13–41.
Block Ahas a mass and is attached to a spring having a
stiffness kand unstretched length . If another block B,
having a mass , is pressed against Aso that the spring
deforms a distance d, determine the distance both blocks
slide on the smooth surface before they begin to separate.
What is their velocity at this instant?
mB
l0
mA
SOLUTION
Block B:
Since ,
Ans.
Ans.y=
C
kd2
(mA+mB)
1
2y2=k
(mA+mB)
B
(d)x–1
2x2
R
d
0
=1
2
kd2
(mA+mB)
Lv
0
ydv=Ld
0
k(d–x)
(mA+mB)dx
ydy=adx
N=0 when
d–x=0,
or
x=d
a=k(d–x)
(mA+mB)N=kmB(d–x)
(mA+mB)
–k(x–d)–mBa=mAa
aA=aB=a
:
+©Fx=max;N=mBaB
:
A
k
B
286
13–42.
SOLUTION
Block B:
Since ,
, then for separation. Ans.
At the moment of separation:
Require , so that
Thus,
Q.E.D.d72mkg
k(mA+mB)
kd 72mkg(mA+mB)
kd2–2mkg(mA+mB)d70
v70
v=Bkd2–2mkg(mA+mB)d
(mA+mB)
1
2v2=k
(mA+mB)
B
(d)x–1
2x2–mkgx
R
d
0
Lv
0
vdv=Ld
0
B
k(d–x)
(mA+mB)–mkg
R
dx
vdv=adx
x=dN =0
N=kmB(d–x)
(mA+mB)
a=k(d–x)–mkg(mA+mB)
(mA+mB)=k(d–x)
(mA+mB)–mkg
aA=aB=a
:
+©Fx=max;N–mkmBg=mBaB
:
Block Ahas a mass and is attached to a spring having a
stiffness kand unstretched length . If another block B,
having a mass , is pressed against Aso that the spring
deforms a distance d, show that for separation to occur it is
necessary that , where is the
coefficient of kinetic friction between the blocks and the
ground. Also, what is the distance the blocks slide on the
surface before they separate?
mk
d72mkg(mA+mB)>k
mB
l0
mA
A
k
B
13–43.
A parachutist having a mass m opens his parachute from an
at-rest position at a very high altitude. If the atmospheric
drag resistance is FD=kv2, where k is a constant,
determine his velocity when he has fallen for a time t. What
is his velocity when he lands on the ground? This velocity is
referred to as the terminal velocity, which is found by letting
the time of fall tS∞.
FD
v
288
*13–44.
SOLUTION
a
Ans.
Ans.
Ans. :
+©Fx=0; Ax=0
Ay=2108.15 N =2.11 kN
+c©Fy=0; Ay–1765.8 –(2)(1,131) +1919.65 =0
By=1,919.65 N =1.92 kN
+©MA=0;
By (6) –(1765.8 +1,131)3 –(1,131)(2.5) =0
T=1,131 N
+c©Fy=may
2T–1962 =200(1.5)
ac=1.5 m>s2
2ac=3 m>s2
2ac=ap
2yc=yp
Sc+(Sc–Sp)
If the motor draws in the cable with an acceleration of
, determine the reactions at the supports Aand B.The
beam has a uniform mass of 30 kg m, and the crate has a
mass of 200 kg. Neglect the mass of the motor and pulleys.
>
3 m>s2
AB
2.5 m3 m
0.5 m
3 m/s2
13–45.
If the force exerted on cable AB by the motor is
N, where tis in seconds,determine the 50-kg
crate’s velocity when .The coefficients of static and
kinetic friction between the crate and the ground are
and respectively.Initially the crate is at rest.mk=0.3,
ms=0.4
t=5 s
F=(100t3>2)
SOLUTION
;
Realizing that ,
;
Equilibrium: For the crate to move, force Fmust overcome the static friction of
.Thus, the time required to cause the crate to be
on the verge of moving can be obtained from.
;
Kinematics: Using the result of aand integrating the kinematic equation
with the initial condition at as the lower integration limit,
When ,
Ans.v
=
0.8(5)5>2
–
2.943(5)
+
2.152
=
32.16 ft
>
s
=
32.2 ft
>
s
t=5 s
v=
A
0.8t5>2–2.943t+2.152
B
m>s
v=
A
0.8t5>2–2.943t
B
2t
1.567 s
Lv
0
dv =Lt
1.567 s
A
2t3>2–2.943
B
dt
Ldv =Ladt(:
+)
t=1.567v=0
dv =a dt
t=1.567 s
100t3>2–196.2 =0:
+©Fx=0
Ff=msN=0.4(490.5) =196.2 N
a=
A
2t3>2–2.943
B
m>s
100t3>2–147.15 =50a+c©Fx=max
Ff=mkN=0.3(490.5) =147.15 N
N=490.5 N
N–50(9.81) =50(0)+c©Fy=may
B
A
290
13–46.
