PROBLEM 12.23
1m
A
20°
To unload a bound stack of plywood from a truck, the driver first tilts the
bed of the truck and then accelerates from rest. Knowing that the
coefficients of friction between the bottom sheet of plywood and the bed
are µ s = 0.40 and µ k = 0.30, determine (a) the smallest acceleration of
the truck which will cause the stack of plywood to slide, (b) the
acceleration of the truck which causes corner A of the stack of plywood to
reach the end of the bed in 0.4 s.
SOLUTION
Let a p be the acceleration of the plywood, aT be the acceleration of the
truck, and a p / T be the acceleration of the plywood relative to the truck.
(a) Find the value of aT so that the relative motion of the plywood with
respect to the truck is impending. a p = aT and
F1 = µ s N1 = 0.40 N1
+
ΣFy = m p a y : N1 − W p cos 20° = −m p aT sin 20°
N1 = m p ( g cos 20° − aT sin 20° )
+
ΣFx = max : F1 − W p sin 20° = m p aT cos 20°
F1 = m p ( g sin 20° + aT cos 20° )
m p ( g sin 20° + aT cos 20° ) = 0.40 m p ( g cos 20° − aT sin 20° )
aT =
( 0.40cos 20° − sin 20° ) g
cos 20° + 0.40sin 20°
= ( 0.03145 )( 9.81) = 0.309
aT = 0.309 m/s 2
(
(b) x p / T = x p / T
ap /T =
2x p / T
t
2
)o + ( v p / T ) t + 12 a p / T t 2
=
( 2 )(1)
( 0.4 )2
= 0+0+
= 12.5 m/s 2 ,
+
1
ap /Tt2
2
a p / T = 12.5 m/s 2
a p = aT + a p / T = ( aT → ) + (12.5 m/s 2
Fy = m p a y :
N 2 − W p cos 20° = −m p aT sin 20°
N 2 = m p ( g cos 20° − aT sin 20° )
W
20° )
20°
PROBLEM 12.39
A
C
O
A 1 kg collar C slides without friction along the rod OA and is attached
to rod BC by a frictionless pin. The rods rotate in the horizontal plane. At
the instant shown BC is rotating counterclockwise and the speed of C is
1 m/s, increasing at a rate of 1.2 m/s 2 . Determine at this instant, (a) the
tension in rod BC, (b) the force exerted by the collar on rod OA.
0.6 m
30°
0.3 m
B
SOLUTION
Geometry
A OC 2 = A OB 2 + A BC 2 − 2A OBA OC cos30°
= (0.3)2 + (0.6)2 – (2) (0.3) (0.6) cos 30° = 0.1382 m2
A OC = 0.372 m
A OC
sin 30°
sin β- =
=
A OB
sin β
l OB sin 30° (0.3) sin 30°
= 0.4035
=
l OC
0.372
β = 23.79°
Acceleration components: at = 1.2 m/s2
an =
v2
ρ
=
vC2
A BC
=
( 1)2
(0.6)
= 1.6667 m/s2
Mass m = 1 kg
+
ΣFt = mat : N cos β = (1) (1.2) = 1.2
N =
+
1.2
= 1.31 N
cos 23.79°
ΣFn = man : T − N sin β = (1) (1.6667) = 1.6667
(a) T = 1.6667 + 1.31 sin 23.79 °
(b) Force exerted by rod on collar is 2.2 N
Force exerted by collar on rod:
T = 2.2 N
( 30° + β )
2.2 N
W
= 53.8°
53.8° W
PROBLEM 12.60
t
vB
B
E
A
F
C
C
0.2 m
0.2 m
0.2 m 0.4 m
0.4 m 0.2 m
D
The parallel-link mechanism ABCD is used to transport a component I
between manufacturing processes at stations E, F, and G by picking it up
at a station when θ = 0 and depositing it at the next station when
θ = 180o. Knowing that member BC remains horizontal throughout its
motion and that links AB and CD rotate at a constant rate in a vertical
plane in such a way that vB = 0.7 m/s, determine (a) the minimum value
of the coefficient of static friction between the component and BC if the
component is not to slide on BC while being transferred, (b) the values of
θ for which sliding is impending.
