Download CHAPTER 4 NEWTON`S LAWS • Little bit of history • Forces

Two books laid the foundations for
our understanding of forces and motion ...
CHAPTER 4
NEWTON’S LAWS
• Little bit of history
• Forces
• Newton’s 1st Law
• Newton’s 2nd Law
! mass and weight
• Newton’s 3rd Law
! action reaction pairs
! free body diagrams
! the normal force
! analysis of day-to-day situations
Forces
A force is best described as an action or influence that
when acting on an object, will change its motion.
There are two main types of forces:
• contact forces (push, pull, friction, etc.)
⇒ direct force
• non-contact (or field) forces (gravity, magnetic,
electric, etc.)
⇒ “action” at a distance and long range
... but they are treated in exactly the same way.
Dimension
[M][L]
[T]2
(show later)
Unit: Newton (N)
1N ⇒ weight of a medium size apple, i.e., the downward
force exerted on your hand when you hold it.
Force is a vector, i.e., with both direction and magnitude
and so to find the resultant or the net force when several
forces act on a object, they must be added vectorially.
!
FR
!
F2
!
F2
!
F1
!
F1
!
F2
!
F1
!
FR
Question 4.1: An object on a horizontal table is acted
!
!
!
on by three forces, F1 = 3 N, F2 = 4 N and F3 = 3 N, as
!
F2
!
F1
!
F3
shown below. What is the net force acting on the
object?
!
FR
!
F3
!
F3
!
! ! !
The net force is FR = ∑i Fi = F1 + F2 + ...
y
!
F
!
Fx = F cos θ
!
Fy = F sin θ
The components of the net force are:
FRx = ∑ Fx and FRy = ∑ Fy .
30"
60"
!
F2
!
F
θ
!
F1
x
Example of forces:
y
"
F3 = 3 N
Define x and y axes. Then,
"
F1 = 3 N
the net force in the x
30!
x
60!
"
F2 = 4 N
direction is Fx = ∑i Fxi and
the net force in the y
!
W
direction is Fy = ∑i Fyi .
!
!
F
!
∴Fx = F1 cos30 + F2 cos60 − F3
= 2.60 N + 2.00 N − 3.00 N = 1.60 N,
and
Two forces: the weight of the ball
downward (due to the attraction
of the Earth) is balanced by the
“normal” force of the ground on
the ball.
No net force ⇒ no motion
!
N
Fy = F1 sin 30! + F2 sin 60!
!
W
!
FR
= 1.50 N − 3.46 N = −1.96 N.
Fx ˆi
θ
Fy ˆj "
FR
Two forces: the weight of the ball
downward and the force of the
foot on the ball.
Net resultant force ⇒
change in motion
acceleration
Thus, the net force is
"
F = (1.60 ˆi − 1.96 ˆj) N,
R
with magnitude
"
FR = (Fx )2 + (Fy )2
!
W
= (1.60 N)2 + (−1.96 N)2 = 2.53 N.
⎛F
⎞
−1
!
The angle θ = tan −1 ⎜ y
⎟ = tan (1.225) = 50.8 .
F
x⎠
⎝
One force: the weight of the ball
downward.
Net force ⇒
change in motion
acceleration
Important concept ...
... we cannot neglect the effect of the Earth ... it always
produces a non-contact force!
Analysis of moving objects led to ...
A “simple” example ....
Newton’s 1st Law:
A body at rest, or moving with constant velocity, will
2
1
remain at rest, or moving with constant velocity, unless it
is acted on by a net force. (NB: the reference frame is
the Earth.)
Sometimes called the Law of inertia ...
3
1 The force on you due to the Earth, i.e., your weight,
• Tendancy for a body that’s moving to keep moving.
(downward) is balanced by the force of the seat of the
• Tendancy for a body at rest to remain at rest.
chair on you (upward). No net force.
!
∴ If the sum of forces ∑ i F , i.e., the net force, on an
object is zero it remains at rest or continues moving with
constant velocity. ( ∑ Fx = 0 and ∑ Fy = 0.)
This is called EQUILIBRIUM.
