ENPH 131 Assignment #5
Solutions
Problem 12.200
Determine the speed of cylinder A, if the rope is drawn towards the motor M at a constant rate of 12 m  s.
Concider the sections of rope sA and sM defined below.
3 s A + sM = l
Where l is the total length of the rope. Taking the time- derivative of this expression yeilds
3
âsA
ât
+
âsM
ât
=0
3 v A + vM = 0
vM = 12 m  s
3 vA = -12
2
ENPH 131 assignment 5 solutions.nb
3
âsA
ât
+
âsM
ât
=0
3 v A + vM = 0
vM = 12 m  s
3 vA = -12
vA = -4.00 m  s
Therefore the speed of A is 4.00 m  s.
Problem 12.212
The man pulls the boy up to the tree limb C by walking backward at a constant speed of 1.3 m  s.
Determine the speed at which the boy is being lifted at the instant xA = 4 m. Neglect the size of the limb. When xA = 0,
yB = 0, so that A and B are coincident, i. e., the rope is 16m long.
The total length of the rope, l, can be descibed as
l = lAC + lBC = 16 m
lBC = yB
lAC is given by the hypotenuse of a right-angle triangle with side lengths 8m and xA :
lAC = I82 + xA 2 M
12
= I64 + xA 2 M
Therefore
l = I64 + xA 2 M
12
+ yB = 16
Taking the time-derivative of this expression yeilds
xA vA
12
Ixa 2 +64M
+ vB = 0
vA is a constant, 1.3m/s. xA = 4 m:
vB = vB =
xA vA
12
Ixa 2 +64M
H4L H1.3L
IH4L2 +64M
12
vB = -0.581 m  s
12
ENPH 131 assignment 5 solutions.nb
vB = vB =
3
xA vA
12
Ixa 2 +64M
H4L H1.3L
IH4L2 +64M
12
vB = -0.581 m  s
Therefore the speed at which the boy is being lifted is 0.581m/s.
Relative-Motion Analysis of Two Particles Using Translating
Axes
a) A cruise ship is traveling at a speed of v1 = 22.4 ft  s. A speedboat with a late passenger is heading toward the cruise ship
at an angle of Θ = 60.0 °; its speed is v2 = 36.0 ft  s. (Part A figure) What is v, the magnitude of the speedboat’s velocity
relative to the cruise ship?
Using the stationary axes given on the diagram velocities v1 and v2 can be written in terms of their x- and y-components:
`
`
v1 = v1 j = 22.4 j
`
`
`
`
v2 = -v2 cosΘ i + v2 sinΘ j = -36.0 cosH60 °L i + 36.0 sinH60.0 °L j
The velocity of the speedboat wrt the cruise ship, v21 , is given by
v21 = v2 - v1
`
`
`
v21 = -36.0 cosH60 °L i + 36.0 sinH60.0 °L j - 22.4 j
`
`
v21 = -18.0 i + 8.78 j
The magnitude of v21 is
12
v21 = v21 = IH-18.0L2 + H8.78L2 M
v21 = 20.0 ft  s
b) As the speedboat approaches the cruise ship, it encounters an ocean current that is moving at v3 = 8.90 ft  s and that is
oriented at an angle Φ = 43.0 °. (Part B figure) What are vx and vy , the scalar components of the speedboat’s velocity
relative to the cruise ship as the speedboat crosses this ocean current?
4
ENPH 131 assignment 5 solutions.nb
With respect to the stationary frame the speed boat’s new velocity, v2 ', is
v2 ' = v2 + v3
`
`
`
`
v2 ' = -v2 cosHΘL i + v2 sinHΘL j - v3 cosHΦL i - v3 sinHΦL j
`
`
`
`
v2 ' = -36.0 cosH60 °L i + 36.0 sinH60.0 °L j - 8.90 cosH43.0 °L i - 8.90 sinH43.0 °L j
`
`
v2 ' = -21.51 i + 25.11 j
The velocity of the speedboat wrt the cruise ship, v21 ’, is given by
v21 ' = v2 ' - v1
`
`
`
v21 ' = -21.51 i + 25.13 j - 22.4 j
`
`
v21 ' = -24.5 i + 2.71 j
vx = - 24.5 ft  s
vy = 2.71 ft  s
c) After crossing the ocean current, the speedboat continues to approach the cruise ship at v2 = 36.0 ft  s and at an angle
Θ = 60.0 °. The cruise ship sees the speedboat and slows down by 5ft/s. At the same time, the speedboat docks with the cruise
ship by turning with a radius of curvature of r1 = 40.0 ft and by reducing its speed to match that of the cruise ship once it is
parallel to the cruise ship. (Part C figure) What are ax and , the scalar components of the speedboat’s acceleration relative to
the cruise ship’s when the speedboat is docked parallel to the cruise ship? Assume that the cruise ship’s acceleration and the
speedboat’s transverse acceleration are constant.
