Transcript Chapter 5 - UCF Physics

Chapter 5 – Force and Motion I
I.
Newton’s first law.
II. Newton’s second law.
III. Particular forces:
- Gravitational
- Weight
- Normal
- Friction
- Tension
IV. Newton’s third law.
Force
• Forces are what cause any change in the
velocity of an object
– A force is that which causes an acceleration
• The net force is the vector sum of all the forces
acting on an object
– Also called total force, resultant force, or
unbalanced force
Zero Net Force
• When the net force is equal to zero:
– The acceleration is equal to zero
– The velocity is constant
• Equilibrium occurs when the net force is equal
to zero
– The object, if at rest, will remain at rest
– If the object is moving, it will continue to move
at a constant velocity
Classes of Forces
• Contact forces involve
physical contact
between two objects
• Field forces act
through empty space
– No physical contact is
required
Fundamental Forces
• Gravitational force
– Between two objects
• Electromagnetic forces
– Between two charges
• Nuclear (strong) force
– Between subatomic particles
• Weak forces
– Arise in certain radioactive decay processes
• A spring can be used
to calibrate the
magnitude of a force
• Forces are vectors,
so you must use the
rules for vector
addition to find the net
force acting on an
object
Newton mechanics laws cannot be applied when:
1) The speed of the interacting bodies are a fraction of the
speed of light Einstein’s special theory of relativity.
2) The interacting bodies are on the scale of the atomic
structure  Quantum mechanics
I. Newton’s first law: If no net force acts on a body, then the
body’s velocity cannot change; the body
cannot accelerate  v = constant in
magnitude and direction.
Principle of superposition: when two or more forces act
on a body, the net force can be obtained by adding the
individual forces as vectors.
Newton’s First Law
• If an object does not interact with other
objects, it is possible to identify a
reference frame in which the object has
zero acceleration
– This is also called the law of inertia
– It defines a special set of reference frames
called inertial frames,
• We call this an inertial frame of reference
Inertial reference frame: where Newton’s laws hold.
Inertial Frames
• Any reference frame that moves with constant
velocity relative to an inertial frame is itself an
inertial frame
• A reference frame that moves with constant
velocity relative to the distant stars is the best
approximation of an inertial frame
– We can consider the Earth to be such an inertial
frame although it has a small centripetal acceleration
associated with its motion
Newton’s First Law – Alternative Statement
• In the absence of external forces, when viewed
from an inertial reference frame, an object at rest
remains at rest and an object in motion
continues in motion with a constant velocity
– Newton’s First Law describes what happens in the
absence of a force
– Also tells us that when no force acts on an object, the
acceleration of the object is zero
Inertia and Mass
• The tendency of an object to resist any
attempt to change its velocity is called
inertia
• Mass is that property of an object that
specifies how much resistance an object
exhibits to changes in its velocity
• Mass is an inherent property of an object
• Mass is independent of the object’s
surroundings
• Mass is independent of the method used
to measure it
• Mass is a scalar quantity
• The SI unit of mass is kg
Mass vs. Weight
• Mass and weight are two different
quantities
• Weight is equal to the magnitude of the
gravitational force exerted on the object
– Weight will vary with location
Newton’s Second Law
• When viewed from an inertial frame, the
acceleration of an object is directly
proportional to the net force acting on it
and inversely proportional to its mass
– Force is the cause of change in motion, as
measured by the acceleration
• Algebraically, SF = m a
II. Newton’s second law: The net force on a body is
equal to the product of the body’s mass and its
acceleration.


Fnet  ma
Fnet , x  max , Fnet , y  ma y , Fnet , z  maz
The acceleration component along a given axis is caused
only by the sum of the force components along the same
axis, and not by force components along any other axis.
Units of Force
System: collection of bodies.
External force: any force on the bodies inside the system.
III. Particular forces:
Gravitational: pull directed towards a second body, normally
the Earth 


