04 Newton S Laws Of Motion

Dynamics describes the relationship between force and motion.

Force? What is it?
Put in simple terms, a force is a push or a pull. It pertains to any influence that causes a change in an object’s state of motion.
• Contact Force
A contact force is produced when there is direct contact between two interacting bodies.
• Long-Range Force
A long-range force is produced when one body influences the state of motion of another body even if these two bodies are separated by empty space. • Concurrent Forces
Concurrent forces are forces whose lines of action intersect at a common point. These forces typically bring about rectilinear motion.

• Nonconcurrent Forces
Nonconcurrent forces are forces whose lines of action do not intersect at a common point. These forces typically bring about rotary motion.

The Laws of Motion form the foundation of dynamics.

First Law of Motion
An object will remain at rest or continue to move at a constant velocity unless acted upon by an external force.

F = net force
If F = 0 ⇒ v = constant

Third Law of Motion
For every action there is an equal but opposite reaction. These two forces
(action & reaction) act on different bodies. Freaction
Faction = − Freaction Faction

Second Law of Motion
When a net external force acts on a body, the body accelerates in the direction of the force.

F = ma where, m = mass of the body a = acceleration due to F
Unit : N = Newton = kg • m / s 2

Acceleration depends upon the force producing it, as well as on the mass of the body upon which the force has been applied. It is proportional to the force but inversely proportional to the mass.
Force and mass are not dependent on acceleration.

motion between the two surfaces.
Kinetic friction is constant while static friction increases in response to any increase in the impending motion
Weight(W)
until a maximum value is reached,
Weight is the force of gravitational beyond which there will be motion attraction between the Earth and the between the two surfaces. object. For points on or near the surface of the Earth:
Kinetic friction is independent of the g = −9.8m / s 2 = towards the " ground" relative speed between the moving surfaces. W = mg
Friction is independent of the size of the area of contact between the surfaces. Friction is dependent on the Normal
W2
W1
Force.
Friction (fs, fk)
Friction arises when two surfaces rub against each other. Static Friction opposes impending motion while
Kinetic Friction opposes existing

N = normal force magnitude f = µN µ = coeffience of friction f Typically, 0 ≤ µ ≤ 1 µs > µk

motion

Normal Force (N)
Normal force is produced when a body presses against a surface, trying to prevent the body from going through the surface. Normal force is always perpendicular outwards with respect to the surface.
If the body does not go through the surface, normal force magnitude is then equal to the magnitude of the force by which the body is pressed against the surface.
N2
N1
N3

Tension (T)
Tension arises when a body (typically a string, rope, or cable) is stretched by an external force.
Tension opposes this external force to maintain the body’s size and shape.

T

W

Compression (C)
Compression arises when
C
a body (typically a pillar, rod, lever, or block) is compressed by an
W
external force.
Compression opposes this external force to maintain the body’s size and shape.
If the body being stretched or compressed maintains its size and shape, tension/compression magnitude is equal to that of the deforming force.

A system in equilibrium is one that is not accelerating.

1st Condition of
Equilibrium
A system is said to be in a state of equilibrium if and only if the net external force acting on the system is equal to zero.

Fnet = F1 + F2 + F3 + F4
= ∑ Fi i =0

F1
F4

F2
F3

Static Equilibrium
A system in equilibrium that is not in motion is considered to be in Static
Equilibrium.

Accelerating Systems
A system that is not in a state of equilibrium is an accelerating system.
The system accelerates in the direction of the net external force at a rate defined by Newton’s 2nd Law of
Motion.

Fnet ≠ 0
= ma
= ∑ Fi i (

Example 1
Consider the system shown below.
What is the maximum possible value that mass M can have if the system is to remain in a state of equilibrium? m = 2kg, µs = 0.65

FBD2: (mass M)

µs m

T
M
Free Body Diagram 1 (FBD1):
(mass m)
N
fs

)

Wm = mg = 2kg 9.8m / s 2 = 19.6 N f s = µ s N = max. static friction
Fnet = 0 = N + T + Wm + f s x - comp : 0 = T − f s
T = fs y - comp : 0 = N − Wm
N = Wm = 19.6 N
⇒ T = f s = µ s N = 0.65(19.6 N )
⇒ T = 12.74 N

T
Wm

WM

Fnet = 0 = T + WM x: 0 = 0+0 y : 0 = T − WM
WM = T = 12.74 N
Mg = 12.74 N
12.74 N
M=
9.8m / s 2
M = 1.3kg

Example 2
Consider the system shown below.
What should µs be for the system to remain in equilibrium? m1 = 0.75kg, m2 = 0.50kg m1 µs

m2

10º
FBD1: (mass m1)
N
T fs 10º
W1

W1 = m1 g = 7.35 N
W2 = m2 g = 4.90 N f s = µs N

N fs W1

T
10º

Fnet = 0 = N + T + W1 + f s x : 0 = T − W1 sin 10° − f s
T = W1 sin 10° + f s = W1 sin 10° + µ s N
T = 1.276314 N + µ s N y : 0 = N − W1 cos10°
N = W1 cos10° = 7.238337 N
⇒ T = 1.276314 N + µ s (7.238337 N )
FBD2: (mass m2)

