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# Newtons Laws - solutions

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```									AP homework on Newton’s Laws
WS #1 - #1 Ch.5 4th ed. - Newton's Laws
The Problem
1.        A force, F, applied to an object of mass m1
produces an acceleration of 3.0 m/s². The
same force applied to a second object of
mass m2 produces an acceleration of 1.0
m/s².
A.     What is the value of the ratio m1/m2?

F
Draw the                              M1
diagram
M2
F
WS #1 - #1 Ch.5 4th ed. - Newton's Laws
The Solution

a1  3.0 m         a2  1.0 m
s²                 s²
m1 a2         1.0 m
        s²  .33
m2 a1 3.0 m
s²
If m1 and m2 are combined, find their acceleration under the
action of the force F.
a2 m1  m2     m
       1 1
a12   m2       m2

a12  a2   1 

m1 
        
  1.0 m s²  1  .33 
m2 
a12  .75 m
s²
WS #2 - #26 Ch.5 4th ed - Newton's Laws

2.   Find the tension in each cord for the systems
described below. (Neglect the mass of the
cords.)

40.°= 1           = 50.°    60.°
2
T                           T2
1                               T1

90.°
T3                                T2
T3

5.0 kg
10 kg
(a)                       (b)
WS #2 - #26 Ch.5 4th ed - Newton's Laws
Free body Diagrams

T1                    T2           T
1
 = 40.°         = 50.°            1= 60.°    2 = 0.0°   T2
1            2

T3                              T3
WS #2 - #26 Ch.5 4th ed - Newton's Laws
The Working Equations

T3  mg ˆ j
ˆ
T1  T1 cos1 i  T1 sin1 ˆ
j
ˆ
T2  T2 cos  2 i  T2 sin  2 ˆ
j

F    x    ma x
T1 cos1  T2 cos 2  0

T1 cos1  T2 cos 2
cos1
T2  T1
cos 2
WS #2 - #26 Ch.5 4th ed - Newton's Laws
The Working Equations

F   y    ma y

T1 sin 1  T2 sin  2  T3  0
cos1
T1 sin1  T1        sin 2  T3
cos 2
          cos1 sin  2 
T1  sin 1                  T3
            cos 2 

 sin1 cos 2  cos1 sin  2 
T1                                 T3
           cos 2             
WS #2 - #26 Ch.5 4th ed - Newton's Laws
The Working Equations

T3 cos 2
T1 
sin1 cos 2  cos1 sin 2

T3 cos 2
T1 
sin 1   2 

From Symmetry
T3 cos 1
T2 
sin  2  1 
WS #2 - #26 Ch.5 4th ed - Newton's Laws
The Solution Part a
 1  40     2  50     m  5.0 kg


T3  mg   5.0 kg  9.80 m
s²   49 N
T3 cos 2         49 N  cos  50 
T1                                            31.5 N
sin 1   2       sin  40  50 

T3 cos1          49 N  cos  40 
T2                                            37.5 N
sin  2  1       sin  50  40 
WS #2 - #26 Ch.5 4th ed - Newton's Laws
The Solution Part b
1  60      2  0    m  10. kg


T3  mg   10 kg  9.80 m
s²   98 N
T3 cos 2         98 N  cos  0 
T1                                           110 N
sin 1   2       sin  60  0 

T3 cos1          49 N  cos  60 
T2                                            57 N
sin  2  1        sin  0  60 
WS #3 - #28 Ch.5 4th ed - Newton's Laws
The Problem
3.        A 225 N weight is tied to the middle of a strong rope,
and two people pull at opposite ends of the rope in an
attempt to lift the weight.
a.     What is the magnitude F of the force that each person must
apply in order to suspend the weight as shown in the Figure
(P5.28)?
b.     Can they pull in such a way as to make the rope horizontal?
Explain.

T1                    T2

10.°= 1            2= 10.°
T3

225 N
WS #3 - #28 Ch.5 4th ed - Newton's Laws
The Diagram

T1                        T2

10.°= 1              2= 10.°
T3
225 N
WS #3 - #28 Ch.5 4th ed - Newton's Laws
The Solution - The equations come from problem 26
 1  10     2  10   T3  225 N

T3 cos 2         225 N  cos10
T1                  
sin 1   2     sin  10  10 

T1  650 N

T3 cos1          225 N  cos10
T2                  
sin  2  1     sin  10  10 

T2  650 N
WS #4 - #31 Ch.5 4th ed - Newton's Laws
The Problem
4.   A bag of cement hangs from three wires as
shown on page 134 figure p5.31. Two of the
wires make angels of q1 and q2 with the
horizontal. If the system is in equilibrium.

 w cos  2
Show that T2 
sin 1   2 

See problem 26
WS #4 - #31 Ch.5 4th ed - Newton's Laws
The Problem
4.   Given that w = 325 N, q1 = 10°,and q2 = 25°, find the tensions
T1, T2, and T3 in the wires.