Implication ⇒ to change motion we have to apply a net
force ... Newton’s 2nd Law gives us the relationship
between the force and the change in motion.
2 The force on the ice cream cone due to the Earth
(downward) is balanced by the force of your hand on the
cone (upward). No net force.
3 The total weight of you, the ice cream cone and the
chair (downward) is balanced by the force of the ground
on the chair legs (upward). No net force.
But don’t get fooled!!
some common misconceptions ....
Are you “thrown” backwards
when a car accelerates from
rest? NO!! There is no
“throwing” force; you tend to
remain at rest as the seat moves
forward!
DISCUSSION PROBLEM [4.1]:
Why does a wire stretched between two posts always sag
Are you “thrown”
!
F
outwards when a car turns
a corner? NO!! There is
no “throwing” force; as
the car turns, you tend to
continue to move in a
straight line; so the side of
the car exerts an inwards
force on you!
in the middle no matter how tightly it is stretched?
So, to produce a change in motion, there must be a net
force ...
Newton’s 2nd Law:
In the Principia Newton stated that
The change in motion is proportional to the motive
force impressed; and is made in the direction of
!
F
the right line in which that force is impressed.
!
v
!
F
!
v
... to move an object from rest, i.e., to change its velocity,
!
v
!
F
frictional force opposing motion
or slow down a puck or change the direction of a ball.
!
F
!
v1
Today, we write the 2nd Law as
! dp!
F= ,
dt
!
where p = mv , which today is called the momentum;
Newton called it the quantity of motion.
!
! dp! d(mv! )
dv ! dm
! ! dm
∴F =
=
=m +v
= ma + v .
dt
dt
dt
dt
dt
However, most textbooks ignore the full Newtonian
expression and assume that m is constant. So, you will
find Newton’s 2nd Law usually stated this way:
!
v2
If a net force acts on an object it accelerates (i.e.,
there is a change in motion). The direction of the
acceleration is the same as the net force,
!
!
i.e., ∑ i Fi = ma .
And in component form,
∑ Fx = ma x : ∑ Fy = ma y : ∑ Fz = ma z
where m is a constant. Remember, however, these
expressions are only true if the mass remains constant.
So, this form of the Law does not apply to a rocket that is
losing mass by burning fuel.
"
F3
NOTE:
!
• The product ma is not itself a force ... it is the net
!
force that is put equal to ma .
!
!
• The acceleration a is parallel to the net force F.
ˆj
30!
60!
ˆi
"
F2
• The net force has to be an external force, i.e., not
inside the object or system, e.g., you can’t lift yourself!
"
F1
Question 4.2: An object on a horizontal table is acted on
by three forces, F1 = 3 N, F2 = 4 N and F3 = 3 N, as shown
above. If the mass of the object is 2.0 kg, and it is initially
at rest at the origin,
(a) what is its acceleration?
(b) What is its velocity after 4.0 s?
(c) What is its displacement after 4.0 s?
The dimension of force is the same as
[L] [M][L]
mass × acceleration ⇒ [M] 2 =
.
[T]
[T]2
The unit of force is the Newton (N).
y
"
F3 = 3 N
"
F1 = 3 N
30!
x
60!
"
F2 = 4 N
(a) From before (Q4.1) we found the resultant force was:
"
FR = (1.60 ˆi − 1.96 ˆj) N.
"
"
Using Newton’s 2nd Law FR = ma .
"
" FR (1.60ˆi − 1.96ˆj) N
∴a =
=
m
2.0 kg
= (0.80 ˆi − 0.98 ˆj) m/s 2 ,
and the magnitude of the acceleration is
"
a = (0.80 m/s2 )2 + (−0.98 m/s2 )2 = 1.27 m/s2
"
and it is parallel to FR.
" "
"
(b) v = v ! + at = (0.80 ˆi − 0.98 ˆj) m/s 2 × 4.0 s
= (3.20 ˆi − 3.92 ˆj) m/s.
"
∴ v = (3.20 m/s) 2 + (−3.92 m/s)2 = 5.06 m/s.
1!
1
! ! !
