Transcript Slide 1 - Phy 2048-0002

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.
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 vectorially.
Inertial reference frame: where Newton’s laws hold.
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.
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: When two bodies interact,
the forces on the bodies from each other are always
equal in magnitude and opposite in direction.


FBC   FCB
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.
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
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
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   s N  ma  0
F
y
 N  mg cos   0
mg sin    s N   s mg cos 
sin 
s 
 tan 
cos 
‘Equilibrium’ problems (F = 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-m2g=m2a 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?
T = 1000N
θ = 250
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 
T = 1000N
θ = 600
a
Connected masses
• Add in friction: μ = 0.4
• What are T* and a now?
F
F
1
 m1a  T  T *  m1 g sin    k m1 g cos 
2
 m2 a  T * m2 g sin    k m2 g cos 
T*
a
Motion over pulleys
a
• We know that the
magnitude and sign
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 sign
of the acceleration
should be the same
for the whole
system
F
F
T
a
10kg
1
 m1a  T  m1 g sin    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   k N 2   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