1  10        2  25       T3  325 N

T3 cos 2         325 N  cos 25
T1                  
sin 1   2     sin  10  25 

T1  510 N
T3 cos1          325 N  cos10
T2                  
sin  2  1     sin  10  25 
T2  560 N
WS #5 - #17 Ch.5 3th ed - Newton's Laws
The Problem
5.     Two forces F1 = (3i - 5j) N and F2 = (2i + j) N
act on a 1.50 kg body.
1. What are the magnitude and direction of the
acceleration?
2. If the body is initially resting at the origin,
what are the coordinates of its position at t =
4.00 s?
WS #5 - #17 Ch.5 3th ed - Newton's Laws
The Solution

 ˆ
F1  3i  5 ˆ N
j             ˆ j
F2  2i  ˆ N              m  1.50kg

 F  ma
a
F       ˆ     j  
3i  5 ˆ N  2i  ˆ N
ˆ j                  
    ˆ     
5i  4 ˆ N
j
m                 1.50kg                            1.50kg


a  3.33i  2.67 ˆ
ˆ        j       m
s2

a  4.27 m s2 at -38.7
WS #5 - #17 Ch.5 3th ed - Newton's Laws
The Problem and The Solution
B.     If the body is initially resting at the origin, what
are the coordinates of its position at t = 4.00 s?


a  3.33i  2.67 ˆ
ˆ        j           m
s2
vi  0   so  0                t  4.00s
s  so  vi t  at   1
2
2

s  0   0 t    1
2          ˆ
3.33i  2.67 ˆ
j       m
s2    4.00s 
2


s  26.6i  21.4 ˆ m
ˆ        j          
WS #6 - #18 Ch.5 4th ed - Newton's Laws
The Problem and Diagrams
6.   Two forces F1 and F2 act on a 5.0 kg mass. If F1 = 20. N and F2
= 15 N, Find the acceleration in both of the following figures.
WS #6 - #18 Ch.5 4th ed - Newton's Laws
The Solution to diagram 1

m  5.0kg                 ˆ
F1  20.iN       F2  15 ˆN
j

a
F  F        1  F2
m         m

a
        ˆ        
20.i  15 ˆ N
j
5.0kg


a  4.0i  3.0 ˆ
ˆ       j        m
s2
WS #6 - #18 Ch.5 4th ed - Newton's Laws
The Solution to diagram 2

m  5.0kg                ˆ
F1  20.iN


F2  15cos 60 i  15sin 60 ˆ N
ˆ             j           
a
F  F      1    F2
m           m

a
 20.iˆ  15cos 60iˆ  15sin 60 ˆj  N
5.0kg


a  5.5i  2.6 ˆ
ˆ       j      m
s2
WS #7 - #67 Ch.5 3rd ed - Newton's Laws
The Problem
7.   The largest caliber antiaircraft gun
operated by the Luftwaffe during World
War II was the 12.8-cm Flak 40. This
weapon fired a 25.8 kg shell with a
muzzle velocity of 880 m/s. What
propulsive force was necessary to attain
the muzzle velocity within the 6.0 m
barrel? (Assume constant acceleration
and neglect the earth’s gravitational
effect.) (Physics for Scientists and Engineers by Serway, 3rd Edition, Ch 5# 67)
WS #7 - #67 Ch.5 3rd ed - Newton's Laws
The Diagram

6.0 m
25.8 kg              F=?
V = 880 m/s
WS #7 - #67 Ch.5 3rd ed - Newton's Laws
The Solution
d  6.0 m         v f  880 m/s
m  25.8 kg           vo  0    F ?
 F  ma
v 2  vi2  2ad
f

v 2  vi2
a
f

2d
v 2  vi2                  880 m/s   0
2

F m                 25.8 kg 
f

2d                         2  6.0 m 
F  1.7  106 N
WS #8 - #43 Ch.5 4th ed - Newton's Laws
The Problem
8.        A 72.0-kg man stands on a spring scale in an
elevator. Starting from rest, the elevator
ascends, attaining its maximum speed of 1.20
m/s in 0.800s. It travels with this constant
speed for the next 5.00 s. The elevator then
undergoes a uniform acceleration in the
negative y direction for 1.50 s and comes to
rest. What does the spring scale register:
a.     before the elevator starts to move?
b.     during the first 0.80 s.?
c.     while the elevator is traveling at constant speed?
d.     during the time it is slowing down?
WS #8 - #43 Ch.5 4th ed - Newton's Laws
The Diagram