(c) r − r" = v " t + at 2 = (0.80 ˆi − 0.98 ˆj) m/s2 × (4.0 s)2
2
2
= (6.40 ˆi − 7.84 ˆj) m.
! !
∴ r − r" = (6.40 m) 2 + (−7.84 m)2 = 10.1 m.
The constant, m, in the expression
F = ma
is really a measure of how difficult it is to change the
motion of an object. We call it the inertial mass, or
simply, the mass. Note: mass is a scalar quantity.
To understand that concept, think about how you check
the “weight” of something ...
DISCUSSION PROBLEM [4.2]:
We have just seen that the way we choose between the
weights of two objects we move them up-and-down and
determine how difficult it is to move them.
m2
m1
m1
m2
... you check to see “how easy or difficult it is to make
them move”! For the same acceleration ...
F
F
a = 1m = 2m ,
1
2
i.e.,
m1
F1
m2 = F2
So, in fact, you are comparing the masses !
Will this idea work in a weightless environment?
OK ... so, a force changes motion. But when you throw
a baseball, what keeps it moving to the right after it
leaves your hand?
!
w
Question 4.3: An object of mass m is traveling at an
initial speed v ! = 25.0 m/s. A net force frictional force
of 15.0 N acting on the object in a direction opposing the
motion, causes it to come to a stop after a distance of
The answer is ... inertia ... remember Newton’s 1st Law.
62.5 m. (a) What is the mass of the object? (b) How
Neglecting air resistance, there are no horizontal forces
long does it take to stop?
to change its horizontal motion, so it keeps going to the
right! Note, inertia is NOT a force.
There is a force downward (its weight), which causes the
ball to “fall”. These effects combine so the ball follows
a parabolic trajectory.
v ! = 25.0 m/s
v=0
x!
x
F = 15.0 N
+ ve direction
(x − x! ) = 62.5 m
Define the positive direction to the right. First, we need
to find the (negative) acceleration.
2
2
(a) We have: v = v ! + 2a(x − x ! )
∴a =
v 2 − v !2
(25.0 m)2
=−
= −5.00 m/s 2 .
2(x − x ! )
2 × 62.5 m
Question 4.4: Two forces, F1 and F2, act on an object on
a frictionless surface, as shown. In which case is the
acceleration greatest?
F2 = 3 N
Note that F opposes the motion, i.e., it acts in the −ve
i.e., m =
F
−15.0 N
=
= 3.00 kg.
a −5.00 m/s2
(b) We have v = v ! + at,
v − v ! 0 − 25.0 m
i.e., t =
=
= 5.00 s.
a
−5.00 m/s2
F1 = 12 N
F2 = 4 N
A
direction (same as the acceleration). Using Newton’s 2nd
Law, F = ma ,
6 kg
F2 = 2 N
3 kg
C
4 kg
F1 = 8 N
B
F1 = 8 N
F2 = 2 N
2 kg
D
F1 = 8 N
F2 = 3 N
6 kg
F1 = 12 N
F2 = 4 N
A
F2 = 2 N
3 kg
F1 = 8 N
4 kg
B
F1 = 8 N
F2 = 2 N
2 kg
C
F1 = 8 N
D
The net force on a block in each case is
What’s the difference between mass and weight ?
• MASS: inertial property
larger mass ⇒ larger force required to accelerate.
F1 − F2 = F.
By Newton’s 2nd law, that force produces an
acceleration given by F = ma , i.e., a = F m.
9N
4N
∴a A =
= 1.5 m/s2 : a B =
= 1 m/s2
6 kg
4 kg
aC =
6N
6N
= 2 m/s2 : a D =
= 3 m/s2
3 kg
2 kg
So, the answer is D.
• WEIGHT: the force exerted by the Earth on an object.
Example:
Bowling ball ⇒ difficult to throw (mass)
⇒ difficult to lift (weight)
If you analyze a situation involving forces you’ll find
that forces ALWAYS occur in pairs:
!
FBT
Who pulls harder, i.e., who applies the greater force, in
the scenario shown here?
!
FAB
!
FFB
!
FTB
!
FBA
A
!
FBF
(Spring)
Books on table (down)
Table on books (up)
!