Fg  mg
Weight: magnitude of the upward force needed to balance
the gravitational force on the body due to an astronomical
body 
W  mg
Normal force: perpendicular force on a body from a surface
against which the body presses.
N  mg
Frictional force: force on a
body when the body
attempts to slide along a
surface. It is parallel to the
surface and opposite to the
motion.
Tension: pull on a body directed away from the body
along a massless cord.
Newton’s Third Law
• If two objects interact, the force F12
exerted by object 1 on object 2 is equal in
magnitude and opposite in direction to the
force F21 exerted by object 2 on object 1
• F12 = - F21
– Note on notation: FAB is the force exerted by
A on B
Newton’s Third Law, Alternative
Statements
• Forces always occur in pairs
• A single isolated force cannot exist
• The action force is equal in magnitude to the
reaction force and opposite in direction
– One of the forces is the action force, the other is the
reaction force
– It doesn’t matter which is considered the action and
which the reaction
– The action and reaction forces must act on different
objects and be of the same type
Action-Reaction Examples
• The force F12 exerted
by object 1 on object
2 is equal in
magnitude and
opposite in direction
to F21 exerted by
object 2 on object 1
• F12 = - F21
Newton’s third law: When two bodies interact,
the forces on the bodies from each other are always
equal in magnitude and opposite in direction.


FBC   FCB
QUESTIONS
Q2. Two horizontal forces,


F1  (3N )iˆ  (4 N ) ˆj and F2  (1N )iˆ  (2 N ) ˆj
pull a banana split across a frictionless counter. Without using
a calculator, determine which of the vectors in the free body
diagram below best represent: a) F1, F2; b) What is the net
force component along (c) the x-axis, (d) the y-axis? Into
which quadrant do (e) the net-force vector and (f) the split’s
acceleration vector point?
QUESTIONS
Q2. Two horizontal forces,

F1  (3 N )iˆ  ( 4 N ) ˆj and

F2  (1N )iˆ  ( 2 N ) ˆj
pull a banana split across a frictionless counter. Without using
a calculator, determine which of the vectors in the free body
diagram below best represent: a) F1, F2; b) What is the net
force component along (c) the x-axis, (d) the y-axis? Into
which quadrant do (e) the net-force vector and (f) the split’s
acceleration vector point?

 
Fnet  F1  F2  (2 N )iˆ  (6 Nˆj )
Same quadrant, 4
Q1. The figure below shows overhead views of four situations
in which forces act on a block that lies on a frictionless floor. If
the force magnitudes are chosen properly, in which situation it
is possible that the block is (a) stationary and (b) moving with
constant velocity?
Q1. The figure below shows overhead views of four situations
in which forces act on a block that lies on a frictionless floor. If
the force magnitudes are chosen properly, in which situation it
is possible that the block is (a) stationary and (b) moving with
constant velocity?
ay≠0
ay≠0
a=0
a=0
Q5. In which situations does the object acceleration have
(a) an x-component, (b) a y component? (c) give the direction
of a.
Q5. In which situations does the object acceleration have
(a) an x-component, (b) a y component? (c) give the direction
of a.
Fnet
Fnet
Q. A body suspended by a rope has a weigh W of 75N. Is T
equal to, greater than, or less than 75N when the body is
moving downward at (a) increasing speed and (b)
decreasing speed?
Fg
Q. A body suspended by a rope has a weigh W of 75N. Is T
equal to, greater than, or less than 75N when the body is
moving downward at (a) increasing speed and (b)
decreasing speed?
(a) Increasing speed:
vf >v0 a>0




