Download Unit 3 - HKU Physics

Unit 3
Newton’s Laws of Motion
3.1
Force and motions
3.2
The apparent weight of an object in an elevator
3.3
A block on a movable wedge
3.4
Forces of friction
3.1
Force and motions
Sir Isaac Newton (1647 – 1727)
(1)
Mathematical Principles of Natural Philosophy
Newton’s first law
Consider a body on which no net force acts. If the body is at rest, it will remain at rest. If the
body is moving along a direction with certain speed, it will continue to move without
changing its speed and direction.
Inertial frame of reference: A non-accelerating frame of reference in which Newton’s first
law is valid.
(2)
Newton’s second law
The net force F on a body is equal to the product of the body’s mass m and the acceleration
of the body a. The direction of acceleration is along the direction of net force. It is important
to know that force is a vector and has a SI unit Newton (N). A well- known formula to
express the idea is ∑ F = ma . One Newton is defined as the force which gives a mass of 1.0
kilogram an acceleration of 1.0 meter per second per second. A more general description of
Newton’s second law is the net force is given by the rate of change of momentum,
1
e.g. ∑ =
F
dp d (mv )
, where p is the momentum of the body. When the mass of the body
=
dt
dt
dv
is fixed, we have ∑
=
F m= ma . Note that Σ F is the vector sum of all forces (resultant
dt
forces)
 ∑ Fx = ma x

∑ Fy = ma y
 F = ma
z
∑ z
Unit: International system of units, or metric system SI
F: Newton (N), m: kilogram (kg) and a: meter per second square (m/s2)
When ΣF = 0 and m ≠ 0 , we have a = 0 or v = const. ⇒ Newton’s first law.
Remark: Are you puzzled to the presentation of Newton’s laws which employs overlapping
idea?
(3)
Newton’s third law:
Forces come in pairs, if a hammer A exerts a force on a nail B; the nail exerts an equal but
oppositely directed force on the hammer. i.e. FA on B
=
Action force
=
−FB on A
Reaction force
Note that they act on different bodies, so they will not cancel. Newton’s third law gives the
conservation of momentum and will be discussed in the next chapter.
Remark:
(1)
When a man of mass m stands on a platform scale, the reaction force R exerted on the
man from a scale shows the apparent weight of the man. When the man is at equilibrium, the
apparent weight has the same magnitude as his weight. Remember that the weight of man W
is the gravitational force exerted on the man due to the Earth and
the forces on the right figure are not action and reaction force pair.
R
Weight W = mg and the reaction R = −mg.
Mass m:
a scalar, measured in kg
Weight W:
a vector, measured in N
scale
W = mg
2
(2)
A mass is placed on a table. Here, the mass is a TV set. Figure (a) shows two pairs of
forces due to Newton’s third law: (i) between the mass and the table, (ii) between the mass
and the Earth. The forces shown in figure (b) are not force pair mentioned by Newton’s third
law, because there are three objects involved, namely, table, mass and the Earth.
Figure (a)
Figure (b)
Example
Two blocks are connected by a string as shown in the figure. The
m1 = 6.7 kg
o
smooth inclined surface makes an angle of 42 with the horizontal,
and the block on the incline has a mass of m1 = 6.7 kg. Find the
m2
mass of the hanging block m2 that will cause the system to be in
42o
equilibrium.
Answer
T
N
m1g
T
T
N
m1g sin42o
m2g
o
m1g 42
m1g cos 42o
The free-body diagrams of the blocks m1 and m2 are shown as follows.
The down-plane component of the 6.7-kg mass is given by
=
m1 g sin 42o (6.7
=
kg )(9.8 ms −2 ) sin 42o 43.9 N
For equilibrium, this force is balanced if the tension has the same magnitude. On the other
hand, the hanging block is also in equilibrium, the weight of it, m2g, must balance the tension
3
force, hence we can write m2 g = 43.9 . The mass of the hanging block is solved as (43.9 N)/g
= 4.48 kg.
Example (Challenging)
Three masses m1, m2 and m3 are connected by a string as shown in the figure. All masses and
pulleys are smooth and the string is light. Find the acceleration of m3 when the system is
released.
m2
m1
m3
Answer
Consider the total length of all segments on the string.
x1=
+ x2 + 2 x3 length
=
of string constant ,
where x1, x2 and x3 are the length of segments. The measurement of x1, x2 and x3 is along the
direction shown in the following figure. Differentiating both sides of the equation, we
obtain x1 + x2 + 2 x3 =
0 , where xi is the velocity of mass mi with i = 1, 2 and 3.
Differentiating again, we have 
x1 + 
x2 + 2 
x3 =
0 , which is an equation to relate the
acceleration of each mass, i.e.
a1 + a2 + 2a3 =
0.
(1)
4
x2
x1
m2
m1
x3
m3
We should notice that the measurement of a1, a2 and a3 is in accord with the measurement of
x1, x2 and x3 respectively. The free-body diagrams of each mass are shown below. The
reaction forces R1 and R2 are balance by the weight of blocks. The tension of string is T.
R1
R2
T
m1
T
T
T
m2
m1g
m3g
m2g
For m1:
−T =
m1 a1
(2)
For m2:
−T =
m2 a2
(3)
For m3:
m3 g − 2T =
m3 a3
(4)
Solving the set of four equations, we obtain a3.
Details:
Substitute (2), (3) and (4) into (1), we have −

