Download Homework 4

Answer, Key { Homework 4 { Rubin H Landau
1
Which of the following best describes a free
This print-out should have 18 questions.
Check that it is complete before leaving the body diagram for the book?
printer. Also, multiple-choice questions may
1.
normal
continue on the next column or page: nd all
tension
choices before making your selection.
Note that only a few (usually 4) of the
frictional
problems will have their scores kept for a
gravitational
grade. You may make multiple tries to get
a problem right, although it's worth less each
2.
normal
time. Worked solutions to a number of these
tension
problems (even some of the scored ones) may
be found on the Ph211 home page.
frictional
Falling Apple
02:05, trigonometry, numeric, > 1 min.
001
gravitational
An apple falls from a tree and hits the ground
10:5 m below.
With what speed will it hit the ground?
Correct answer: 14:3457 m=s.
correct
3.
tension
frictional
Explanation:
When an object undergoes free fall from
rest, its nal speed is given by
p
v = 2gh
Algorithm
6 1
h = 10:5 m 15
g = 9p
:8 m=s2
v = p2:0 g h
= 2:0 h9:8i h10:5i
= 14
:3457 m=s
q
hm=si = hi hm=s2i hmi
:
(1)
(2)
(3)
gravitational
4.
gravitational
tension
frictional
normal
Explanation:
normal
tension
units
P303k conc forces1
05:01, noArithmetic, multiple choice, < 1 min.
002
A string is tied to a book and pulled at an angle as shown in the gure. The book remains
in contact with the table and does not move.
String
normal
frictional
gravitational
The gravitational force points down. The
frictional forces points to the left. The normal
force points up. The tension in the string
points in the direction the string is pulling.
003
Which forces on the book would change if the
string were pulled twice as hard?
1. normal
2. gravitational
Book
Table
Answer, Key { Homework 4 { Rubin H Landau
2
A 89:8 kg boxer has his rst match in the
3. tension in string
Canal Zone with gravitational acceleration
9:782 m=s2 and his second match at the
4. friction
North Pole with gravitational acceleration
9:832 m=s2.
5. normal and gravitational
a) What is his mass in the Canal Zone?
Correct answer: 89:8 kg.
6. normal, tension and friction correct
7. tension and friction
8. normal and friction
9. gravitational, tension and friction
Explanation:
The force the table exerts on the book (the
normal force) is less when the string is taut.
The book will slide easier since the normal
force has decreased. Thus the frictional force
has also decreased.
The string is taut indicating a tension that
is not there when the string is not there.
Pulling Two Blocks
05:04, calculus, numeric, > 1 min.
004
Two blocks on a frictionless horizontal surface
are connected by a light string.
m2
m1
F
m1 = 14:7 kg and m2 = 20 kg. A force
of 49:8 N is applied toward the right on the
20 kg block.
Find the acceleration of the system.
Correct answer: 1:43516 m=s2.
Explanation:
Let the tension in the string between the
blocks be T , and apply Newton's second law
to the each block:
F1 = m1 a = T
(1);
F2 = m2 a = F , T
(2);
Adding these equations gives us
(m1 + m2 )a = F
F
a= m +
m2
1
net
Explanation:
Basic Concept
An object's mass is constant, regardless of the
gravitational acceleration.
Solution
An object's weight varies with gravitational
position and is given by
W = mg
009
b) What is his weight in the Canal Zone?
Correct answer: 878:424 N.
Explanation:
W = mg
010
c) What is his mass at the North Pole?
Correct answer: 89:8 kg.
Explanation:
m = Wg
Weight of a Boxer
05:05, calculus, numeric, > 1 min.
011
d) What is his weight at the North Pole?
Correct answer: 882:914 N.
Explanation:
W = mg
net
008
Algorithm
80
m = 89:8 kg 100
g = 9:782 m=s2
g = 9:832 m=s2
W =mg
= h89:8i h9:782i
C
N
C
C
(1)
(2)
(3)
(4)
Answer, Key { Homework 4 { Rubin H Landau
3
that the spring scale reads 309:5 N. His true
= 878:424 N
2
hNi = hkgi hm=s i
units weight is 370 N, and the chair weighs 238 N.
m =m
(5)
= 89:8 kg
W =mg
(6)
= h89:8i h9:832i
= 882:914 N
hNi = hkgi hm=s2i
units
What is the magnitude of the acceleration
of the system?