Fnet  Fg  T  ma  T  m( g  a)
T < Fg
(b) Decreasing speed: vf < v0 a<0




Fnet  Fg  T  ma  T  m( g  a)
T > Fg
Fg
Q8. The figure below shows a train of four blocks being pulled
across a frictionless floor by force F. What total mass is
accelerated to the right by (a) F, (b) cord 3 (c) cord 1?
(d) Rank the blocks according to their accelerations, greatest
first. (e) Rank the cords according to their tension, greatest
first.
Q8. The figure below shows a train of four blocks being pulled
across a frictionless floor by force F. What total mass is
accelerated to the right by (a) F, (b) cord 3 (c) cord 1?
(d) Rank the blocks according to their accelerations, greatest
first. (e) Rank the cords according to their tension, greatest
first.
T2
T3
T1
(a) F pulls mtotal= (10+3+5+2)kg = 20kg
(b) Cord 3  T3  m=(10+3+5)kg = 18kg
(c) Cord 1  T1  m= 10kg
(d) F=ma  All tie, same acceleration
(e) T3 > T2 >T1
Q. A toy box is on top of a heavier dog house, which sits on a wood floor.
These objects are represented by dots at the corresponding heights, and six
vertical vectors (not to scale) are shown. Which of the vectors best
represents (a) the gravitational force on the dog house, (b) on the toy box,
(c) the force on the toy box from the dog house, (d) the force on the dog
house from the toy box, (e) force on the dog house from the floor, (f) the
force on the floor from the dog house? (g) Which of the forces are equal in
magnitude? Which are (h) greatest and (i) least in magnitude?
10
9
8
•
>
7
Q. A toy box is on top of a heavier dog house, which sits on a wood floor.
These objects are represented by dots at the corresponding heights, and six
vertical vectors (not to scale) are shown. Which of the vectors best
represents (a) the gravitational force on the dog house, (b) on the toy box,
(c) the force on the toy box from the dog house, (d) the force on the dog
house from the toy box, (e) force on the dog house from the floor, (f) the
force on the floor from the dog house?
(a) Fg on dog house: 4 or 5
(b) Fg on toy box: 2
10
9
8
•
>
7
(c) Ftoy from dog house: 10
(d) Fdog-house from toy box: 9
(e) Fdog-house from floor: 8
(f) Ffloor from dog house: 6
5. There are two forces on the 2 kg box in the overhead view of
the figure below but only one is shown. The figure also shows
the acceleration of the box. Find the second force (a) in unitvector notation and as (b) magnitude and (c) direction.
F2
5. There are two forces on the 2 kg box in the overhead view of the figure
below but only one is shown. The figure also shows the acceleration of the
box. Find the second force (a) in unit-vector notation and as (b) magnitude
and (c) direction.
F2
5. There are two forces on the 2 kg box in the overhead view of the figure
below but only one is shown. The figure also shows the acceleration of the
box. Find the second force (a) in unit-vector notation and as (b) magnitude
and (c) direction.
F2

a  (12 cos 240 iˆ  12 sin 240 ˆj )m / s 2  (6iˆ  10.39 ˆj )m / s 2


Fnet  ma  2kg(6iˆ  10.39 ˆj )m / s 2  (12iˆ  20.78 ˆj ) N

 

F  F  F  20iˆ  F
net
1
2
2
5. There are two forces on the 2 kg box in the overhead view of the figure
below but only one is shown. The figure also shows the acceleration of the
box. Find the second force (a) in unit-vector notation and as (b) magnitude
and (c) direction.
Fnet
 

 F1  F2  20iˆ  F2
Fnet , x  12 N  F2 x  20 N
F2
F2 x  32 N
Fnet , y  20.78 N  F2 y
F2  (32iˆ  20.78 ˆj ) N
F2  32  21  38.27 N
2
2
 20.78