Fnet = 0 = T + W2 x: 0 = 0+0 y : 0 = T − W2
W2
T = W2 = 4.90 N
⇒ 4.90 N = 1.276314 N
+ µ s (7.238337 N ) µ s (7.238337 N ) = 3.623686 N
3.623686 N µs =
7.238337 N
T

µs = 0.501

40,082 N
W
= sin 55° sin 55°
40,082 N
⎛ 40,082 N ⎞
⇒ T1 = ⎜
⎟ cos 55° = tan 55°
⎝ sin 55° ⎠
T2 =

Example 3

A large wrecking ball is held in place
T1 = 28,065.719N
T2 = 48,931.087N by two light steel cables as shown.
Determine all the forces in the
FBD2: (top end of crane boom) system. Frxn
Fnet = 0 = T2 + C + Frxn
C
x : 0 = −T2 cos10° + C
10º
35º
C = T2 cos10°
T
2
45º

y : 0 = −T2 sin 10° + Frxn
Frxn = T2 sin 10°

m = 4,090kg
FBD1: (wrecking ball)
T1

T2
55º
W

W = mg = 40,082 N
Fnet = 0 = T1 + T2 + W x : 0 = −T1 + T2 cos 55°
T1 = T2 cos 55° y : 0 = T2 sin 55° − W

10º
T2

Frxn
C

C = 48,187.714N
Frxn = 8,496.794N

FBD1: (bricks)
T

Example 4
A 15.0-kg load of bricks hangs from one end of a thin steel cable that has been passed over an ideal pulley. A
28.0-kg weight is suspended from the other end of the cable. What is the acceleration of the system if it is released from rest?
Since the system is no longer in equilibrium, the direction of the
28.0kg acceleration must be assumed (and then later verified). Assuming that
15.0kg the bricks will be accelerating upwards.

a
W1

Fnet = m1a = T + W1
⇒ m1a = T − W1
⇒ m1a = T − m1 g

FBD2: (counterweight)
T
W2

Fnet = m2 a = T + W2
⇒ − m2 a = T − W2 a ⇒ − m2 a = T − m2 g
T = −m2 a + m2 g

m1a = (− m2 a + m2 g ) − m1 g
⇒ m1a + m2 a = m2 g − m1 g
⇒ (m1 + m2 )a = (m2 − m1 )g
(m − m1 ) g = (28kg − 15kg ) g a= 2
(m1 + m2 ) (15kg + 28kg ) a = 0.302326g = 2.963m/s2

⇒ m1a = − µ k N1 + T
T = m1a + µ k N1 y : 0 = N1 − W1
⇒ N1 = W1
T = m1a + µ kW1
⎛ 1.4 N ⎞
2
1 m / s + (0.3)(1.4 N )
T =⎜
⎟
2
⎝ 9.8m / s ⎠
T = 0.562857 N
FBD2: (counterweight)

Example 5
Consider the system below. Mass A has a weight of 1.40N while mass B has a weight of 4.20N The coefficient of kinetic friction of all surfaces is 0.3. Find the force necessary such that the acceleration of mass B is 1m/s2.
A
B

F

(

Fnet = m2 a
T + fk1 + fk2
= f k1 + f k 2 + T + W1
+ W2 + N 2 + F
W1 + W2 y : 0 = −W1 − W2 + N 2
N2

F a FBD1: (mass A) a = 1m / s 2
N1
fk1

T
W1

a

)

µ k = 0.3

Fnet = m1a = f k1 + T + W1 + N1 x : m1a = − f k1 + T

⇒
⇒
x:
⇒
⇒

N 2 = W1 + W2 = 1.4 N + 4.2 N
N 2 = 5.6 N
− m2 a = f k1 + f k 2 + T − F
F = m2 a + f k1 + f k 2 + T
F = m2 a + µ k N1 + µ k N 2 + T

Example 5 (continued)

(

)

⎛ 4. 2 N ⎞
2
F =⎜
1
/ m s
+ (0.3)(1.4 N )
⎟
2
⎝ 9.8m / s ⎠
+ (0.3)(5.6 N ) + 0.562857 N
F = 3.091N

Some problems involving Newton’s
Laws of Motion combine Kinematics with the Second Law of Motion.
These problems pose situations similar to the examples just discussed but instead of asking for the acceleration of the system or for the applied force, they ask for displacement, velocity, or time.
In such problems, Newton’s Laws of
Motion and vector analysis to calculate the acceleration. The acceleration is then used with the kinematical equations.