13–5.
The water-park ride consists of an 800-lb sled which slides from rest down down the incline incline and then into the pool. pool. If the frictional resistance on the incline is Fr = 30 lb, and in the pool for a short distance Fr = 80 lb, determine how fast the sled is traveling when s = 5 ft. 100 ft
SOLUTION
a F x
+ b
=
ma x;
800 sin 45° a
2 v1 =
v1 =
+
a F x
= ma x;
0
+
=
2ac(s
800 a 32.2
-
s
s0)
2(21.561)(100 2 2
-
0))
78.093 ft> s
- 80 =
a
=
21.561 ft> s2
=
2 2 v1 = v0 +
;
30
-
= - 3.22
2 2 v2 = v1 +
800 a 32.2 ft> s2 2ac(s2
2 v2 =
(78.093)2
v2 =
77.9 ft> s
+
-
s1)
2( - 3.22)(5
-
0) Ans.
100 ft
*13–112.
The pilot of an airplane executes a vert ical loop which in part follows the path of a card ioid, r = 600(1 + cos u) ft. If his speed at A (u = 0°) is a constant vP = 80 ft s , determine the vertical force the seat belt must exert on h im to hold him to his seat when the plane is upside down at A. He weighs 150 lb.
A
>
r u
SOLUTION r = 600(1 + cos u)|u = 0° = 1200 ft
#
#
$
$
r = - 600 sin uu u = 0° = 0
#
#
r = - 600 sin uu - 600 cos uu2 u = 0° = - 600u2
a # b # = 0 + a 1200 b # $ # = 2 + 2a b a 2
#
v2p = r2 +
(80)2 2vpvp
ru
2
u
rr
ru
#
u
= 0.06667 $
ru + ru
b
$ $ u = 0 0 = 0 + 0 + 2r2 uu # $ ar = r - ru2 = - 600(0.06667) 2 - 1200(0 .06667) 2 = - 8 ft s2 $ ## au = ru + 2ru = 0 + 0 = 0
>
+ c © Fr = mar;
N - 150 =
a 321502 b ( - 8) .
N = 113 lb
Ans.
600 (1 + cos u ) ft
13–15.
A freight elevator, including its load, has a mass of 500 kg. It is prevented from rotating by the track and wheels mounted along its sides.When t = 2 s, the motor M draws in the cable with a speed of 6 m> s, measured relative to the elevator. If it starts from rest, determine the constant acceleration of the elevator and the tension in the cable. Neglect the mass of the pulleys, motor, and cables.
M
SOLUTION 3sE + sP = l 3vE = - vP
A+TB
vP
= vE + vP>E
- 3vE = vE + 6 vE
A+cB
v
6 4
= - = - 1.5 m> s = 1.5 m> s c
= v0 + ac t
1.5 = 0 + aE (2) aE
= 0.75 m> s2 c
+ c © Fy = may ;
Ans.
4T - 500(9.81) = 500(0.75) T
= 1320 N = 1.32 kN
Ans.
*13–44.
When the blocks are released, determine their acceleration and the tension of the cable. Neglect the mass of the pulley.
SOLUTION Free-Body Diagram: The free-body diagram of blocks A and B are shown in Figs. b and c, respectively. Here, a A and aB are assumed to be directed downwards so that they are consistent with the positive sense of position coordinates s A and sB of blocks A and B, Fig. a. Since the cable passes over the smooth pulleys, the tension in the cable remains constant throughout.
2T - 10(9.81) = - 10aA
(1)
and
+ c © Fy = may;
T
- 30(9.81) = - 30aB
(2)
Kinematics: We can express the length of the cable in term s of s A and sB by referring to Fig. a.
2sA + sB = l The second derivative of the above equation gives 2aA + aB = 0
(3)
Solving Eqs. (1), (2), and (3) yields aA
= - 3.773 m> s2 = 3.77 m> s2 c T
= 67.92 N = 67.9 N
aB
= 7.546 m> s2 = 7.55 m> s2 T
10 kg B
30 kg
Equations of Motion: By referring to Figs. b and c,
+ c © Fy = may;
A
Ans. Ans.
*13–68.