T
T
2T
−
+ 2 g −
m1 m2
m3


0 . After simplification,
=

 1
1
4 
+
+
2 g and thus an expression for T, where
we obtain T 
=
m
m
m
2
3 
 1
5
−1
 1
1
4 
=
+
+
T 2g 
 .
 m1 m2 m3 
−1
2T
4g  1
1
4 
From (4) and the result of T, we obtain a3 =g −
=g −
+
+

 , that is
m3
m3  m1 m2 m3 
a3 =
3.2
g (m2 m3 + m3 m1 )
.
m2 m3 + m3 m1 + 4m1 m2
The apparent weight of an object in an elevator
Example
A man of mass 50 kg rides on an elevator. He is standing on a scale fastened to the floor of an
elevator. The elevator has the following motions.
(i) moving upward with an acceleration 2 ms-2,
(ii) moving upward with a deceleration 2 ms-2,
(iii) moving downward with an acceleration 2 ms-2,
(iv) moving downward with a deceleration 2 ms-2.
R
Calculate the scale’s reading in each case.
mg
Answer
Assuming upward is positive and by Newton’s second law, we have
R − mg = ma.
Hence, the apparent weight R (scale’s reading) is represented by
R = mg + ma.
(i)
Upward acceleration, a is positive: the apparent weight is increased (feel heavier).
(ii)
Upward deceleration, a is negative: the apparent weight is decreased (feel lighter).
(iii)
Downward acceleration, a is negative: the apparent weight is decreased (feel lighter).
(iv)
Downward deceleration, a is positive: the apparent weight is increased (feel heavier).
Remarks
“Weight of an object” in physical science is different from the everyday phrase that we are
familiar with. Unlike mass, weight is not an inherent property of an object; rather, it is a
measure of the gravitational force between the object and the Earth (or other planet). In this
example, the man has mass m and weight mg. His apparent weight is given by R.
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Example
Two blocks are in contact on a frictionless table. A horizontal force F is applied to one block,
as shown in figure.
F
m1
m2
(a) Find the force of contact between the two blocks.
(b) Suppose m1 > m2 and you are going to move the two blocks to the right by an applied
force F such that they move together with a constant acceleration. Which block do
you apply the force, if you are allowed to relocate the blocks?
Answer
F
m1
N
N
m2
(a)
Consider mass m1
F−N =
m1a
(1)
Consider mass m2
N = m2 a
(2)
(1) + (2) implies
=
F
( m1 + m2 ) a , which gives a =
Therefore, N =
(b)
m2 F
m1 + m2
F
.
m1 + m2
(3)
From (3) and the fact that m2 < m1 , we note that N is smaller if the applied force is
through m1. In order to reduce damage on both blocks, the force should be applied
through the heavier block.
7
Example
A block of mass m slides down on the frictionless inclined surface of a wedge which is fixed
on a scale. If the mass of the wedge is also m, find the weight of the system recorded by the
scale.
m
Fixed
wedge
m
Slides down
Frictionless
surface
θ
Scale
A.
mg + mg cos θ
B.
mg + mg cos θ sin θ
C.
mg + mg cos2θ
D.
mg + mg sin2θ
E.
2 mg
Answer
Since the wedge is fixed on the scale, the block slides down the wedge with an
acceleration g sin θ . On the other hand, we notice that the block has no acceleration in the
direction perpendicular to the inclined plane, hence the normal reaction, N given by the
wedge to the block is mg cos θ . The same force acts on the wedge by the block. As the scale
measures the reaction force R which is given by mg + N cosθ, that is =
R mg + mg cos 2 θ .
N sinθ
θ
N
N
N cosθ
Frictionless
surface
g sinθ
mg
N
mg
The block slides down
along the inclined
R θ
f
Fixed wedge
8
3.3
A block on a movable wedge
Example
A right triangular wedge of mass M and inclination angle θ , supporting a small block of
mass m on its side, rests on a horizontal frictionless table, as shown in figure. (a) Assuming
all surfaces are frictionless, what horizontal acceleration a must M have relative to the table
to keep m stationary relative to the wedge? (b) What horizontal force F must be applied to the
system to achieve this result? (c) Suppose no force is supplied to M, describe the resulting
motion.
m
F
M
θ
Answer
(a) and (b)
In this case M and m are moving with the same acceleration a. Note that a is
N1
N2
N2
F
mg
Mg
along x-direction only.
for M and m:
 F − N 2 sin θ =
Ma