Elevator

Normal
72 kg Force of the scale
1.2m/s

Force
of
Gravity
Scale
WS #8 - #43 Ch.5 4th ed - Newton's Laws
The Solution
a.   What does the spring scale register: before the
elevator starts to move?

m  72.0kg            g  9.80 m s 2          ay  0

F     y    ma y
N  Fg  0
N  Fg   mg    72.0kg   9.8 0 m s 2   706 N
WS #8 - #43 Ch.5 4th ed - Newton's Laws
The Solution
b.   What does the spring scale register: during the first
0.80 s.?

m  72.0kg     g  9.80 m s2          v f  1.20 m
s
t  0.80 s   vi  0
v f  v i  at
v f  vi     1.20 m  0
a                      s
 1.50 m s2
t           0.80s
 Fy  ma y
N  Fg  ma y
N  Fg  ma y

              
N  mg  ma y  m g  a y   72.0kg   9.80 m s2  1.50 m s2   814N
WS #8 - #43 Ch.5 4th ed - Newton's Laws
The Solution
c.     What does the spring scale register: while the
elevator is traveling at constant speed

m  72.0kg              g  9.80 m s2           ay  0

F     y    ma y
N  Fg  0
N  Fg  mg   72.0kg   9.80 m s2   706N
WS #8 - #43 Ch.5 4th ed - Newton's Laws
The Solution
d.    What does the spring scale register: during the time
it is slowing down?

m  72.0kg       g  9.80 m s 2          vf  0      t  1.50s   v i  1.2 0 m
s

v f  v i  at
v f  vi     0  1.20 m
a                        s
  0.800 m s 2
t          1.50s
 F y  ma y
N  Fg  ma y
N  Fg  ma y

          
N  mg  ma y  m g  a y   72.0kg   9.80 m s 2  0.8 00 m s 2   648 N
WS #9 - #45 Ch.5 4th ed - Newton's Laws
The Problem and Solution
9.      A net horizontal force F = A + Bt3 acts on a 3.50-kg object, where A
= 8.60 N and B = 2.50 N/s3. What is the horizontal speed of this
object 3.00 s after it starts from rest?
F = A + Bt 3    A  8.60N          B  2.50 N s3
m  3.50kg      t  3.00s     vi  0       vf  ?
 F  ma
A + Bt 3  ma
A + Bt 3
a
m
3.00       3.00               3.00       3.00
A + Bt 3           A          Bt 3
v f   a  dt               dt        dt          dt
0          0
m            0
m      0
m

 2.50   3.00s         A         B   0  
3.00                                                  4                          4
A    Bt 
4
8.60N                      N
vf   t                          3.00s               s3
    0              
m    4m  0
             3.50kg                  4  3.50kg            m           4m  
                                                                      
v f  7.37 m  14.5 m  21.8 m
s        s        s
WS #10 - #48 Ch.5 4th ed - Newton's Laws
The Problem
10.   A 25.0 kg block is initially at rest on a horizontal
surface. A horizontal force of 75.0 N is required to set
the block in motion. After it is in motion, a horizontal
force of 60.0 N is required to keep the block moving
with constant speed. Find the coefficients of static and
kinetic friction from this information.
WS #10 - #48 Ch.5 4th ed - Newton's Laws
The Solution
m  25kg     Fs  75N        Fk  60.N  s  ?   k  ?
N
 Fy  ma y  0
f             F
 Fy  N  mg  0  N  mg                    25kg

F  x    ma x  0
F  x    F  f 0 f  F                      mg
fs  F       75N
s        s                      .306
N mg  25kg   9.80 m s2 
fk   Fk      60N
k                            .245
N m g  25kg   9.80 s2 
m
WS #11 - #72 Ch.5 4th ed - Newton's Laws
The Problem
11.     In the figure, the man and the
platform together weigh 750
N. Determine how hard the
man would have to pull to hold
himself off the ground. (Or is
it impossible? If so, explain
why.)

