PROBLEM 13.46
A chair–lift is designed to transport 1000 skiers per hour from the base A to
the summit B. The average mass of a skier is 70 kg and the average speed of
the lift is 75 m/min. Determine (a) the average power required, (b) the
required capacity of the motor if the mechanical efficiency is 85 percent and
if a 300 percent overload is to be allowed.
SOLUTION
0.85
PROBLEM 13.47
It takes 15 s to raise a 1200-kg car and the supporting 300-kg hydraulic
car–lift platform to a height of 2.8 m. Determine (a) the average output
power delivered by the hydraulic pump to lift the system, (b) the average
power electric required, knowing that the overall conversion efficiency
from electric to mechanical power for the system is 82 percent.
PROBLEM 13.48
The velocity of the lift of Problem 13.47 increases uniformly from zero to
its maximum value at mid–height 7.5 s and then decreases uniformly to
zero in 7.5 s. Knowing that the peak power output of the hydraulic pump is
6 kW when the velocity is maximum, determine the maximum life force
provided by the pump.
PROBLEM 13.47 It takes 15 s to raise a 1200-kg car and the supporting
300-kg hydraulic car-lift platform to a height of 2.8 m. Determine
(a) the average output power delivered by the hydraulic pump to lift the
system, (b) the average power electric required, knowing that the overall
conversion efficiency from electric to mechanical power for the system is
82 percent.
SOLUTION
PROBLEM 13.49
(a) A 120–lb woman rides a 15-lb bicycle up a 3–percent slope at a constant speed of 5 ft/s. How much power
must be developed by the woman? (b) A 180–lb man on an 18-lb bicycle starts down the same slope and
maintains a constant speed of 20 ft/s by braking. How much power is dissipated by the brakes? Ignore air
resistance and rolling resistance.
SOLUTION
(b)
PROBLEM 13.50
A power specification formula is to be
derived for electric motors which drive
conveyor belts moving solid material at
different rates to different heights and
distances. Denoting the efficiency of the
motors by
η
and neglecting the power
needed to drive the belt itself, derive a
formula (a) in the SI system of units for
the power P in kW, in terms of the mass
flow rate m in kg/h, the height b and
horizontal distance l in meters, and (b) in
U.S. customary units, for the power in hp,
in terms of the material flow rate
w
in
tons/h, and the height b and horizontal
distance l in feet.
SOLUTION
2
η
PROBLEM 13.51
A 1400-kg automobile starts from rest and travels 400 m during a
performance test. The motion of the automobile is defined by the relation
4000ln(cosh 0.03 ),xt=
where x and t are expressed in meters and seconds,
respectively. The magnitude of the aerodynamic drag is
2
0.35 ,Dv=
where
D and
v
are expressed in newtons and m/s, respectively. Determine the
power dissipated by the aerodynamic drag when (a) t = 10 s, (b) t = 15 s.
SOLUTION
PROBLEM 13.52
A 1400-kg automobile starts from rest and travels 400 m during a
performance test. The motion of the automobile is defined by the relation
0.0005
3.6
x
ae
−
=
, where a and x are expressed in
2
m/s
and meters,
respectively. The magnitude of the aerodynamic drag is
2
0.35 ,Dv=
where D and
v
are expressed in newtons and m/s, respectively. Determine
the power dissipated by the aerodynamic drag when (a) x = 200 m,
(b) x = 400 m.
SOLUTION
Motion is defined by the following function:
0.0005
3.6
x
ae
dv
vdx
−
=
=
0.0005
00
0.0005
0
3.6
3.6
0.0005
vx
x
xu
vdv e dx
e du
−
−
=
−
=
∫∫
∫
( )
20.0005
7200 1
2
x
ve−
=−−
( )
2 0.0005
14400 1
x
ve
−
= −
( )
1
0.0005 2
120 1
x
ve
−
= −
(a)
200 m,x=
( )
3
17.75 10 17.75 kWPW= =
(b)
400 m,x=
( )
3
46.7 10 46.7 kWPW= =
3
Power dissipated 0.35P Dv v= = =
( )
3
30.0005 2
604.8 10 1
x
Pe
−
= −
PROBLEM 13.53
The fluid transmission of a 15-Mg truck allows the engine to deliver an essentially constant power of 50 kW
to the driving wheels. Determine the time required and the distance traveled as the speed of the truck is
increased (a) from 36 km/h to 54 km/h, (b) from 54 km/h to 72 km/h.
