the spool. Since
EvD=O D EvD EvO
and
OutD O{
, using the second of Eqs. (2) and the last of Eqs. (4), Eq. (5)
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permission of McGraw-Hill, is prohibited.
Dynamics 2e 1241
Problem 6.69
At the instant shown, the center
O
of a spool with inner and outer radii
rD3ft
and
RD7ft
, respectively, is moving down the incline at a
speed
vOD12:2 ft=s
. If the spool does not slip relative to the rope
and if the rope is fixed at one end, determine the velocity of point
C
(the point on the spool that is in contact with the incline), as well as
the rope’s unwinding rate, that is, the length of rope being unwound
per unit time.
of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the
permission of McGraw-Hill, is prohibited.
Observe that the definition in Eq. (6) allows
P
`
to be positive, equal to zero, or negative. Given the direction of
Out
, if
P
`>0
, then the rope is unwinding from the spool, whereas if
P
`<0
, the rope is being wound onto the
spool. Since
OutD O{
and recalling that
EvA=O D EvA EvO
, substituting Eq. (1) and the first of Eqs. (3) into
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permission of McGraw-Hill, is prohibited.
Dynamics 2e 1243
Problem 6.70
The bucket of a backhoe is the element
AB
of the four-bar linkage
system
ABCD
. Assume that the points
A
and
D
are fixed and
that, at the instant shown, point
B
is vertically aligned with point
A
, point
C
is horizontally aligned with point
B
, and point
B
is
moving to the right with a speed
vBD1:2 ft=s
. Determine the
velocity of point
C
at the instant shown, along with the angular
velocities of elements
BC
and
CD
. Let
hD0:66 ft
,
eD0:46 ft
,
`D0:9 ft, and wD1:0 ft.
of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the
permission of McGraw-Hill, is prohibited.
Problem 6.71
Bar
AB
is rotating counterclockwise with an angular velocity of
15 rad=s
.
Letting
LD1:25
m, determine the angular velocity of bar
CD
when
D
45ı.
Solution
Bar
AB
is rotating about the fixed point at
A
and we know its angular velocity, so we can write the velocity
of Bin the given component system as
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permission of McGraw-Hill, is prohibited.
Dynamics 2e 1245
Problem 6.72
At the instant shown, bars
AB
and
CD
are vertical and point
C
is moving to the left
with a speed of
35 ft=s
. Letting
LD1:5 ft
and
HD0:6 ft
, determine the velocity
of point B.
of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the
permission of McGraw-Hill, is prohibited.
Problem 6.73
Collars
A
and
B
are constrained to slide along the guides shown and are
connected by a bar with length
LD0:75
m. Letting
D45ı
, determine
the angular velocity of the bar
AB
at the instant shown if, at this instant,
vBD2:7 m=s.
Solution
Using the given component system, we see that
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permission of McGraw-Hill, is prohibited.
Dynamics 2e 1247
Problem 6.74
At the instant shown, an overhead garage door is being shut with point
B
moving to the left within the horizontal part of the door guide at a speed
of
5ft=s
, while point
A
is moving vertically downward. Determine the
angular velocity of the door and the velocity of the counterweight
C
at this
instant if LD6ft and dD1:5 ft.
Solution
Referring to the figure at the left, we determined the IC for the garage door
and can use the concept of instantaneous center of rotation to find the angular
speed of the door as
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permission of McGraw-Hill, is prohibited.
the solution of which is
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permission of McGraw-Hill, is prohibited.
Dynamics 2e 1249
Problem 6.75
In the four-bar linkage system shown, the lengths of the bars AB
and
CD
are
LAB D46 mm
and
LCD D25 mm
, respectively. In
addition, the distance between points
A
and
D
is
dAD D43 mm
.
The dimensions of the mechanism are such that when the angle
D132ı
, the angle
D69ı
. For
D132ı
and
P
D27 rad=s
,
determine the angular velocity of bars
BC
and
CD
, as well as
the velocity of the point
E
, the midpoint of bar
BC
. Note that
the figure is drawn to scale and that bars
BC
and
CD
are not
collinear.
27 rad=s
, and
D132ı
to obtain the numerical result. We now
relate the velocity of Cto points Band Dusing:
EvCD EvBC E!BC ErC =B D E!CD ErC =D;(3)
where
EvB
is given by Eq. (1) and we have used the fact that
D
is fixed. Using the cartesian frame shown
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permission of McGraw-Hill, is prohibited.