tan  
 33 or 180  33  213
 32
Rules to solve Dynamic problems
- Select a reference system.
- Make a drawing of the particle system.
- Isolate the particles within the system.
- Draw the forces that act on each of the isolated bodies.
- Find the components of the forces present.
- Apply Newton’s second law (F=ma) to each isolated particle.
23. An electron with a speed of 1.2x107m/s moves horizontally into a
region where a constant vertical force of 4.5x10-16N acts on it. The mass
of the electron is m=9.11x10-31kg. Determine the vertical distance the
electron is deflected during the time it has moved 30 mm horizontally.
F
Fg
dy
v0
dx=0.03m
23. An electron with a speed of 1.2x107m/s moves horizontally into a
region where a constant vertical force of 4.5x10-16N acts on it. The mass
of the electron is m=9.11x10-31kg. Determine the vertical distance the
electron is deflected during the time it has moved 30 mm horizontally.
F
Fg
dy
v0
dx=0.03m
d x  vx t  0.03m  (1.2 107 m / s)t  t  2.5ns
Fnet  may  F  Fg  4.5 1016 N  (9.111031kg)(9.8m / s 2 )
Fnet  (9.111031kg)a y  a y  4.94 1014 m / s 2
d y  voyt  0.5a y t 2  0.5  (4.94 1014 m / s 2 )  (2.5 109 s) 2  0.0015m
13. In the figure below, mblock=8.5kg and θ=30º. Find (a) Tension
in the cord. (b) Normal force acting on the block. (c) If the cord is
cut, find the magnitude of the block’s acceleration.
N
T
Fg
13. In the figure below, mblock=8.5kg and θ=30º. Find (a) Tension
in the cord. (b) Normal force acting on the block. (c) If the cord is
cut, find the magnitude of the block’s acceleration.
N
T
Fg
(a) a  0  T  Fgx  mg sin 30  (8.5kg)(9.8m / s 2 )0.5  41.65N
(b) N  Fgy  mg cos sin 30  72.14 N
(c) T  0  Fgx  ma  41.65 N  8.5a  a  4.9m / s 2
Friction
• Related to microscopic interactions of surfaces
• Friction is related to force holding surfaces
together
• Frictional forces are different depending on
whether surfaces are static or moving
Kinetic Friction
• When a body slides over a surface there is a
force exerted by the surface on the body in the
opposite direction to the motion of the body
• This force is called kinetic friction
• The magnitude of the force depends on the
nature of the two touching surfaces
• For a given pair of surfaces, the magnitude of
the kinetic frictional force is proportional to the
normal force exerted by the surface on the body.
Kinetic Friction
The kinetic frictional
force can be written as
F fr  m k FN
Where mk is a
constant called the
coefficient of kinetic
friction
The value of mk
depends on the
surfaces involved
+
+
FN = mg
Fp
m=
Ffr = mFN
20.0 kg
Fg = -mg
Static Friction
A frictional force can arise even if the body
remains stationary
– A block on the floor - no frictional force
FN = -mg
m
Fg = mg
Static Friction
– If someone pushes the desk (but it does not
move) then a static frictional force is exerted
by the floor on the desk to balance the force
of the person on the desk
FN = -mg
Fp
m
Ffr = mFN
Fg = mg
Static Friction
– If the person pushes with a greater force and
still does not manage to move the block, the
static frictional force still balances it
FN = -mg
Fp
m
Ffr = mFN
Fg = mg
Static Friction
– If the person pushes hard enough, the block
will move.
• The maximum force of static friction has been
exceeded
FN = -mg
Fp
m
Ffr = mFN
Fg = mg
Static Friction
• The maximum force of static friction is
given by
F fr (max)  m s FN
• Static friction can take any value from zero
to msFN
– In other words
F fr  m s FN
Forces on an incline
• Often when solving
problems involving
Newton’s laws we
will need to deal
with resolving
acceleration due to
gravity on an
inclined surface
Forces on an incline
What normal force
does the surface exert?
mgsin
mgcos
W = mg
y

x
Forces on an incline
F
F
x
 mg sin 
y
 N  mg cos 
Forces on an incline
F
F
x
 mg sin   ma
y
 N  mg cos   0
Equilibrium
Forces on an incline
• If the car is just
stationary on the
incline what is the
(max) coefficient of
static friction?
 Fx  mg sin   ms N  ma  0
F
y
 N  mg cos   0
mg sin   m s N  m s mg cos 
sin 
ms 
 tan 
cos 
‘Equilibrium’ problems (SF = 0)
0
0
F

T
cos
53

T
cos
37
0
 x 2
1
0
0
F

T
sin
37

T
sin
53
 T3  0
 y 1
2
Connected Object problems
• One problem often posed is how to work
out acceleration of a system of masses
connected via strings and/or pulleys
– for example - Two blocks are fastened to the
ceiling of an elevator.
Connected masses
• Two 10 kg blocks are
strung from an elevator
roof, which is accelerating
up at 2 m/s2.
T1
m1=10kg
T2
m2=10kg
a
2 m/s2
Connected masses
• Two 10 kg blocks are
strung from an elevator
roof, which is accelerating
up at 2 m/s2.
T1
m1=10kg
T2