012
answer: 0:177303 m=s2.
Find the weight in pounds of M = 694 grams Correct
Explanation:
of salami.
T
T
Correct answer: 1:52891 lb.
N
N
N
Explanation:
W = Mg
1 kg 9:8 m=s2
= 694 grams 1000
g
= 6:8012 N
= 6:8012 N 0:2248 lb=N
= 1:52891 lb
Algorithm
hkggi = 0:001 kg=g
hlbNi = 0:2248 lb=N
WBrian + Wchair
Brian
(1)
(2)
(3)
(4)
(5)
g = 9:8 m=s2 400
M = 694 grams 1000
W = M hkggi g
= h694i h0:001i h9:8i
= 6:8012 N
hNi = hgramsi hkg=gi hm=s2i units
W = W hlbNi
(6)
= h6:8012i h0:2248i
= 1:52891 lb
hlbi = hNi hlb=Ni
units
N
lb
N
014
Consider the system of Brian and the chair.
Note that two ropes support the system. Let
T be the tension in the ropes. Hence applying
Newton's law to the system gives
X
F = 2T , W
, W = Ma
where M represents the total mass of the
system.
+W
= 62:0408 kg
M=W
g
Hence solving for a,
a = 2T , W M , W
N , 370 N , 238 N)
= (2 309:5 62
:0408 kg
= 0:177303 m=s2
An inventive child named Brian wants to
reach an apple in a tree without climbing the
tree. Sitting in a chair connected to a rope
that passes over a frictionless pulley, he pulls
on the loose end of the rope with such a force
Brian
chair
chair
Brian
chair
015
What is the force that Brian exerts on the
chair?
Correct answer: 67:1941 N.
Explanation:
T
N
WBrian
Answer, Key { Homework 4 { Rubin H Landau
4
N =m
a+W
,T
(9)
Applying Newton's law to Brian only yields
X
= h37:7551i h0:177303i + h370i , h309:5i
F =T +N ,W
=m
a
= 67:1941 N
where N is the force exerted on Brian by
hNi = hkgi hm=s2i + hNi , hNi
units
the chair. Since m
= W
=g =
37:7551 kg, we obtain for N ,
Suspended in an Elevator
N =m a+W
,T
05:05,
calculus,
numeric, > 1 min.
2
= 37:7551 kg (0:177303 m=s )
016
) + 370 N , 309:5 N
The tension in a string from which a 7:9 kg
= 67:1941 N
object is suspended in an elevator is equal to
N is the force the chair exerts on Brian, but by 46 N.
Newton's third law, N is also the magnitude What is magnitude of the acceleration a of
the elevator?
of the force Brian exerts on the chair.
Correct answer: ,3:97722 m=s2.
Algorithm 2
g = 9:8 m=s (1) Explanation:
Basic Concepts:
W
= 370 N 260
(2)
512
X
64 W = 238 N 256
(3)
F
=
m
a
=
F
10
= 11 20
(4)
W = mg
T = 0:5 (W
+ W + )
(5)
= 0:5 (h370i + h238i + h11i)
Solution:
The weight W = mg of the object is less
= 309:5 N
than
the tension in the string, so assume the
hNi = hi (hNi + hNi + hi)
units
acceleration is upward.
+W
(6)
M=W
T
g
h238i
a
= h370ih+
9:8i
= 62:0408 kg
Mg
h
N
i
+
h
N
i
hkgi = hm=s2i
units
=) a = T ,mmg
a = 2:0 T , W M , W
(7)
Suspended in an Elevator
2
:
0
h
309
:
5
i
,
h
370
i
,
h
238
i
05:07, trigonometry, numeric, > 1 min.
