G
= m aGT ⇒ VA + W = m α rG / O ⇒ VA + 441.45 = 45 × 8 × 17.5
Consider the segment of the blade to the right of A.
Free Body Diagram
•A
NA
NA = ?
VA = ?
MA = ?
7.5m
1
1
m A 2 = × 45 × 152 = 843.75 kg ⋅ m 2
12
12
4) Equilibrium of Moments about A: Recall that aGT = α rG / O
IG α
IG =
•G
m aGT
∑M
∑ FN = m aGN ⇒ N A = m ω 2 rG / O = 45 × 152 ×17.5
N A = 177188 N = 177.2 kN
A
= ∑ ( Mk
)
A
⇒ M A + W rG / A = m(α rG / O )rG / A + I Gα
M A + 441.45 × 7.5 = 45 × (8 × 17.5) × 7.5 + 843.75 × 8
M A = 50689 N ⋅ m = 50.7 kN ⋅ m
19
3. The cord is wrapped around the inner core of the spool. If
a 5 lb block B is suspended from the cord and released from
rest, determine the spool's angular velocity when t = 3 s.
Neglect the mass of the cord. The spool has a weight of 180 lb
and the radius of gyration about the axle A is k A = 1.25 ft . Solve
the problem in two ways, first by considering the "system"
consisting of the block and spool, and then by considering the
block and spool separately.
20
1) Consider the whole system:
= ∑ ( Mk ) A
⇓
WB rB / A = mB rC2/ Aα + I Aα
∑M
A
5 × 1.5 = 0.15528 × 1.52 α + 8.7345α
I Aα
A• an C
at
A•
C
1.5m
at = aB
5lb
5lb
mB aB
α = 0.82564 rad / s 2
SOLUTION
First calculate the Moment
of Inertia about A:
Data:
I A = mA k A2 = 5.5901 × 1.252
I A = 8.7345 slug ⋅ ft 2
WA = 180 lb
WB = 5 lb
mA = 5.5901 slug
mB = 0.15528 slug
k A = 1.25 ft
ω (3s) = ? 21
ω = ω0 + α t = 0.82564 × 3 ⇒ ω = 2.48 rad / s
2) Consider each particle separate.
a) Spool
∑M
A
= I Aα
T × 1.5 = 8.7345α ⇒ T = 5.823α
Ay
Ax
•
1.5m
T
22
b) Weight B:
∑F
y
= mB aBy ⇒ 5 − T = 0.15528 × (1.5 × α )
T
T = 5 − 0.23292 × α
W
Solving the system:
α = 0.826
0 826 radd / s 2
Same as before
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