F    y    ma  0                  T T

T  T  T  750N  0
T  750N                      750N      T down
on the man’s
arms
WS #12 - #53 Ch.5 4th ed – Newton’s Law
The Problem
11.        A boy drags his 60.0 N sled at constant speed up a 15°
hill. He does so by pulling with a 25 N force on a rope
attached to the sled. If the rope is inclined at 35° to the
horizontal
1.     what is the coefficient of kinetic friction between the sled and the
snow?
2.     At the top of the hill, he jumps on the sled and slides down the
hill. What is the magnitude of his acceleration down the slope?
(Physics for Scientists and Engineers by Serway, 4th Edition, Ch 5# 53)
WS #12 - #53 Ch.5 4th ed – Newton’s Law
The Diagram
WS #12 - #53 Ch.5 4th ed – Newton’s Law
The Solution

mg  60.N       F  25N
 =20°        15
F   y    ma y  0
N  F sin   mg cos   0
N  mg cos   F sin 
F   x    ma x  0
F cos   mg sin   F f  0
F cos   mg sin    N  0
WS #12 - #53 Ch.5 4th ed – Newton’s Law
The Solution

F cos   mg sin     mg cos   F sin    0
   mg cos   F sin    mg sin   F cos 
  mg cos   F sin    F cos   mg sin 
F cos   m g sin 

mg cos   F sin 


 25N  cos 20°   60.N  sin 15  .16
 60.N  cos 15   25N  sin 20°
WS #12 - #53 Ch.5 4th ed – Newton’s Law
The solution Part b
1.    At the top of the hill, he jumps on the sled and slides down the
hill. What is the magnitude of his acceleration down the slope?
(Physics for Scientists and Engineers by Serway, 4th Edition, Ch 5# 53)

F     y    ma y  0
N  m g cos   0
N  m g cos 
WS #12 - #53 Ch.5 4th ed – Newton’s Law
The solution Part b
1.     At the top of the hill, he jumps on the sled and slides down the
hill. What is the magnitude of his acceleration down the slope?
(Physics for Scientists and Engineers by Serway, 4th Edition, Ch 5# 53)

F      x    ma x
mg sin   F f  ma x
mg sin    N  ma x
mg sin    m g cos   ma x
a x  g sin    g cos   g  sin    cos  

a x   9.80 m s 2   sin 15   .16  cos 15 
a x  1.0 m s 2
WS #13 - #39 Ch.5 4th ed - Newton's Laws
The Problem
13.       A small bug is placed between two blocks of masses m1 and m2
(m1 > m2) on a frictionless table. A horizontal forcec F, can be
applied to either m1, as in the first figure below or m2 as in the
second figure below. For which of these two cases does the bug
have a greater chance to surviving? Show! (Physics for Scientists and
Engineers by Serway, 4th Edition, Ch 5# 39)

F            m1                                     m1                F
m2                                   m2
WS #13 - #39 Ch.5 4th ed – Newton’s Laws
The Solution

a
F               F
m m            1  m2

When force is on m1                      When force is on m1
m1 exerts a force F2 on m2             m2 exerts a force F1 on m1
F2  m2 a                                  F1  m1a
m2 F                                       m1 F
F2                                        F1 
m1  m2                                    m1  m2
if m2  m1
F2  F1
WS #14 - #50 Ch.5 4th ed – Newton’s Laws
The Problem

14. What force must be exerted on Block A
in order for Block B not to fall. The
coefficient of static friction between
blocks A and B is 0.55, and the horizontal
surface is frictionless. (Physics for Scientists and
Engineers by Serway, 4th Edition, Ch 5# 50)
WS #14 - #50 Ch.5 4th ed – Newton’s Laws
The Diagram

Block B
Block A          10.0 kg

100. kg
WS #14 - #50 Ch.5 4th ed – Newton’s Laws
The Free Body Diagrams
WS #14 - #50 Ch.5 4th ed – Newton’s Laws
The Solution – 10 kg mass

F    y   0             F    x    mb a
F f  mb g  0              N  mb a
F f  mb g               mb g
 mb a
 N  mb g                 
g
N
mb g                  a
                         
WS #14 - #50 Ch.5 4th ed – Newton’s Laws
The Solution – both mass together
g  9.80 m s 2   .55 ma  100 kg m b  10.0 kg
g
a