FAB
!
FBA
A
B !
Pulling force A (to the left) is applied on B ( FAB ).
!
Pulling force B (to the right) is applied on A ( FBA ).
!
FAW
!
FWA
A
!
Pulling force A (to the left) is applied on the wall ( FAW ).
!
The wall applies a force (to the right) on A ( FWA ).
B
FAB
FBA
50
FBA
0
−50
FAB
0
2
4
6
Time (s)
8 10
Ha! They both “pull” equal amounts ... the force exerted
by A on B is equal and opposite to the force exerted by
B on A, i.e.,
!
!
FAB = − FBA and FAB = FBA
at all times!
This leads us to ...
Examples of action/reaction pairs:
!
FBT
!
FTB
Newton’s third Law:
If an object A exerts a force on an object B (an action),
Books on table (down)
Table
! on books
! (up)
FBT = FTB
object B exerts an equal and opposite force on A (the
reaction), i.e.,
!
!
FAB = − FBA
!
!
( FAB = FBA )
Note:
• the two forces act on different objects, sometimes
!
FFB
!
FFE
!
FBF
!
!
FFB = FBF
!
FEF
!
!
FFE = FEF
called an action/reaction pair.
• the Law applies whether the objects are at rest,
moving with constant velocity or accelerating.
That’s weird! How come it seems to hurt your eye more
than it hurts your finger?
But we have to be careful in identifying
action/reaction pairs correctly (later).
Often, Newton’s laws appear in combination look ...
So, if the Earth applies a force on
an apple (its weight), the apple must
apply an equal and opposite force
on the Earth! Does that mean the
Earth moves towards an apple when
Centripetal force
(Action)
Car on you
Reaction
(“Centrifugal
force”)
You on car
we drop the apple? Let’s see ...
!
!
We know FEA = FAE ,
and the 2nd Law tells us:
FEA = m Aa A and FAE = mE a E .
m a
∴mAa A = mE a E ⇒ a E = A A .
mE
First Law: “Every body perseveres in its state of rest, or
of uniform motion in a right [straight] line, unless it is
compelled to change that state by forces impressed
thereon.”
Third Law: “To every action there is always opposed an
equal reaction: or the mutual action of two bodies on
each other are always equal, and directed to contrary
parts.”
Principia Mathematica (1687).
But mE ≈ 6 × 1024 kg, mA ≈ 0.1 kg and a A = g
∴a E ≈ 1.7 × 10 −25 m/s2 .
If the apple takes 1 s to reach Earth ...
1
Δy A = a A t 2 ≈ 4.91 m.
2
1
Δy E = a E t 2 ≈ 0.85 ×10 −25 m.
2
WOW ... that’s small!
!
FEA
!
FAE
Take something as simple as a book or cups on a table on
the Earth (i.e., the floor) ...
~ Let’s identify ALL action-reaction pairs ~
DISCUSSION PROBLEM [4.3]:
You are holding a baseball in
your hand. The reaction force
to the weight of the baseball is
....
!
!
!
• Forces acting ON the book: FTB and FEB (= w B )
These are not an action-reaction pair
Why ??
!
!
!
• Forces acting ON the table: FBT and FET (= w T )
!
!
• Forces acting ON the Earth: FBE and FTE
For action-reaction pairs: !
!
F12 = F21 .
A: the gravitational force of the Earth on the ball,
B: the gravitational force of the ball on the Earth,
C: the force of your hand on the ball,
D: the force of the ball on your hand, or
E: the gravitational force of the Earth on your hand.
Does Newton’s 3rd Law apply if objects are moving?
You betcha!!
You know all about the birds
and bees, but now it’s time to
learn about FBD’s
Believe me, dear, crucial to
solving “force problems” is
the free-body diagram.
W
R
S
Identify all the forces associated with the rope:
(Take L ⇒ R as positive direction)
!
FRS (rope on sled)*
!
FSR (sled on rope)*
!
FWR (woman on rope)#
!
FRW (rope on woman)#
!
!
!
!
But FRS = − FSR and FWR = −FRW .
!
!
The net force on the rope ⇒ FWR − FSR .