SOLUTION
For constant power, P:
P Fv
=
( )
( )
3
3 50 10
×
PROBLEM 13.54
The elevator E has a weight of 6600 lbs when fully loaded and is connected as
shown to a counterweight W of weight of 2200 lb. Determine the power in hp
delivered by the motor (a) when the elevator is moving down at a constant speed
of 1 ft/s, (b) when it has an upward velocity of 1 ft/s and a deceleration of
2
0.18 ft/s .
SOLUTION
PROBLEM 13.54 (Continued)
2
2
(2200 lb)(0.18 ft/s )
(2200 lb) (32.2 ft/s )
W
T= +
2212 lb
W
T=
Elevator
2 ()
E
CW E E
W
F ma T T W a
g
−
Σ= + − =
2
2
(6600 lb)(0.18 ft/s )
2 ( 2212 lb) (6600 lb) (32.2 ft/s )
C
T=−+ −
2 4351 lb
C
T=
2175.6 lb
C
T=
2 ft/s (see part( ))
C
va=
(2175.6 lb)(2 ft/s) 4351.2 lb ft/s
CC
P Tv=⋅= = ⋅
7.911 hp=
7.91 hpP=
PROBLEM 13.55
A force P is slowly applied to a plate that is attached to two springs and causes a deflection
0
.x
In each of the
two cases shown, derive an expression for the constant
,
e
k
in terms of
1
k
and
2
,k
of the single spring
equivalent to the given system, that is, of the single spring which will undergo the same deflection
0
x
when
subjected to the same force P.
SOLUTION
System is in equilibrium in deflected
0
x
position.
1 20 0
e
12e
PROBLEM 13.56
A loaded railroad car of mass m is rolling at a
constant velocity v0 when it couples with a
massless bumper system. Determine the
maximum deflection of the bumper assuming the
two springs are (a) in series (as shown), (b) in
parallel.
SOLUTION
Let position A be at the beginning of contact and position B be at maximum deflection.
2
0
1
2
0 (zero force in springs)
A
A
T mv
V
=
=
0
12
kk
0 1 2 12
PROBLEM 13.56 (Continued)
12 0
0
12
PROBLEM 13.57
A 750–g collar can slide along the horizontal rod shown. It is
attached to an elastic cord with an undeformed length of 300 mm
and a spring constant of 150 N/m. Knowing that the collar is
released from rest at A and neglecting friction, determine the
speed of the collar (a) at B, (b) at E.
SOLUTION
PROBLEM 13.58
A 4–lb collar can slide without friction along a horizontal rod and is
in equilibrium at A when it is pushed 1 in. to the right and released
from rest. The springs are undeformed when the collar is at A and
the constant of each spring is 2800 lb/in. Determine the maximum
velocity of the collar.
2
2
PROBLEM 13.59
A 4–lb collar can slide without friction along a horizontal rod and is
released from rest at A. The undeformed lengths of springs BA and
CA are 10 in. and 9 in., respectively, and the constant of each spring
is 2800 lb/in. Determine the velocity of the collar when it has moved
1 in. to the right.
223.1 ft/sv=
PROBLEM 13.60
A 500–g collar can slide without friction on the curved rod BC in a horizontal
plane. Knowing that the undeformed length of the spring is 80 mm and that
400k=
kN/m, determine (a) the velocity that the collar should be given at A
to reach B with zero velocity, (b) the velocity of the collar when it eventually
reaches C.
SOLUTION
PROBLEM 13.60 (Continued)
C
PROBLEM 13.61
For the adapted shuffleboard device in Prob 13.28, you decide to utilize an elastic
cord instead of a compression spring to propel the puck forward. When the cord is
stretched directly between points A and B, the tension is 20 N. The 425 gram puck
is placed in the center and pulled back through a distance of 400 mm; a force of
100 N is required to hold it at this location. Knowing that the coefficient of friction
is 0.3, determine how far the puck will travel.
SOLUTION
2
PROBLEM 13.61 (Continued)
Find Stiffness of cord, k:
106.25 N/m
k
=
Find stretch in cord at positions 1, 2 and 3:
,1
1
62.5 N
106.25 N/m
0.5882 m
C
F
sk
=
=
,2
23
20 N
106.25 N/m
0.1882 m
C
F
ss k
= =
=
Work Energy from position 1 to 3:
NC
2
1