mg

mg − N 2 cosθ =0 ⇒ N 2 =
cosθ

 N 2 sin θ = ma
(1)
(2)
(3)
From (2) and (3), a = g tan θ
Substitute (3) into (1) obtains F =( M + m ) a =( M + m ) g tan θ
c)
The block m will slide down along the wedge while the wedge will move backwards
9
Example (Challenging)
A wedge with mass M rests on a frictionless horizontal tabletop. A block with mass m is
placed on the wedge. There is no friction between the block and the wedge. The system is
released from rest. Calculate (a) the acceleration of the wedge, (b) the horizontal and vertical
components of the acceleration of the block, check the limit when M → ∞.
m
M
θ
Answer
Before doing this problem, we look at two remarks first.
•
If M is not moving then we have the trajectory
Trajectory
sketched in the right.
•
If M is moving to the left, then the actual trajectory is sketched as below, where aM is
the acceleration of the block relative to the wedge of mass M.
a1
a
aM
θ
Actual trajectory
Now, we look at the force diagram of the wedge and the block.
For M:
N 2 sin θ = Ma1
(1)
For m:
mg − N 2 cosθ =
maM sin θ
(2)
=
N 2 sin θ m(aM cosθ − a1 )
(3)
10
N1
a1
N2
θ
N2
a
θ
θ
mg
Mg
This is a set of simultaneous equations with 3 unknowns. After solving, we obtain
mg sin θ cosθ

a1 = M + m sin 2 θ
∴
a = ( M + m) g sin θ
 M
M + m sin 2 θ
(*)
Since the horizontal and vertical components of a are given by
=
ax aM cosθ − a1 and a y = aM sin θ .
Mg sin θ cosθ

ax = M + m sin 2 θ
After solving, we have 
2
a = ( M + m) g sin θ
y
M + m sin 2 θ

The reaction between the wedge and the block is given by the expression of a1 and equation
(1), i.e.
N2 =
mMg cosθ
.
M + m sin 2 θ
Consider the equation set (*), when M is very large, e.g. M → ∞ , we have
a1 → 0 and aM = g sin θ .
3.4
Forces of friction
N
(a)
v
Rough surface
m
fk: Kinetic frictional force
fk
mg
(b)
N
Fapp
fs
fs: Static frictional force
mg
11
(c)
N
Fapp
fs,max
when Fapp = fs,max
fk
when Fapp > fs,max
fk: Kinetic frictional force
mg
(d)
v
N
Fapp
mg
The laws of friction are stated as follows.
Static 0 ≤ f s ≤ f s,max (fs,max = µsN)
1)
µs : coefficient of static friction

It is independent of the area of contact between the surfaces.

It is parallel to the surface of contact, and in the direction that opposes relative motion.
Kinetic f k = µ k N (N = normal force)
2)
µk: coefficient of kinetic friction

It is independent of the area of contact between the surfaces.

It is independent of the relative speed of the surfaces.
Remarks:
(1)
It should be noted that in most cases, µk < µs, and all these coefficients are positive
number smaller than one.
(2)
In the right figure, the normal reaction is labelled as n.
12
Table: Some coefficients of friction
Materials
µs
µk
Steel on steel
0.74
0.57
Wood on wood
0.25−0.5
0.2
Glass on glass
0.94
0.4
Ice on ice
0.1
0.03
Example
What minimum acceleration amin must the cart X have in order that block A does not fall? The
coefficient of static friction between the block and the cart is µs.
rough surface
amin
A
cart X
Answer
When the cart moves to the front with an acceleration a, an upward frictional force
f ≤ µ s N would act on A, where N is the normal reaction on A by the cart. If the acceleration
is greater enough, the frictional force balances the weight of A. The equations that describe
the situation are
=
f mg ≤ m s N and N = ma , where m is the mass of A.
f
a
A
N
mg
Eliminating the normal reaction N, we obtain mg ≤ m s (ma ) and thus a ≥
g
µs
. The minimum
acceleration amin is g /µs.
13
Example
A 7.96 kg block rests on a plane inclined at 22° to the horizontal, as shown in figure. The
coefficient of static friction is 0.25, while the coefficient of kinetic friction is 0.15. (a) What
is the minimum force F, parallel to the plane, which can prevent the block from slipping
down the plane? (b) What is the minimum force F that will start the block moving up the
plane? (c) What force F is required to move the block up the plane at constant velocity?
Answer
F
22o
(a)
When F is at its minimum, fs is directed up-hill and assumes its maximum
value: f s ,=
mm
=
sN
s mg cos θ .
max
Thus for the block mg sin θ − Fmin − m s mg cosθ =
0 , which gives
=
Fmin mg ( sin θ − m s cos θ )
( 7.96 × 9.8)( sin 22° − 0.25 × cos 22° )
=
= 11.14 N
(b)
To start moving the block up the plane, we must have F − mg sin θ − m s mg cosθ ≥ 0 ,
which gives the minimum value of F required:
Fmin mg ( sin θ + m s cos θ )
=
( 7.96 × 9.8) ( sin 22° + 0.25 × cos 22° )
=
= 47.3 N
(c)
In this case the block is already in motion, so we should replace µ s with µk in
calculation the frictional force. Thus F − mg sin θ − mk mg cosθ =
ma =
0 , which gives
the value of F:
=
F mg ( sin θ + mk cos=
θ)
( 7.96 × 9.8)( sin 22° + 0.15 × cos 22° )
= 40.1 N
14