Blocks Aand Beach have a mass m. Determine the largest
horizontal force Pwhich can be applied to Bso that Awill
not move relative to B. All surfaces are smooth.
SOLUTION
Block A:
Block B:
Ans.P=2mg tan u
P–mg tan u=mg tan u
;
+©Fx=max;P–Nsin u=ma
a=gtan u
;
+©Fx=max;Nsin u=ma
+c©Fy=0; Ncos u–mg =0
A
BP
u
C
13–47.
SOLUTION
Block A:
Block B:
Ans.P=2mgasin u+mscos u
cos u–mssin ub
P–mgasin u+mscos u
cos u–mssin ub=mgasin u+mscos u
cos u–mssin ub
;
+©Fx=max;P–msNcos u–Nsin u=ma
a=gasin u+mscos u
cos u–mssin ub
N=mg
cos u–mssin u
;
+©Fx=max;Nsin u+msNcos u=ma
+c©Fy=0; Ncos u–msNsin u–mg =0
Blocks Aand Beach have a mass m. Determine the largest
horizontal force Pwhich can be applied to Bso that Awill
not slip on B. The coefficient of static friction between Aand
Bis .Neglect any friction between Band C.ms
A
BP
u
C
292
*13–48.
SOLUTION
Thus,
Ans.
Ans.s=1
2a0sin ut2
sB>AC =s=Lt
0
a0sin utdt
vB>AC =a0sin ut
LvB>AC
0
dvB>AC =Lt
0
a0sin udt
aB>AC =a0sin u
0=m(–a0sin u+aB>AC)
Q+
aBsin f=-a0sin u+aB>AC
aB=a0+aB>AC
aB=aAC +aB>AC
Q+©Fx=ma
x;0=ma
Bsin f
The smooth block Bof negligible size has a mass mand
rests on the horizontal plane. If the board AC pushes on the
block at an angle with a constant acceleration
determine the velocity of the block along the board and the
distance sthe block moves along the board as a function of
time t.The block starts from rest when t=0.s=0,
a0,u
θ
A
C
B
s
a0
2
13–49.
SOLUTION
Block B:
(2)
(3)
Solving Eqs. (1)–(3)
Ans.aB=7.59 ft s2
aA=28.3 ft>s2NB=19.2 lb
aB=aAtan 15°
sB=sAtan 15°
+c©Fy=may;NBcos 15° –15 =
¢
15
32.2
≤
aB
32.2
If a horizontal force is applied to block A
determine the acceleration of block B. Neglect friction.
P
=12 lb
P
A
B
15 lb
8lb
15°
13–50.
A freight elevator, including its load, has a mass of 1 Mg. It
is prevented from rotating due to the track and wheels
mounted along its sides. If the motor M develops a constant
tension
T=4 kN
in its attached cable, determine the
velocity of the elevator when it has moved upward 6 m
starting from rest. Neglect the mass of the pulleys and
cables.
SOLUTION
M
13–51.
SOLUTION
Block:
Block and pan
Thus,
Require ,
Since dis downward,
Ans.d=(mA+mB)g
k
kd =-(mA+mB)g
N=0y=d
–(mA+mB)g+kcamA+mB
kbg+yd=(mA+mB)a–mAg+N
mAb
+c©F
y=may;
–(mA+mB)g+k(yeq +y)=(mA+mB)a
+c©F
y=may ;
–mA g+N=mA a
yeq =F
s
k=(mA+mB)g
k
The block Ahas a mass and rests on the pan B,which
has a mass Both are supported by a spring having a
stiffness kthat is attached to the bottom of the pan and to
the ground. Determine the distance dthe pan should be
pushed down from the equilibrium position and then
released from rest so that separation of the block will take
place from the surface of the pan at the instant the spring
becomes unstretched.
mB.
m
AA
B
k
dy
*13–52.
Agirl, having a mass of 15 kg,sits motionless relative to the
surface of a horizontal platform at a distance of from
the platform’s center.If the angular motion of the platform is
slowly increased so that the girl’s tangential component of
acceleration can be neglected, determine the maximum speed
which the girl will have before she begins to slip off the
platform. The coefficient of static friction between the girl and
the platform is m=0.2.
r=5m
SOLUTION
Applying Eq. 13–8, we have
Ans.v=3.13 m>s
©Fn=man; 0.2(147.15) =15av2
5b
©Fb=0; N–15(9.81) =0N=147.15 N
z
5m
13–53.