F  m a
F1  m1aup  T1  T2  m1g
2
2 up
 T2  m2 g
m2=10kg
a
2 m/s2
Connected masses
• Two 10 kg blocks are
strung from an elevator
roof, which is accelerating
up at 2 m/s2.
F  m a
F  m a
1
2
1 up
2
 T1  T2  m1 g
up  T2  m2 g
m1a + m2a = T1 – T2 – m1g + T2 – m2g
T1 = (m1+m2) (a+g)=(10+10)(2+9.8)=236N
T2 = m2(a+g)=10(2+9.8)=118N
T1
m1=10kg
T2
m2=10kg
a
2 m/s2
Connected masses
• What is the acceleration of the system
below, if T is 1000 N?
• What is T*?
a
T*
m2=10kg
m1=1000kg
T
Connected masses
• What is the acceleration of the system
below, if T is 1000 N?
• What is T*?
a
T*
m2=10kg
F
F
1
 m1a  T  T *
2
 m2 a  T *
m1a=T-m2a
m1=1000kg
T
a = T / (m1+m2)=.99m/s2
T* = m2a=9.9N
Connected masses
• What are T* and a now?
a
Connected masses
• What are T* and a now?
F
F
1
 m1a  T  T *  m1 g sin 
2
 m2 a  T * m2 g sin 
a
Connected masses
• Add in friction: μ = 0.4
• What are T* and a now?
T*
a
Connected masses
• Add in friction: μ = 0.4
• What are T* and a now?
F
F
1
 m1a  T  T *  m1 g sin   m k m1 g cos 
2
 m2 a  T * m2 g sin   m k m2 g cos 
T*
a
Motion over pulleys
a
• We know that the
magnitude and
sense of the
acceleration should
be the same for the
whole system
T
a
10kg
Θ = 300
μ = 0.4
Motion over pulleys
a
• We know that the
magnitude and
sense of the
acceleration should
be the same for the
whole system
F
F
T
a
10kg
1
 m1a  T  m1 g sin   m k m1 g cos 
2
 m2 a  m2 g  T
Θ = 300
μ = 0.4
55. The figure below gives as a function of time t, the force component
Fx that acts on a 3kg ice block, which can move only along the x axis.
At t=0, the block is moving in the positive direction of the axis, with a
speed of 3m/s. What are (a) its speed and (b) direction of travel at
t=11s?
55. The figure below gives as a function of time t, the force component
Fx that acts on a 3kg ice block, which can move only along the x axis.
At t=0, the block is moving in the positive direction of the axis, with a
speed of 3m/s. What are (a) its speed and (b) direction of travel at
t=11s?
t  0  v0  3m / s
t  11s  v f  ?
t
F
dv
dv
a x  x  x   x dt  v f  v0 
m
dt
dt
0
11s
Fx
0 m dt
11s
Total graph area  15 Ns 
 F dt  (v
x
0
15kgm / s
 vf 
 3m / s  8m / s
3kg
f
 v0 )m  (v f  3m / s )3kg
Two bodies, m1= 1kg and m2=2kg are connected over a
massless pulley. The coefficient of kinetic friction between
m2 and the incline is 0.1. The angle θ of the incline is 20º.
Calculate: a) Acceleration of the blocks. (b) Tension of the
cord.
N
f
T
m2
T
m1
20º
m2g
m1g
Two bodies, m1= 1kg and m2=2kg are connected over a massless
pulley. The coefficient of kinetic friction between m2 and the incline
is 0.1. The angle θ of the incline is 20º. Calculate: a) Acceleration of
the blocks. (b) Tension of the cord.
F2 g , x  m2 g sin 20  6.7 N
N 2  F2 g , y  m2 g cos 20  18.42 N
f  m k N 2  m k m2 g cos 20  1.84 N
Block 1 : m1 g  T  m1a
 9.8  T  a
Block 2 : T  f  F2 g , x  m2 a
 T  1.84  6.7  2a
Adding 3T  28.14 N
 T  9.38 N ,
a  0.42m / s 2
N
f
T
m2
T
m1
20º
m2g
m1g
The three blocks in the figure below are connected by massless cords
and pulleys. Data: m1=5kg, m2=3kg, m3=2kg. Assume that the incline plane
is frictionless.
(i) Show all the forces that act on each block.
N
T2
(ii) Calculate the acceleration of m1, m2, m3. T
m2
Fg2x
2
m3
(iii) Calculate the tensions on the cords.
(iv) Calculate the normal force acting on
m2
Fg2y
30º
T1
T1
m3g
m2g
m1
m1g
The three blocks in the figure below are connected by massless cords
and pulleys. Data: m1=5kg, m2=3kg, m3=2kg. Assume that the incline plane
is frictionless.
(i) Show all the forces that act on each block.
N
T2
(ii) Calculate the acceleration of m1, m2, m3. T
m2
Fg2x
2
m3
(iii) Calculate the tensions on the cords.
(iv) Calculate the normal force acting on
m2
Fg2y=m2gcos
Fg2x=m2gsin30º
Fg2y
30º
T1
T1
m3g
m2g
m1g
m1
The three blocks in the figure below are connected by massless cords
and pulleys. Data: m1=5kg, m2=3kg, m3=2kg. Assume that the incline
plane is frictionless.
N
T2
(i) Show all the forces that act on each block.
m2
(ii) Calculate the acceleration of m1, m2, m3.T2
Fg2x
T1
(iii) Calculate the tensions on the cords.
Fg2y 30º
m3
(iv) Calculate the normal force acting on
T1
m2
m3g
Block 1: m1g-T1=m1a
Block 2: m2g(sin30º) +T1-T2=m2a
Block 3: T2-m3g = m3a
m2g
m1g
Adding (1)+(2)+(3):
g(m1+0.5m2-m3)=a(m1+m2+m3)  a=4.41m/s2
(ii) T1=m1(g-a)= 5kg(9.8 m/s2-4.41 m/s2) = 26.95N
(iii) T2=m3(g+a)= 2kg(9.8 m/s2+4.41 m/s2)= 28.42N
(iv) N2= Fg2y= m2gcos30º = 25.46N
m1
1B. (a) What should be the magnitude of F in the figure below if the body of
mass m=10kg is to slide up along a frictionless incline plane with constant
acceleration a=1.98 m/s2? (b) What is the
y
magnitude of the Normal force?
N
x
20º F
30º
Fg
1B. (a) What should be the magnitude of F in the figure below if the body of
mass m=10kg is to slide up along a frictionless incline plane with constant
acceleration a=1.98 m/s2? (b) What is the
y
magnitude of the Normal force?
N
x
20º F
30º
Fg
m(a  0.5g )
F cos 20  mg sin 30  ma  F 
 73.21N