SOLUTION
For constant speed, at = 0
an =
vB 2
ρ
with vB = 0.7 m/s, ρ = 0.2 m
+ ΣF = ma : F = ma cosθ
x
x
n
ΣFy = ma y : N − W = −man sin θ
Ratio
With
N = mg − man sin θ
F
man cosθ
cosθ
cosθ
=
=
=
g
g
ρ
N
mg − man sin θ
− sin θ
− sin θ
an
vB 2
g ρ ( 9.81)( 0.2 )
F
cosθ
=
= 4.0041, the ratio becomes
=
2
2
N 4.0041 − sin θ
vB
( 0.7 )
For no impending slide, µ S ≥
F
cosθ
=
N
4.0041 − sin θ
To find the value of θ for which the ratio is maximum set the derivative
with respect to θ equal to zero.
± cosθ
1 − 4.0041 sin θ
d 

= 0

 = ±
dθ  4.0041 − sin θ 
( 4.0041 − sinθ )2
sin θ =
θ = 14.446°,
F
N
1
= 0.24974
4.0041
=
cos14.446°
4.0041 − 0.24974
θ = 180° − 14.446° = 165.554°,
(a) Minimum value of µ s for no slip.
(b) Corresponding values of θ .
= 0.258
F
= 0.258
N
( µs )min
= 0.258 W
θ = 14.5° and 165.5° W
PROBLEM 12.71
The horizontal rod OA rotates about a vertical shaft according to the
relation θ = 10t, where θ and t are expressed in rad/s and seconds,
respectively. A 250-g collar B is held by a cord with a breaking strength
of 18 N. Neglecting friction, determine, immediately after the cord
breaks, (a) the relative acceleration of the collar with respect to the rod,
(b) the magnitude of the horizontal force exerted on the collar by the rod.
SOLUTION
θ = 10t rad/s, θ = 10 rad/s 2
m = 250 g = 0.250 kg
Before cable breaks: Fr = −T and r = 0.
(
Fr = mar : − T = m r − rθ 2
mrθ 2 = mr + T
or
θ 2 =
)
0 − 18
mr + T
=
= 144 rad 2 /s 2
mr
( 0.25)( 0.5)
θ = 12 rad/s
Immediately after the cable breaks: Fr = 0, r = 0
(a) Acceleration of B relative to the rod.
(
)
m r − rθ 2 = 0
or
r = rθ 2 = ( 0.5 )(12 ) = 72 m/s2
2
a B / rod = 72 m/s 2 radially outward W
(b) Transverse component of the force.
(
Fθ = maθ : Fθ = m rθ + 2rθ
)
Fθ = ( 0.250 ) ( 0.5 )(10 ) + ( 2 )( 0 )(12 )  = 1.25
Fθ = 1.25 N W
PROBLEM 12.89
A space shuttle S and a satellite A are in the circular orbits shown. In order
for the shuttle to recover the satellite, the shuttle is first placed in an
elliptic path BC by increasing its speed by ∆vB = 85 m/s as it passes
through B. As the shuttle approaches C, its speed is increased by
∆vC = 79 m/s to insert it into a second elliptic transfer orbit CD.
Knowing that the distance from O to C is 6900 km, determine the amount
by which the speed of the shuttle should be increased as it approaches D
to insert it into the circular orbit of the satellite.
SOLUTION
R = 6370 km = 6.37 × 106 m, rA = rD = 6370 + 610 = 6980 km = 6.98 × 106 m
rB = 6370 + 290 = 6660 km = 6.66 × 106 m,
( vA )circ
( vB )circ
=
gR 2
=
rA
=
gR 2
=
rB
( 9.81) ( 6.37 × 106 )
6.98 × 10
rC = 6900 km = 6.90 × 106 m
2
= 7.55173 × 103 m/s
6
( 9.81) ( 6.37 × 106 )
6.66 × 10
2
= 7.73102 × 103 m/s
6
For path BC.
vB = ( vB )circ + ( ∆v ) B = 7.73102 × 103 + 85 = 7.81602 × 103 m/s
mrBvB = mrC ( vC )1
( vC )1 =
(
)(
)
6.66 × 106 7.81602 × 103
rBvB
=
= 7.54416 × 103 m/s
6
rC
6.90 × 10
For path CD.
( vC )2
= ( vC )1 + ( ∆v )C = 7.54416 × 103 + 79 = 7.62316 × 106 m/s
mrC ( vC )2 = mrDvD
vD =
( ∆v ) D
rC ( vC )2
rD
( 6.90 × 10 )( 7.62316 × 10 ) = 7.53579 × 10
6
=
6
3
6.98 × 106
m/s
= ( v A )circ − vD = 7.55173 × 103 − 7.53579 × 103 = 15.94 m/s
( ∆v )D
= 15.94 m/s W