ENPH 131 assignment 5 solutions.nb
5
The cruise ship slows by 5ft/s in some time, T, and so has an acceleration
a1 = -
v1 '-v1
T
`
i=
5
T
`
i
Since the speed boat is travelling in a circular path its acceleration is given by
a2 = an + at
When the speedboat and cruise ship are parallel, this becomes
`
`
a2 = -an i - at j
At this time, the speed boat’s speed is equal to that of the cruise ship
an =
v2 '2
r1
=
v1 '2
r1
=
H22.4-5L2
40
= 7.569 ft ‘ s2
The change in the speed boat’s speed, at , can be found using eqn 12-6:
Hv2 'L2 = v2 2 + 2 at Hs - so L
Using the fact that the speed boat is initially travelling at an angle of Θ = 60 ° and finishes at an angle of Θ=90° (parallel to
cruise ship), we can claculate the total distance, s - s0 , travelled by the speed boat by taking the angular fraction I
90 °-60 °
Mof
360 °
the circumference of its circular path:
30 °
s - s0 = 2 Π r1 I 360 ° M
s - s0 =
1
6
Π r1
at = 3
Hv2 'L2 -v2 2
Π r1
at = 3
H22.4-5L2 -H36.0L2
ΠH40L
`
at = -23.712 ft ‘ s2 j
`
`
a2 = -7.569 i - 23.712 j
Also, we may use at to determine the time T, and thus a1 . at is the change in the speed of the speed boat and can also be
written as
at =
T =
T =
v2 '-v2
T
v2 '-v2
at
22.4-5-36
-23.712
T = 0.784 s
6
ENPH 131 assignment 5 solutions.nb
at =
T =
T =
v2 '-v2
T
v2 '-v2
at
22.4-5-36
-23.712
T = 0.784 s
`
5 `
5 `
a1 = - T j = - 0.784 j = -6.374 j
The acceleration of the speedboat wrt the cruise ship, a21 , is given by
a21 = a2 - a1
`
`
`
a21 = -7.569 i - 23.712 j + 6.374 j
`
`
a21 = 7.57 i - 17.3 j
ax = 7.57 ft ‘ s2
ay = - 17.3 ft ‘ s2
Newton’
s Laws of Motion
a) Three identical train locomotives, on separate tracks, apply the maximum force needed to begin moving. The first train
consists of a single empty freight car, the second train consists of 50 empty freight cars, and the third train consists of 50
equally loaded freight cars. Rank the accelerations of the trains. All of the freight cars, when empty, are identical.
Since the force applied to each train is equal a µ
1
.
m
Therefore the train with the smallest mass has the largest acceleration.
Thus, from largest to smallest acceleration, the trains rank as folows: 1 empty car, 50 empty cars, 50 laoded cars.
b) A crate sliding down a ramp reaches the bottom of the ramp and slides across a flat floor. At the instant shown, the crate
has a speed of v = 5.10 m  s. The crate comes to a stop after a distance x = 18.5 m. (Part B figure) What is Μk , the coefficient
of kinetic friction between the crate and the floor?
The coefficient of kinetic friction is given by
Ff
Μk =
FN
Where F f is the force of friction, resisting motion, and FN = m g is the normal force acting on the object. The crate slows
from a velocity of 5.10 m/s to rest over a distance of 18.5m. Use eqn 12-6 to determine its acceleration:
v2 = v0 2 + 2 aHx - xo L
a=
v2 -v0 2
2 Hx-x0 L
Ff = m a
v2 -v
2
F f = mJ 2 Hx-x0 L N
0
ENPH 131 assignment 5 solutions.nb
Ff = m a
v2 -v
7
2
F f = mJ 2 Hx-x0 L N
0
Μk =
Ff
FN
mK
=
v2 -v0 2
2 Hx-x0 L
O
mg
Μk =
v2 -v0 2
2 gHx-x0 L
Μk =
-5.102
2 H9.81L H18.5L
Μk = 0.0717
c) Two players hit an air hockey puck simultaneously. One player hits the puck with a force F1 = 4.20 N at an angle
Θ = 21.8 °. The other player hits the puck with a force F2 = 7.35 N at an angle Φ = 66.6 °. (Part C figure) If the puck has a
mass m = 43.0, what is a, the acceleration of the puck, and Γ, the angle of the puck’s motion? The friction between the puck
and the table’s surface is negligible.
Fnet = F1 +F2 = m a
a = Fnet  m
`
`
`
`
Fnet = F1 cosΘ i - F1 sinΘ j + F2 cosΦ i + F2 sinΦ j
`
`
`
`
Fnet = 4.20 cosH21.8 °L i - 4.20 sinH21.8 °L j + 7.35 cosH66.6 °L i + 7.35 sinH66.6 °L j
`
`
Fnet = 6.819 i + 5.186 j
Fnet = I6.8192 + 5.1862 M
12
Fnet = 8.567 N
a = Fnet  m
m = 43 g = 0.043 kg
a = 8.567  0.043
a = 199 m ‘ s2
tan Γ = ay ‘ ax = m Fy ‘ m Fx
Fy
Γ = arctanJ F N
x
Γ = arctanH5.186  6.819L
Γ = 0.6502 rad = 37.3 °
8
ENPH 131 assignment 5 solutions.nb
Fundamental Problem 13.4
The 2Mg car is being towed by a winch. If the winch exerts a force of T = H100 sL N on the cable, where s is the displacement of the car in meters, determine the speed of the car when s = 11 m, starting from rest. Neglect rolling resistance of the
car.
The acceleration of the car can be determined using the force exerted by the winch:
a = Fm =
100 s
m
Since the acceleration depends on position, use eqn 12-3 to determine an expression for the car’s velocity:
a âs = v âv
100
m
s
v
Ù s âs = Ù v âv
0
100 s2
m 2
v=J
0
=
v2
2
100 s2 12
N
m
m = 2 Mg = 2000 kg
v=J
100 H11L2 12
N
2000
v = 2.46 m  s