cos 20


N  mg cos 30  F sin 20  0  N  109.9 N




Three forces, given by , F1  2.00ˆi  2.00 ˆj N , F2  5.00 ˆi  3.00 ˆj N


and , F3  45.0ˆi N act on an object to give it an acceleration of
magnitude 3.75 m/s2. (a) What is the direction of the acceleration?
(b) What is the mass of the object? (c) If the object is initially at rest,
what is its speed after 10.0 s? (d) What are the velocity components of
the object after 10.0 s?




Three forces, given by , F1  2.00ˆi  2.00 ˆj N , F2  5.00 ˆi  3.00 ˆj N


and , F3  45.0ˆi N act on an object to give it an acceleration of
magnitude 3.75 m/s2. (a) What is the direction of the acceleration?
(b) What is the mass of the object? (c) If the object is initially at rest,
what is its speed after 10.0 s? (d) What are the velocity components of
the object after 10.0 s?

F  m a



ˆ
2.00ˆ
i 2.00ˆ
j 5.00ˆ
i 3.00ˆ
j 45.0ˆ
i N  m 3.75 m s2 a
where

â
represents the direction of a



ˆ
42.0ˆ
i 1.00ˆ
j N  m 3.75 m s2 a
 F   42.0  1.00
2
 F  42.0 N
2
N
00
1  1.
at tan 

42.0

2

ˆ
at181  m 3.75 m s a




Three forces, given by , F1  2.00ˆi  2.00 ˆj N , F2  5.00 ˆi  3.00 ˆj N


and , F3  45.0ˆi N act on an object to give it an acceleration of
magnitude 3.75 m/s2. (a) What is the direction of the acceleration?
(b) What is the mass of the object? (c) If the object is initially at rest,
what is its speed after 10.0 s? (d) What are the velocity components of
the object after 10.0 s?
ˆ is at181
(a)  a
(b)
counterclockwise from the x-axis
42.0 N
m 
 11.2 kg
2
3.75 m s