SOLUTION
circular path (positive naxis).
Equations of Motion: Realizing that and referring to Fig. (a),
Ans.r=1.36 m
©Fn=man; 147.15 =2a102
rb
an=v2
r=102
r
The 2-kg block Band 15-kg cylinder Aare connected to a
light cord that passes through a hole in the center of the
smooth table. If the block is given a speed of ,
determine the radius rof the circular path along which it
travels.
v=10 m>s
r
v
B
13–54.
The 2-kg block Band 15-kg cylinder Aare connected to a
light cord that passes through a hole in the center of the
smooth table.If the block travels along a circular path of
radius ,determine the speed of the block.r=1.5 m
SOLUTION
circular path (positive naxis).
Equations of Motion:Realizing that and referring to Fig. (a),
Ans.v=10.5 m>s
©Fn=man; 147.15 =2av2
1.5 b
an=v2
r=v2
1.5
r
v
B
13–55.
Determine the maximum constant speed at which the pilot
can travel around the vertical curve having a radius of
curvature
r=
800 m, so that he experiences a maximum
acceleration an
=
8g
=
78.5 m
>
s
2
. If he has a mass of 70 kg,
determine the normal force he exerts on the seat of the
airplane when the plane is traveling at this speed and is at its
lowest point.
SOLUTION
v
2
v
2
r 800 m
*13–56.
Cartons having a mass of 5 kg are required to move along
the assembly line at a constant speed of 8 m/s. Determine
the smallest radius of curvature, for the conveyor so the
cartons do not slip.The coefficients of static and kinetic
friction between a carton and the conveyor are and
respectively.mk=0.5,
ms=0.7
r,
SOLUTION
Ans.r=9.32 m
;
+©Fn=ma
n; 0.7W=W
9.81 (82
r)
Fx=0.7W
N=W
+c©Fb=ma
b;N–W=0
8 m/s
ρ
301
13–57.
SOLUTION
Ans.
Ans. N=2(7.36)2+(0) =7.36 N
n=1.63 m>s
©F
t=mat ;
N
t=0
©F
b=0;
N
b–0.75(9.81) =0
N
b=7.36
The collar A,having a mass of 0.75 kg, is attached to a
spring having a stiffness of When rod BC
rotates about the vertical axis, the collar slides outward
along the smooth rod DE. If the spring is unstretched when
determine the constant speed of the collar in order
that Also, what is the normal force of the rod
on the collar? Neglect the size of the collar.
s=100 mm.
s=0,
k=200 N>m.
k 200 N/m
A
C
D
Ans:
v=1.63 m>s
N=7.36
N
13–58.
The 2-kg spool Sfits loosely on the inclined rod for which the
coefficient of static friction is If the spool is located
0.25 m from A,determine the minimum constant speed the
spool can have so that it does not slip down the rod.
ms=0.2.
SOLUTION
Ans. v=0.969 m>s
N
s=21.3 N
+c©F
b=m ab;
N
s a4
5b+0.2N
s a3
5b–2(9.81) =0
;
+©Fn=m an;
Ns a3
5b–0.2Ns a4
5b=2av2
0.2 b
r=0.25a4
5b=0.2 m
z
S
A
0.25 m
3
4
5
303
13–59.
The 2-kg spool Sfits loosely on the inclined rod for which
the coefficient of static friction is If the spool is
located 0.25 m from A, determine the maximum constant
speed the spool can have so that it does not slip up the rod.
ms=0.2.
SOLUTION
Ans.v=1.48 m s
Ns=28.85 N
+c©Fb=ma
b;Ns(4
5)–0.2Ns(3
5)–2(9.81) =0
;
+©Fn=ma
n;Ns(3
5)+0.2Ns(4
5)=2( v2
0.2)
r=0.25(4
5)=0.2 m
z
S
A
0.25 m
3
4
5
Ans:
v
=
1.48
m>s
*13–60.
SOLUTION
Ans.Q+©Fn=man ;
2T– 60 sin 60° =60
32.2a152
10 b
T=46.9 lb
32.2 at
At the instant the boy’scenter of mass Ghas a
downward speed Determine the rate of
increase inhisspeed and the tension ineach of the two
supporting cords of the swing at thisinstant.Theboy has a
weight of 60 lb.Neglect hissize and the mass of the seat
and cords.
vG=15 ft>s.
u
=60°,
10 ft
G
u