F   x    a m
g
F          ma  mb 

9.80 m s 2
F             100 kg  10.0 kg 
.5 5
F  1960 N
WS #15 - #71 Ch.5 4th ed – Newton’s Laws
The Problem
15.        Three baggage carts of masses m1, m2, and m3 are
towed by a tractor of mass M along an airport apron.
The wheels of the tractor exert a total frictional force F
on the ground as shown. In the following, express your
answers in terms of F, M, m1, m2, m3 and g.
a.     What are the magnitude and direction of the horizontal force
exerted on the tractor by the ground?
b.     What is the smallest value of the coefficient of static friction
that will prevent the wheels from slipping? Assume that each
of the two drive wheels on the tractor bears 1/3 of the tractor’s
weight.
c.     What is the acceleration, a, of the system (tractor plus baggage
carts)?
d.     What are the tensions T1, T2, and T3 in the connecting cables?
e.     What is the net force on the cart of mass m2?
(Physics for Scientists and Engineers by Serway, 4th Edition, Ch 5# 71)
WS #15 - #71 Ch.5 4th ed – Newton’s Laws
The Solution Part a
a.    What are the magnitude and direction of the horizontal
force exerted on the tractor by the ground?
a.   From Newton’s Third Law since the force of the tractor on
the ground is F, the magnitude of the force of the ground on
the tractor must also be F but in the opposite direction.
Therefore the force exerted on the tractor is F to the right.
WS #15 - #71 Ch.5 4th ed – Newton’s Laws
The Solution Part b
b.   What is the smallest value of the coefficient of static friction
that will prevent the wheels from slipping? Assume that each of
the two drive wheels on the tractor bears 1/3 of the tractor’s
weight.
N
F   y    ma y  0
N  2 mg  0
3

N  2 mg                                                                F
3                      F  f friction   N                  2/3 mg
F     F
    2
N 3 mg
3F

2m g
WS #15 - #71 Ch.5 4th ed – Newton’s Laws
The Solution Part c
c.   What is the acceleration, a, of the system (tractor plus
baggage carts)?

a
F                external

m
F
a
M  m1  m2  m3
WS #15 - #71 Ch.5 4th ed – Newton’s Laws
The Solution Part d
d.   What are the tensions T1, T2, and T3 in the connecting
cables?

There is one force on m1          T2 is accelerating two masses m1 and m2
F  1
 m1a
 F  m     1    m2  a
F
T1  m1
M  m1  m2  m3                      T2 
 m2  m1  F
M  m1  m2  m3
m1 F
T1 
M  m1  m2  m3
WS #15 - #71 Ch.5 4th ed – Newton’s Laws
The Solution Part d

T3 is accelerating three masses m1 and m2 and m3
 F  m   1    m2  m3  a

T3 
 m1  m2  m3  F
M  m1  m2  m3
WS #15 - #71 Ch.5 4th ed – Newton’s Laws
The Solution Part e
e.   What is the net force on the cart of mass m2?

The net force is  F
F    2
 m2a
F
F2  m2 M  m  m  m
1   2   3

m2F
F2  M  m  m  m
1   2   3
WS #16 - #37 Ch.5 4th ed – Newton’s Laws
The Problem
1.   In the system on the next slide, a
horizontal force Fx acts on a 8.00 kg
mass.
a. For what values of Fx does the 2.00 kg mass
accelerate upward?
b. For what values of Fx is the tension in the cord
zero?
c. Plot the acceleration of the 8.00 kg mass versus
Fx. Include values of Fx from -100 N to +100 N.
(Physics for Scientists and Engineers by Serway, 4th Edition, Ch 5# 37)
WS #16 - #37 Ch.5 4th ed – Newton’s Laws
The Diagram
WS #16 - #37 Ch.5 4th ed – Newton’s Laws
The Diagram
y

x

x

y
WS #16 - #37 Ch.5 4th ed – Newton’s Laws
The Free Body Diagram

y                       y

x
x
WS #16 - #37 Ch.5 4th ed – Newton’s Laws
Find F such that a > 0

m  2.0 kg        4m  8.0 kg
F    x    ma x
F  mg   m  4m  a
a  0 if F  mg  0
F  mg
F   2.0 kg   9.80 m s2   19.6 N
WS #16 - #37 Ch.5 4th ed – Newton’s Laws
Find F such that T = 0

For m                             For 4m

 F  ma                          F  ma
F  T  4ma
T  mg  m a
F  0  4ma
0  mg  ma
F  4m   g   4 2.0 kg  9.80 m s2 
a  g
F  78.4 N
WS #16 - #37 Ch.5 4th ed – Newton’s Laws
Part C – The Plot
WS #17 - #47 Ch.5 4th ed – Newton’s Laws
The Problem
17.   Two blocks of masses m and
4m are connected by a light
string passing over a fixed
pulley as shown at the right.
The 4m mass is on a smooth
frictionless surface and is
pushed toward the left by a
constant horizontal force F.
a.     What value of F will give the
system an acceleration of 0.400 g
b.     Find the tension in the string
when F has the value determined
in (A).
c.     What happens when F>4.00 mg?
(Physics for Scientists and Engineers
by Serway, 3rd Edition, Ch 5# 47)
WS #17 - #47 Ch.5 4th ed – Newton’s Laws
The Diagram
y

x

x

y
WS #17 - #47 Ch.5 4th ed – Newton’s Laws
The Solution
(A) What value of F will give the system an acceleration of
0.400 g

a system   
F
m
 F  mg
0.400 g 
4m  m
0.400 g  4m  m    F  mg
F  0.400 g  5m   mg  0.400 g  5m   mg  2 m g  mg
F  mg
WS #17 - #47 Ch.5 4th ed – Newton’s Laws
The Solution
(b) Find the tension in the string when F has the value
determined in (A).