2
(d) v f  vi  at 0  3.75 m s at181 10.0 s
vf 
(c)
 37.5iˆ 0.893ˆj m
s
v f  37.52  0.8932 m s  37.5 m s
A bag of cement of weight 325 N hangs from three wires as
suggested in Figure. Two of the wires make angles 1 = 60.0
and 2 = 25.0 with the horizontal. If the system is in
equilibrium, find the tensions T1, T2, and T3 in the wires.
A bag of cement of weight 325 N hangs from three wires as
suggested in Figure. Two of the wires make angles 1 = 60.0
and 2 = 25.0 with the horizontal. If the system is in
equilibrium, find the tensions T1, T2, and T3 in the wires.
2
1
T3  Fg
T1 sin1  T2 sin2  Fg
T1 cos1  T2 cos 2
Fg
T1
1
2
T3
T2
T1 
Fg cos 2
 sin 1 cos 2  cos1 sin  2 

Fg cos 2
sin 1   2 
T3  Fg  325 N
 cos25.0 
T1  Fg 
 296 N

 sin 85.0 
 cos1 
T2  T1 
 296 N

 cos 
2
 cos60.0 

  163 N
cos25.0
A block is given an initial velocity of 5.00 m/s up a
frictionless 20.0° incline. How far up the incline does the
block slide before coming to rest?
A block is given an initial velocity of 5.00 m/s up a
frictionless 20.0° incline. How far up the incline does
the block slide before coming to rest?
After it leaves your hand, the
block’s speed changes only
because of one component of
its weight:
 Fx  m ax
 m g sin 20.0  m a


v2f  vi2  2a xf  xi .
vf  0 vi  5.00 m s
a   g sin  20.0


0   5.00  2 9.80 sin  20.0 xf  0
2
25.0
xf 
 3.73 m
2 9.80 sin  20.0
A woman at an airport is towing her 20.0-kg suitcase at
constant speed by pulling on a strap at an angle  above
the horizontal. She pulls on the strap with a 35.0-N force,
and the friction force on the suitcase is 20.0 N. Draw a
free-body diagram of the suitcase. (a) What angle does
the strap make with the horizontal? (b) What normal force
does the ground exert on the suitcase?
A woman at an airport is towing her 20.0-kg suitcase at
constant speed by pulling on a strap at an angle  above
the horizontal. She pulls on the strap with a 35.0-N force,
and the friction force on the suitcase is 20.0 N. Draw a
free-body diagram of the suitcase. (a) What angle does
the strap make with the horizontal? (b) What normal force
does the ground exert on the suitcase?
m suitcase  20.0 kg F  35.0 N
 Fx  m ax :
 Fy  m ay :
(a)
20.0 N  F cos  0
 n  F sin   Fg  0
F cos  20.0 N
20.0 N
cos 
 0.571
35.0 N
  55.2
A woman at an airport is towing her 20.0-kg suitcase at
constant speed by pulling on a strap at an angle  above
the horizontal. She pulls on the strap with a 35.0-N force,
and the friction force on the suitcase is 20.0 N. Draw a
free-body diagram of the suitcase. (a) What angle does
the strap make with the horizontal? (b) What normal force
does the ground exert on the suitcase?
m suitcase  20.0 kg F  35.0 N
 Fx  m ax :
 Fy  m ay :
20.0 N  F cos  0
 n  F sin   Fg  0
(b)
n  Fg  F sin   196  35.0 0.821  N
n  167 N
Three objects are connected on the table as shown in
Figure. The table is rough and has a coefficient of kinetic
friction of 0.350. The objects have masses 4.00 kg, 1.00
kg and 2.00 kg, as shown, and the pulleys are frictionless.
Draw free-body diagrams of each of the objects. (a)
Determine the acceleration of each object and their
directions. (b) Determine the tensions in the two cords.
Three objects are connected on the table as shown in Figure. The
table is rough and has a coefficient of kinetic friction of 0.350. The
objects have masses 4.00 kg, 1.00 kg and 2.00 kg, as shown, and the
pulleys are frictionless. Draw free-body diagrams of each of the
objects. (a) Determine the acceleration of each object and their
directions. (b) Determine the tensions in the two cords.
n
For m1
T12
T23
f = m kn
m2 g
T12
T23
m1 g
m3 g
m1
,
 Fy  m ay
T12  m 1g  m 1a