!
!
So, if FWR > FSR
+
A free-body diagram (FBD) shows all of the forces acting
−
ON an object. For example, the forces acting on a ball
+
−
the rope and everything attached to it, moves to the right,
but the equations above still hold!
that’s been kicked ...
!
F
!
N
!
W
⇒
!
W
!
W
⇒
⇒
The Normal Force
In mechanics, the normal force FN (sometimes n or N) is
the component of the contact force exerted on an object ,
DISCUSSION PROBLEM [4.4]:
perpendicular to the surface of contact by, for example,
Which of the free-body diagrams below best represents
the surface of a floor or incline, preventing the object
the forces acting on an object sliding down a frictionless
from penetrating the surface. In the case of an object
FN
incline?
resting on a horizontal surface
(e.g., a table top, the ground),
!
v
the normal force is equal and
mg
opposite to the weight of the
object (Newton’s 3rd Law), i.e., FN = mg .
When an object rests on an incline, the normal force is
perpendicular to the plane the
FN
object rests on. In that case, the
A
B
C
D
mgcos θ
mg
θ
mgsin θ
normal force is equal and
opposite to the component of
the weight perpendicular to the
surface, i.e., FN = mg cosθ.
There is a force acting parallel to the incline, i.e., the
component of the weight parallel to the surface, mg sin θ.
FN
mg cos θ
mg
STRATEGY for analyzing “force” problems:
FN
mgsin θ
θ
Draw free body diagram
mgcos θ
FR
mg
(identify the forces acting ON an object)
Newton’s third Law
mgsin θ
θ
Therefore, when an object sits on a frictionless incline, the
resultant (net) force,
FR = mg sin θ
produces motion down the incline. By Newton’s 2nd Law
the acceleration of the object down the slope is given by
F
a = R = gsin θ,
m
a result I showed in Chapter 2.
Equilibrium
Motion
(i.e., at rest or constant velocity)
Newton’s first Law
(i.e., accelerating)
Newton’s second Law
Separate the force components in two (perpendicular)
directions.
F2
F1
f
Draw the FBD for the blocks. Since the blocks are not
Question 4.5: Two forces, F1 and F2, act on an object on
accelerating, Newton’s 2nd Law tells us the net force on
a surface, as shown. If the objects are all moving with a
each block is zero, i.e.,
constant velocity, in which case is the frictional force
F1 − F2 − f = 0
greatest?
∴f = F1 − F2 .
( F1 − F2 ) is largest for block A ( = 9 N to the right). Since
F2 = 3 N
6 kg
F1 = 12 N
F2 = 4 N
A
F2 = 2 N
3 kg
C
4 kg
F1 = 8 N
the net force must be zero, the frictional force of the
ground on block A must be 9 N (to the left).
B
F1 = 8 N
F2 = 2 N
2 kg
F1 = 8 N
F2 = 3 N
f
6 kg
F1 = 12 N
F2 = 4 N
A
4 kg
F1 = 8 N
B
D
F2 = 2 N
3 kg
C
F1 = 8 N
F2 = 2 N
2 kg
D
F1 = 8 N
DISCUSSION PROBLEM [4.5]:
A
Question 4.6: To prevent a box from sliding down a
frictionless inclined plane, physics student Anna pushes
on the box horizontally with just enough force so that the
box is stationary. If the mass of the box is 2.00 kg and
the slope of the incline is 35!, what is the magnitude of
B
the force she exerts?
You and a friend have identical cars. You want to rip an
old pair of Levi’s apart; there are two possible
Anna
arrangements, as shown above. Which is more likely to
do the job, i.e., in which is the tension in the jeans
35!
greater?
A: A
B: B
C: They’re the same
Draw the free body diagram for the box:
FN
y
FA
x
mgcos θ
mgsin θ
θ
Question 4.7: The system shown below is in equilibrium.
mg
Find the three tensions T1, T2 and T3 in the string, and the
The components of the weight force along x and along y
unknown mass, M.
are (−mg sin θ) and (−mg cosθ), respectively. The
components of the force Anna applies (FA ) along x and y
are FA cosθ and −FA sin θ,
FA sin θ
respectively. Summing the
FA
θ
FA cosθ
forces in the x-direction, we
have
θ
∑ Fx = 0
= FA cosθ − mg sin θ,
i.e., FA = mg tan θ.