a
F                                         4m
y
m
F  T
0.400 g 
4m                                            x
 F  T   0.400 g  4m
T  4m  0.400 g   F
T  4m  0.400 g   mg
T  .600mg
WS #17 - #47 Ch.5 4th ed – Newton’s Laws
The Solution
(c) What happens when F>4.00 mg?

a
F                             4m
m                                 y
F  T
a
4m
 F  T  4ma                                x
T  4ma  F
T  4ma  4mg  4m  a  g 
but amax  g  T  0
The String goes slack
WS #18 - #46 Ch.5 4th ed – Newton’s Laws
The Problem
18. Mass m1 on a smooth horizontal table is
connected to mass m2 through a very
light pulley P1 and a light fixed pulley P2
as shown.
a.   If a1 and a2 are the accelerations of m1 and m2,
respectively, what is the relationship between these
accelerations?
b.   Express the tensions in the strings, and
c.   the accelerations a1 and a2 in terms of the masses m1, m2,
and g.
(Physics for Scientists and Engineers by Serway, 4th Edition, Ch 5# 46)
WS #18 - #46 Ch.5 4th ed – Newton’s Laws
The Diagram
T’              P1
T     P2
m1      T’

T

m2
m2g
WS #18 - #46 Ch.5 4th ed – Newton’s Laws
If a1 and a2 are the accelerations of m1 and m2, respectively,
what is the relationship between these acceleration

A) Find a1 in terms of a 2 .
2d1           2d 2
a1 = 2        a2 = 2        But when m 2 moves a distance d 2 ,
t             t
Pulley P1 moves a distance d 2 ,
which means that the string
between m1 and P1 gets a distance
d 2 shorter. This means that m1 moves 2d 2 .
2d1 2  2d 2     2d
a1 = 2 =            = 2 2 2 = 2a 2
t      t2         t
a1 = 2a 2
WS #18 - #46 Ch.5 4th ed – Newton’s Laws
Express the (B) tensions in the strings,
For m1
F     1   = m 1 a1
T1 = m1a1
T1
a1 =
m1
For P1
F  p1    = m p1 ap1 = 0
T2 - T1 - T1 = 0
T2
T1 =
2
WS #18 - #46 Ch.5 4th ed – Newton’s Laws
Express the (B) tensions in the strings,
For m 2
F    2   = m 2a2
T1
a1            m1
m 2 g - T2 = m 2 a 2 = m 2     = m2
2             2
T2
T1                  m 2 T2
m 2 g - T2 = m 2        = m2 2 =
2m1           2m1     4m1
m 2 T2
m 2 g - T2 =
4m1
4m1m 2 g - 4m1 T2 = m 2 T2
4m1m 2 g
T2 =
m 2 + 4m1
WS #18 - #46 Ch.5 4th ed – Newton’s Laws
Express the (B) tensions in the strings,

4m1m 2 g
T2   m 2 + 4m1   2m1m 2 g
T1 =    =           =
2        2       m 2 + 4m1
WS #18 - #46 Ch.5 4th ed – Newton’s Laws
(c) the accelerations a1 and a2 in terms of the masses
m1, m2, and g

 2m1m 2 g 
           
a1 =
T1
=  m 2 + 4m1  = 2m 2 g
m1         m1       m 2 + 4m1
2m 2 g
a1   m 2 + 4m1      m2g
a2 =    =           =
2        2       m 2 + 4m1
WS #19 - #85 Ch.5 4th ed - Newtons Laws
The Problem
19.   The three blocks to the right are connected by mass-
less strings that pass over massless frictionless pulleys.
The acceleration of the system is 2.35 m/s2 to the left
and the surfaces are rough. Mass 1 is 10.0 kg, mass 2
is 5.00 kg, and mass 3 is 3.00 kg. Find:
a.   the tensions in the strings and
b.   the coefficient of kinetic friction between blocks and surfaces.
(Assume the same m for both blocks)

(Physics for Scientists and Engineers by Serway, 4th Edition, Ch 5# 85)