The 0.8-Mg car travels over the hill having the shape of a parabola. If the driver maintains a constant speed of 9 m s, determine both the resultant normal force and the resultant frictional force that all the wheels of the car exert on the road at the instant it reaches point A. Neglect the size of the car.
y
>
y
20 (1
x2 )
6400 A
SOLUTION Geometry : Here,
dy dx
= - 0.00625 x
and
d2y dx2
80 m = - 0.00625 . The
slope angle u at point
A is given by
tan u
=
dy dx
2
= - 0.00625(80)
u = - 26.57°
x = 80 m
and the radius of curvature at point A is r =
[1
> >
>
(dy dx)2]3 2 |d2y dx2|
+
=
[1
+
Equat io ns of Mot io n : Here, at
>
( - 0.00625 x)2]3 2 | - 0.00625| =
2
=
223.61 m
x = 80 m
0. Applying Eq. 13–8 with u
=
26.57° and
r = 223.61 m , we have © Ft =
mat;
800(9.81) sin 26.57° F f
=
3509.73 N
=
-
F f
=
800(0)
3.51 kN
Ans. 2
© Fn =
man;
800(9.81) cos 26.57° N
=
6729.67 N
=
-
N
=
6.73 kN
800
9 a 223.61 b Ans.
x
13–81.
A 5-Mg airplane is flying at a constant speed of 350 km h along a horizontal circular path. If the banking angle u = 15° , determine the uplift force L acting on the airplane and the radius r of the circular path. Neglect the size of the airplane.
L
>
u
r
SOLUTION Free-Body Di a gram : The free-body diagram of the airplane is shown in Fig. (a). Here, a must be directed towards the center of curvature (positive n axis). n
m 1h a350 kmh b a 1000 b a b 1 km 3600 s
Equat io ns of Mot io n : The speed of the airplane is v =
>
= 97.22 m s. Realizing that an = + c © Fb = 0;
+ ©F = n
;
man;
v
2
r
=
97.222 r
and referring to Fig. (a),
L
cos 15° - 5000(9.81) = 0
L
= 50780.30 N = 50.8 kN
50780.30 sin 15° = 5000 r
¢
Ans.
97.222
= 3595.92 m = 3.60 km
r
≤ Ans.
13–90.
The boy of mass 40 kg is sl iding down the sp iral slide at a constant speed such that h is position, measured from the top of the chute, has components r = 1.5 m, u = 10.7t2 rad, and z = 1 - 0.5t2 m, where t is in seconds. Determine the components of force Fr , Fu , and Fz which the sl ide exerts on him at the i nstant t = 2 s. Neglect the size of the boy.
z
u
r
SOLUTION r =
1.5
#
$
r = r = $ u =
0
0.7t
z = - 0.5t
# u =
0. 7
z = - 0.5
$
0 #
$
2
=
0
z
#
0
z =
ar = r - r (u) au
u =
-
1.5(0.7)2
= - 0.735
$ # # = r u + 2r u = 0 $
az = z = 0 © F r = ma r ;
F r =
40( - 0.735)
© F u = mau;
F u =
0
© F z = ma z;
F z - 40(9.81) =
F z = 392
N
= - 29.4
N
1.5 m
Ans. Ans.
0 Ans.
13–95.
The ball has a mass of 2 kg and a negligible size. It is originally traveling around the horizontal circular path of radius r# 0 = 0.5 m such that the angular rate of rotation is u0 = 1 rad s. If the attached cord ABC is drawn down throughthe hole ata constant speed of 0.2 m s, determinethe tension the cord exerts on the ball at the instant r = 0.25 m. Also, compute the angular velocity of the ball at this instant. Neglect the effects of friction between the ball and horizontal plane. Hint: First show that$ the equation of motion in the u # # # direction yields a u = ru + 2ru = 1 r d r2u dt = 0. # When integrated, r 2u = c, where the constant c is determined from the problem data.
>
A r B
>
r 0
0.2 m/ s
1 > 21 1 2> 2
C F
SOLUTION
a Fu
=
mau;
$
0
=
m[ru
+
# #
2ru]
=
m
c 1r dtd (r ) d #
2 u
=
0
Thus, #
d(r2u)
=
0
r2u
=
C
(0.5)2(1)
=
C
#
#
#
=
u = 4.00 #
Since r $
= #
-
r(u)2
a Fr
=
mar;
=
>
>
0
= -
T
T
Ans. $
r
0.2 m s,
r
ar
(0.25)2u rad s
-
0
0.25(4.00)2
= =
=
= -
>
4 m s2
2( - 4) 8N
u
Ans.
·
u0