(
)
∴FA = (2.00 kg ) 9.81 m/s 2 (0.700) = 13.7 N.
60! T3
T1
M
T2
6 kg
60!
First draw the free body diagram for the 6 kg
T2
mg
mass; the two forces are the weight (mg) and
T2. Since the system is in equilibrium,
T2 = mg = 6.00 kg × 9.81 m/s2 = 58.9 N.
60! T3
T1
M
60!
60! T3
T1
T2
M
The tensions all pass through a single point (circled); draw
the free body diagram at that point. Then, we have
T2
∑ Fh = 0 and ∑ Fv = 0.
!
−T1 sin 30 + T3 sin 30 = 0.
∴T1 = T3
(v)
T1 cos30! + T3 cos30! − T2 = 0,
T2
58.9 N
! = 0.866 = 68.0 N.
cos30
∴T1 = T3 = 34.0 N.
Since the pulley only changes the direction of T1, it does
not affect its magnitude.
!
(h)
i.e., T1 + T3 =
T2
6 kg
6 kg
T1 30! 30! T3
60!
i.e.,
∴T1 = Mg,
T
34.0 N
M= 1 =
= 3.47 kg.
g 9.81 m/s2
Sometimes, we may not know the angles as in this
problem ...
Question 4.8: Isobel Newton hangs motionless by one
Identify all of the forces acting through Isobel’s hand and
draw the vector force diagram.
!
F2
hand from a clothes line, as shown. If the clothes line is
!
F1
!
w
!
F2
!
w
!
F1
about to break, which side is most likely to break?
Since she is not moving, the net force on her hand is zero,
!
! !
!
i.e.,
Fnet = F1 + F2 + w = 0.
So the vector diagram shown on the right must be closed.
!
!
Clearly, F1 > F2 and so the part of the line on the right
has the greater tension and so is more likely to break!
A: The left side.
B: The right side.
C: Equal chance of either side breaking.
Hence, even if the angles are not known, you can use the
fact that the vector diagram of forces must be a closed
figure to solve a problem.
N (Upward force due to spring
balance, i.e., apparent weight.
Question 4.9: A 2 kg mass hangs from a spring scale,
calibrated in Newtons, that hangs from the ceiling of an
elevator. What does the scale read
when the elevator is (a) moving up
with a constant velocity of 30 m/s,
(b) moving down with a constant
velocity of 30 m/s, and (c) is
ascending at 20 m/s and gaining
Draw the free body
diagram for the 2 kg
block. In parts (a)
av
w = mg (Force due to Earth)
and (b) the block is
not accelerating.
!
∴a v = 0 so ∑ Fv = 0.
Take up direction as positive:
∴ N +(−mg ) = 0,
i.e., N = mg = (2 kg)(9.81 m/s 2 )
= 19.8 N ... (2 kg).
speed at the rate of 3 m/s 2 ?
elevator moves up at 10 m/s. Its velocity is then reduced
(c) Since the block is accelerating upward we have:
!
∑ Fv = ma v ( a v > 0).
uniformly to zero in the next 4 s, so that it comes to rest at
Then N +(−mg ) = ma v
(d) From t = 0 to t = 5 s, the
t = 9 s. What does the scale read during the time interval
0 < t < 9 s?
i.e., N = m(g + a v ) = (2 kg)(12.81 m/s 2 )
= 25.6 N ... (2.61 kg).
(d) When 0 < t ≤ 5 s, a v = 0, so N = 19.8 N.
This is ME ... !
Draw the free body diagram for the forces
on me:
N (Upward force from scale)
w = mg (force due to Earth)
When 5 s < t ≤ 9s, a v =
Δv (0 − 10 m/s)
=
= −2.5 m/s2 .
Δt
4s
Since N +(−mg ) = ma v
∑ v Fv = N −mg = ma v .
My apparent weight is the force I exert on the scale
⇒ N = m(g + a v ).