5.00
kg
T1
T2       3.00
kg
10.0
kg
WS #19 - #85 Ch.5 4th ed - Newtons Laws
The Diagram
WS #19 - #85 Ch.5 4th ed - Newtons Laws
The Free body Diagram

They all acceleration as a unit. The tensions
are internal.
WS #19 - #85 Ch.5 4th ed - Newtons Laws
(b) the coefficient of kinetic friction between blocks and surfaces.

m1  10.0kg             m2  5.00kg               m3  3.00kg
a  2.35 m s2   25.0

For body 2                             For body 3
F    y     ma y  0                 F      y    ma y  0
N 2  m2 g  0                      N 3  m3 g cos   0
N 2  m2 g                          N 3  m3 g cos 
WS #19 - #85 Ch.5 4th ed - Newtons Laws
(b) the coefficient of kinetic friction between blocks and surfaces.

F      x
  ma x
F f 3  m3 g sin   F f 2  m1 g   m1  m2  m3  a
 N 3  m3 g sin    N 2  m1 g   m1  m2  m3  a
  N 2  N 3    m1  m2  m3  a  m1 g  m3 g sin 


 m1  m2  m3  a  m1 g  m3 g sin 
N2  N3


 m1  m2  m3  a  m1 g  m3 g sin 
m3 g cos   m2 g
WS #19 - #85 Ch.5 4th ed - Newtons Laws
(b) the coefficient of kinetic friction between blocks and surfaces.


 m1  m2  m3  a  m1 g  m3 g sin 
m3 g cos   m2 g

  10.0  5.00  3.00    2.35    10.0   9.80    3.00   9.80  sin 25.0   kg m s 2
                                                                                        
 9.8 0 m s 2    3.00kg  cos 25.0    5.0 0kg  
  .57 2
WS #19 - #85 Ch.5 4th ed - Newtons Laws
(a) the tensions in the strings

Find T1 using body 1
F  m a        1

T1  m1 g  m1a
T1  m1a  m1 g  m1  a  g 
T1   10.0kg   2.35 m s 2  9.80 m s 2 
T1  74.5N
WS #19 - #85 Ch.5 4th ed - Newtons Laws
(a) the tensions in the strings

Find T2 using body 3
F  m a      3

m3 g sin   F f 3  T2  m3 a
T2   m3 a  m3 g sin    m3 g cos 
T2  m3   a  g  sin    cos   


T2   3.00kg    2.35 m s 2   9.80 m s 2  sin 25.0   .572  cos 25.0    
T2  34.7N
WS #20 - #65 Ch.5 4th ed - Newtons Laws

20. A mass M is held in place by an applied
force FA and a pulley system as shown at
the right. The pulleys are mass-less and
frictionless. Find the tension in each
section of rope, T1, T2, T3, T4, and T5 and
the applied force FA.
WS #20 - #65 Ch.5 4th ed - Newtons Laws
Starting solutions

Mass M
y

x
On mass M
F m    Ma  0
T5  Mg  0
T5  Mg
WS #20 - #65 Ch.5 4th ed - Newtons Laws
The Solution - T2,T3

On Pulley P2                    Pulley 2
T2  T3  T5  0                     y
T2 and T3 are the same string  T2  T3
T3  T3  T5  2T3  T5  0
x
T5 Mg
T3     
2     2
Mg
T3  T2 
2
WS #20 - #65 Ch.5 4th ed - Newtons Laws
The Solution – T1,T4
On Pulley P1
T1  T2  T3  T4  0                Pulley 1
T1 , T2 and T3 are the same string  T1  T2  T3       y
Mg
T1 
2                                    x
Mg Mg Mg
                  T4  0
2    2       2
3 Mg
T4 
2
Mg
F  T1 
2
Newton’s Law 1998 AP Question - Solution
From the 1998 Physics C Mech AP Test

1.   Block 1 of mass ml is placed
on block 2 of mass m2, which
m
1

is then placed on a table. A
string connecting block 2 to a
m
2

hanging mass M passes over
a pulley attached to one end
of the table, as shown to the
right. The mass and friction
M
of the pulley are negligible.
The coefficients of friction
between blocks 1 and 2 and
between block 2 and the
tabletop are nonzero and are
given in the following table.
Newton’s Law - Solution
From the 1998 Physics C Mech AP Test

Coefficient Between                       Coefficient Between
Blocks 1 and 2                        Block 2 and the Tabletop
Static                         s1                                s2

Kinetic                        k1                                k2

friction, and g, the acceleration due to gravity.
Newton’s Law - Solution
From the 1998 Physics C Mech AP Test