=
h62:0408i
017
= 0:177303 m=s2
What is its direction?
hi
h
N
i
,
h
N
i
,
h
N
i
hm=s2i =
units
1. upward correct
hkgi
m
=W g
(8)
2. downward
i
3. not moving
= hh370
9:8i
Explanation:
= 37:7551 kg
The acceleration, assumed upward, was
h
N
i
positive,
units rect. so the upward assumption was corhkgi = hm=s2i
Brian
Brian
Brian
Brian
Brian
Brian
Brian
Brian
chair
net
Brian
Brian
chair
chair
Brian
Brian
Brian
chair
Brian
Answer, Key { Homework 4 { Rubin H Landau
5
For the mass m1 , T1 acts up and the weight
Algorithm
2
g = 9:8 m=s (1) m1 g acts down, with the acceleration a di4
m = 7:9 kg 8
(2) rected upward:
T = 46 N 20
(3)
F 1 = m1 a = T1 , m1 g
(1)
50
(4) For the mass on the table, a is directed to
a = T ,mm g
the right, T2 acts to the right, T1 acts to the
h
46
i
,
h
7
:
9
i
h
9
:
8
i
left, and the motion is to the right so that the
=
h7:9i
frictional force m2 g acts to the left:
= ,3:97722 m=s2
F 2 = m2 a = T2 , T1 , m2 g (2)
2i
h
N
i
,
h
kg
i
h
m
=
s
2
units For the mass m , T acts up and the weight
hm=s i =
hkgi
3
2
m3 g acts down, with the acceleration a directed downward:
Acceleration with Friction
05:08, calculus, numeric, > 1 min.
F 3 = m3 a = m3 g , T2
(3)
018
a
Adding these equations yields
m
(m1 + m2 + m3 )a = m3 g , m2 g , m1 g
µ
2 , m1 )g
a = (m(m3 ,+m
m2 + m3 )
1
m
m
net
net
net
2
1
3
The suspended 3:95 kg mass on the left is
accelerating up, the 2:4 kg mass slides to the
right on the table, and the suspended mass
7:3 kg on the right is accelerating down. If
the coecient of friction is 0:109, what will be
the acceleration of the system?
Correct answer: 2:21731 m=s2.
Algorithm
g = 9:8 m=s2 (1)
2
m1 = 3:95 kg 4
(2)
1 2 m2 = 2:4 kg 3 5
(3)
6
m3 = 7:3 kg 9
(4)
0 1 = 0:109 0 2
(5)
(m3 , m2 , m1 ) g
Explanation:
(6)
a
=
m1 + m2 + m3
Basic Concepts:
109i h2:4i , h3:95i) h9:8i
= (h7:3i , hh30::95
F = ma 6= 0
i + h2:4i + h7:3i
2
= 2:21731 m=s
The acceleration a of each mass is the same,
but the tensions in the two strings will be hm=s2i = (hkgi , hi hkgi , hkgi) hm=s2i units
dierent.
hkgi + hkgi + hkgi
:
:
:
:
net
Solution:
Let T1 be the tension in the left string and
T2 be the tension in the right string.
Consider the free body diagrams for each
mass:
T1
T2
a
a
T1
a
T2
µm2g
m1g
m3g
021
A rider in a \barrel of fun" nds herself stuck
with her back to the wall.
Answer, Key { Homework 4 { Rubin H Landau
Which diagram correctly shows the forces
m v2 sin 2.
F
=
acting on her?
r
2
1.
3. F = mrv tan 2
4. F = rmcosv 2.
2
5. F = rmsinv 2
6. F = rmtanv 3.
7. F = m g tan 6
8. F = m g cos correct
4.
5.
9. F = mrv cos correct
2
r
10. F = m g2 + vr2
Explanation:
Basic Concepts: To keep an object mov4
ing in a circle requires a force directed toward
the center of the circle; the magnitude of the
force is
2
F = m a = mrv
Also remember:
c
6. None of the others
Explanation:
The normal force of the wall on the rider
provides the centripetal acceleration necessary to keep her going around in a circle. The
downward force of gravity is equal and opposite to the upward frictional force on her.
Car on a Banked Curve
022
c
X
F~ = F~
i
i
Solution: Solution in an Inertial Frame:
Watching from the Point of View of Someone
Standing on the Ground
05:08, trigonometry, numeric, > 1 min.
A curve of radius r is banked at angle so
that a car traveling with uniform speed v can
round the curve without relying on friction to
keep it from slipping to its left or right.