... but a v < 0.
2
∴ N = m(g + a v ) = (2 kg)(7.31 m/s )
= 14.6 N ... (1.49 kg)
4.0
av
(m/s2 )
acceleration
upward
2.0
0
acceleration
downward
−2.0
−4.0
80
120
160
200
240
Apparent weight (pounds)
So, if you know your true weight, you can find the
acceleration of an elevator!
Note that the apparent weight of someone in an elevator
if the cable breaks is zero. Draw the FBD for the forces
acting on the person ...
N (Upward force from scale)
w = mg (force due to Earth)
Yikes!
From the free-body diagram
N −mg = ma v ,
so apparent weight
N = m( g + a v ) = m( g + (−g))
=0
Conceptually what’s happening is ... the floor and
weighing scales are falling at the same rate as the person
inside the car so there’s no net force on the weighing
scale.
This is what is popularly known as weightlessness except,
of course, the person still has real weight!
Question 4.10: Two objects are connected by a massless
string, as shown above. The incline and pulleys are
frictionless. (a) Show that the acceleration of the objects
is
a=
(m2 − m1 sin θ)g
.
(m1 + m2 )
(b) Find the acceleration and tension in the string when
m1 = m2 = 5 kg and θ = 30! .
y
y
x
N
T
T
y
x
y
m2 g
m1g
x
T
N
m1g
(a) Treat each object separately. Note that T, the
tension, is common to both free-body diagrams as the
pulley only changes the direction of T. Assume m2
(b) With m1 = m2 = 5 kg and θ = 30! ,
a=
descends (i.e., m1 ascends the slope).
[#1] along x: T + (−m1gsin θ) = m1a
[#2] along y: T + (−m2 g) = −m2a
... (i)
... ... (ii)
Subtract (ii) from (i):
(m2 − m1 sin θ)g = (m1 + m2 )a
∴a =
(m2 − m1 sin θ)g
.
(m1 + m2 )
Note: if m2 > m1 sin θ then a > 0,
if m2 < m1 sin θ then a < 0,
if m2 = m1 sin θ then a = 0.
=
( m2 − m1 sin θ)g
( m1 + m2 )
(5 kg − (5 kg × 0.50))(9.81 m/s2 )
(5 kg + 5 kg)
= 2.45 m/s2 .
Since a > 0, m1 moves up the slope.
Also, using equation (ii),
T = m2 (g − a) = (5 kg)(7.36 m/s 2 )
= 36.8 N.
T
x
m2 g
N1
F21
N2
F
m1g
F12
m2 g
Draw the free-body diagrams for each object. It is the
contact force of m1 on m2 that causes m2 to accelerate.
(a) Consider only horizontal forces (why??) ...
#1 ... F − F21 = m1a.
Question 4.11: Two blocks are in contact on a
!
frictionless, horizontal surface. A horizontal force F is
applied to one of them, as shown. Find the acceleration
and the contact force for (a) general values of F, m1 and
m2, and (b) when F = 3.2 N, m1 = 2 kg and m2 = 6 kg.
#2 ... F12 = m2a.
But F12 = F21.
∴a =
F
.
m1 + m2
The contact force is: F12 (= F21 ) =
(b) a =
m2
F.
m1 + m2
F
3.2
=
= 0.400 m/s2 .
m1 + m2 2 + 6
F12 =
6
× 3.2 = 2.4 N.
2+6
!
F
!
F
(a)
m1 < m2
(b)
!
Do the results change if the force F is applied from the
right? Conceptually, the magnitude of the acceleration
!
will be the same because the total mass (m1 + m2 ) and F
are the same. But, the contact force will be smaller since
a smaller force will be required to produce the same
acceleration for m1. We get the same result analytically.
N1
F21
F12
F
m1g
We have
N2
m2 g
F − F12 = m2a,
and
F21 = m1a.
Eliminating a, and putting F12 = F21, we find
m1
F21 =
F,
(m1 + m2 )
i.e., the contact force of m2 on m1, F21, depends on the
mass of the second block, m1, which in this case is smaller
(2 kg) than in the problem (6 kg).