(a)      Suppose that the value of M is small enough that                          m1

the blocks remain at rest when released. For each                         m2

of the following forces, determine the magnitude
of the force and draw a vector on the block
provided to indicate the direction of the force if it                          M
is nonzero.
i.       The normal force N1 exerted on block 1 by block 2

F
Forces
on
m1      N1
N1                             y   =0
N1 -m1g = 0
m1g
m         1
N 1 = m 1g
Newton’s Law - Solution
From the 1998 Physics C Mech AP Test

ii.       The friction force f1 exerted                                     m
1

on block 1 by block 2                                             m
2

F    x   = max = m1 (0) = 0                                                    M

f1 = 0
m      1

This is true since there is no motion and friction only opposes motion, there is
no horizontal force for the friction to oppose.
Newton’s Law - Solution
From the 1998 Physics C Mech AP Test

iii.    The force T exerted on block                                        m1

2 by the string                                                     m2

M
Forces
on
m2          N2
F      = ma x = 0
f2       T           m         2
T                  x

Mg - T = 0
N1
m2g
T = Mg
Newton’s Law - Solution
From the 1998 Physics C Mech AP Test

m
1

iv.   The normal force N2 exerted                                m
2
on block 2 by the tabletop

Forces
on
m2          N2
F           y    =0                                M

f2      T        N 2 - N1 - m 2 g = 0
N1
m2g      N 2 = N1 + m 2 g
N 2 = m1g + m 2 g                             N2

N 2 = g(m1 + m 2 )
m
2
Newton’s Law - Solution
From the 1998 Physics C Mech AP Test
m
1

m
2
v.    The friction force f2 exerted
on block 2 by the tabletop
Forces
on
F             x   =0       M
m2          N2
T - f2 = 0
f2       T
N1
m2g
f 2 = T = Mg

f2                        m          2
Newton’s Law - Solution
From the 1998 Physics C Mech AP Test

m  1

(b)   Determine the largest value
of M for which the blocks
m  2

can remain at rest.
From (a) v. above
M
f 2max = Mg
f 2max              s2 N 2
M max =                    =
g                       g
but from (a) iv. above N 2 = g(m1 + m 2 )
s2 g(m1 + m 2 )
M max =                                           = s2 (m1 + m 2 )
g
Newton’s Law - Solution                                m
From the 1998 Physics C Mech AP Test                  1

m   2

(c)     Now suppose that M is large
enough that the hanging block Forces
descends when the blocks are                                        Forces
on                                    on              M
released. Assume that blocks 1 m                              N2
2                                   M
and 2 are moving as a unit (no
slippage). Determine the
f2                        T
magnitude a of their                 N1                                 T        Mg
acceleration.                                                 m2g

F    2x    a2 x  M
Mg  f 2  a2 x  m1  m2  M                                 M  2k  m1  m2 
a2 x  g
Mg  2 k g  m1  m2                                   m1  m2  M
a2 x   
m1  m2  M
Newton’s Law - Solution
From the 1998 Physics C Mech AP Test

(d) Now suppose that M is large enough that as the hanging block descends, block
1 is slipping on block 2. Determine each of the following.

m
1

i.      The magnitude a1 of the                                          m
2

acceleration of block 1

Forces
on
m1          N1                                  Fx  m1a x              M
f1                               f1 m1 g 1k
ax     
m1g                                    m1   m1
a x  g 1k
Newton’s Law - Solution
From the 1998 Physics C Mech AP Test
(d) Now suppose that M is large enough that as the hanging block descends, block 1 is
slipping on block 2. Determine each of the following.

m  1

m  2

ii.        The magnitude a2 of the
acceleration of block 2
Forces                      Forces                                                                M
on                          on
m2              N2          M                                             N 2  m2 g  N 1
f2        T                                                        N 2  m2 g  m1 g
f1                       T               Mg
N 2  g  m2  m1 
N1
m2g
Newton’s Law - Solution
From the 1998 Physics C Mech AP Test
(d) Now suppose that M is large enough that as the hanging block descends, block 1 is
slipping on block 2. Determine each of the following.        m              1

m
2

ii.        The magnitude a2 of the                   Forces                           Forces
acceleration of block 2                     on                               on                M
m2                     N2        M
N 2  g  m2  m1                                        f2             T
f1           N1             T       Mg
F     x    ax  m                                                   m2g

ax    
F    x

Mg  f1  f 2
m            m2  M
Mg  m1 g 1k  2 k g  m1  m2 
Mg  m1 g 1k  2 k N 2             ax 
ax                                                          m2  M
m2  M
g  M  m11k  2 k  m1  m2  
ax 
m2  M

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