Find the component
X of the net force parallel
F~ k .
to the incline
1. F = mg cot N
mg
θ
The car is performing circular motion with
a constant speed, thus its acceleration is just
the centripetal acceleration, a = v2=r. The
net force on the car is:
c
2
v
F = ma = m r
net
c
Answer, Key { Homework 4 { Rubin H Landau
05:09, trigonometry, numeric, > 1 min.
7
024
N
Fnet
F||
mg
θ
The component of this force parallel to the
incline is
X
F~ k = m g sin = F cos 2
= mrv cos In this reference frame, the car is at rest,
which means that the net force on the car
(taking in consideration the centrifugal force)
is zero. Thus the component of the net \real"
force parallel to the incline is equal to the
component of the centrifugal force along that
direction. Now, the magnitude of the cen2
trifugal force is equal to F = m vr , so
With what frictional force must the road push
on a 1200 kg car if the driver exceeds the
speed for which the curve was designed by
v = 11 km=hr?
Correct answer: 1657:11 N.
Explanation:
The speed of the car is greater, so its centripetal acceleration is greater.
N
net
F||
The free body diagram showing the forces
acting on the car is
N
Fnet
F||
θ
c
Fk = F
net
2
cos = F cos = mrv cos c
023
Explanation:
Fk is the component of the weight of the car
parallel to the incline. Thus
2
m
v
m g sin = Fk = r cos 2
tan = gv r
km=hr )2
= (9:8(150
m=s2 )(71:8 m )
m 2 hr 2
1000
km
3600 s
= 2:46733
= arctan(2:46733) = 67:9375
Car on a Banked Curve
mg
Thus, the net force parallel to the incline is
Fk = m g sin + F
where F is the friction force. On the other
hand, the component of the acceleration parallel to the incline is still
2
ak = (v +rv) cos Then,
2
m g sin + F = m ak = m (v +rv) cos or
2
(
v
+
v
)
F =m
cos , g sin r
= 1657:11 N :
Alternative Solution: Solution in a Noninertial Frame: Watching from the Driver's
Point of View.
Note: The centrifugal \force" is a \ctitious force" that comes from the accelerating
car being a non-inertial frame.
fr
fr
If r = 71:8 m and v = 150 km=hr, what is ?
Correct answer: 67:9375 .
mg
θ
fr
fr
Answer, Key { Homework 4 { Rubin H Landau
8
When the velocity is increased to v + v, hm=si = hi hkm=hri
units
the "no sliding" implies that
hi
m = 1200:0 kg
(11)
2
m
(
v
+
v
)
2
0
2
0
(mg)k + F =
cos = (v + v ) , v
(12)
r
20
= (h41:6667i + h3:05556i) , h41:6667i2 0
2
= 263:966 m2=s2
Using: (m g)k = mrv cos ,
hm2=s2 i = (hm=si + hm=si)2 0 , hm=si2 0 units
2
( )
(
v
+
v
)
f = m cos
(13)
F =m
cos
,
g
sin
:
r
r
i cos (h1:18573i)
= h1200i h263:966
Algorithm
h71:8i
(1)
hdeg
rad i = 57:2958 deg=rad
= 1657:11 N
hkgi hm2=s2 i cos (hradi)
g = 9:8 m=s2 (2)
units
h
N
i
=
50
hmi
r = 71:8 m 100
(3)
90 v = 150 km=hr 160
(4)
:0 v
(5)
v = 1000
3600:0
:0 h150i
= 1000
3600:0
= 41:6667 m=s
=hri
hm=si = hi hkm
units
hi
20
(6)
p = vr g
:6667i2 0
= h41
h71:8i h9:8i
= 2:46733
20
units
hi = hhmmi=hsmi =s2i
= arctan (p)
(7)
= arctan (h2:46733i)
= 1:18573 rad
hradi = arctan (hi)
units
= hdeg
(8)
radi
= h1:18573i h57:2958i
= 67:9375
h i = hradi hdeg=radi
units
v = 11 km=hr 10
(9)
20
:0 v
(10)
v = 1000
3600:0
:0 h11i
= 1000
3600:0
= 3:05556 m=s
fr
:
v
u
u
:
u
:
:
v
fr
u
:
u
:
:
r
r
u
r
:
: