1. Unit 4: Newton’s Laws of Motion
14 April 2015
Compiled by Mphiriseni Khwanda
University of Johannesburg

2. Newton’s First Law
Forces are balanced
Resultant force 𝐹 = 0 𝑁
acceleration 𝑎 = 0 𝑚 𝑠−2
Objects at rest Objects in uniform motion
Stay at rest
Stay in uniform motion
Same speed & direction
𝑉 = 0 𝑚/𝑠 𝑉 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 ≠ 0 𝑚/𝑠
What happens to acceleration when
forces are balanced?
What happens to the motion of an object
when acceleration is zero?

3. Misconceptions about newton’s first law
 Newton's first law of motion declares that a force is not
needed to keep an object in motion.
 Slide a book across a table and watch it slide to a rest
position.
 The book in motion on the table top does not come to a
rest position because of the absence of a force; rather it
is the presence of a force - that force being the force
of friction - that brings the book to a rest position.
 In the absence of a force of friction, the book would
continue in motion with the same speed and direction -
forever (or at least to the end of the table top)!
 A force is not required to keep a moving book in
motion

4. Static and kinetic friction
When the two surfaces are not sliding
across one another the friction is called
static friction.
𝝁 𝑺 is called the coefficient of static friction.
𝒇 𝑺 is called the static frictional force.
𝑭 𝑵 is called the normal force which is
perpendicular to the surface.
Ns
MAX
s Ff 
When the two surfaces are sliding across
one another the friction is called kinetic
friction.
𝝁 𝑲 is called the coefficient of static friction.
Nkk Ff 
NOTE:
The frictional force is always in the
opposite direction of the motion

5. Statement of Newton’s first law and inertia
 Every object will remain at rest or
continue moving with uniform motion
in a straight line unless acted upon by
the unbalanced forces
 The continuous state of motion is
caused by inertia
 Inertia is the property of matter that
requires the unbalanced force to
change its state of motion

6. Newton’s Second Law
Forces are unbalanced
Resultant force 𝐹 ≠ 0 𝑁
There is acceleration 𝑎 = constant ≠ 0 𝑚 𝑠−2
indirectly proportional
to the mass of the object
An object changes its state of motion
It changes its velocity (increases or decreases)
OR /and
changes its direction
Directly proportional
to the Resultant Force

7. The Statement of Newton’s second
law
 The acceleration of an object is directly proportional
to the resultant force acting on an object and inversely
proportional to its mass, the object’s acceleration
should be in the same direction as the resultant
force
 Mathematically the relation can be expressed as:
𝐹 = 𝑚 𝑎
NOTE:
 The acceleration depends on force and mass and
not the other way round.
 In other words, you cannot increase acceleration in
order to increase a force

8. Misconception
 The resultant force is not a single
force but the sum of all the forces
acting on the object

9. Free-Body Diagrams
1. A free body diagram is a diagram that shows all of
the forces acting on an object.
2. There are five different ways to draw free body
diagrams: horizontal, vertical, incline plane, free
fall, and tension.
3. Gravity’s effect on an object represented by weight
4. The normal force is the force perpendicular to the
surface

10. Problem Solving Strategies: Newton’s Second Law
1. Draw a Free-Body Diagram for each object.
2. Identify if there is a friction and label the friction force
immediately because one can easily forget it.
3. Note that frictional force 𝒇 = 𝝁𝑭 𝑵 = 𝝁𝒎𝒈 =𝝁𝒎𝒈𝒄𝒐𝒔𝜽
𝒐𝒏 𝒊𝒏𝒄𝒍𝒊𝒏𝒆𝒅 𝒑𝒐𝒔𝒊𝒕𝒊𝒐𝒏
4. Also check if weight will have an impact and label it immediately,
it can be easily forgotten.
5. Choose the direction of the motion: for example if towards the right
is positive, hence all vectors pointing to the right will be positive
and to the left negative.
6. For each object, apply Newton’s second law to formulate an
equation and label the equation 1, 2 etc. for example, object 1:
𝑭 = 𝒎 𝟏 𝒂 and for object 2: 𝑭 = 𝒎 𝟐 𝒂 etc.
7. Note that if objects are attached by string all will have the
same acceleration.
8. Solve the equations simultaneously for unknowns

11. Your Turn
1. Draw a free body diagram of a 10 kg
rock as it is lifted straight up with a
constant force of 148 N. Calculate
the net force acting on the rock and
its acceleration.
2. An 8000 kg Navy jet is accelerating
upward at 4 m/s2. Calculate the
upward force provided by the jet’s
engines to achieve this acceleration

12. Your Turn
Problem117Chapter4:
Thedrawingshowsthreeobjects.Theyareconnectedbystringsthatpassovermasslessand
frictionlesspulleys.Theobjectsmoveandthecoefficientofkineticfrictionbetweenthemiddle
objectandthesurfaceofthetableis0.100.
(a) Whatistheaccelerationofthethreeobjects?
(b) Findthetensionineachofthetwostrings

13. Choose direction of the motion of the whole system
Hence choosing the movement to the Right as Positive, any
movement directed to the left will be negative

14. Considering the 10 kg object
Free-body diagram
Direction of
movement
𝑇1 − 𝑚1 𝑔 = 𝑚1 𝑎
𝑇1
𝑊1 = 𝑚1 𝑔
Applying Newton’s second law
𝐹 = 𝑚1 𝑎
𝑇1 − 10 9.8 = (10)𝑎
𝑇1 − 98 = 10𝑎 ….(1)
Considering the 80 kg
object
𝑇1
𝑓
𝑇2
Direction of movement
Note: since object moving
horizontally hence vertical forces
are balanced and can be ignored
Applying Newton’s second law
𝐹 = 𝑚2 𝑎
𝑇2 − 𝑇1 − 𝒇 = 𝑚2 𝑎
𝑇2 − 𝑇1 − 𝜇𝐹 𝑁 = 𝑚2 𝑎
𝑇2 − 𝑇1 − 𝜇𝑚𝑔 = 𝑚2 𝑎
𝑇2 − 𝑇1 − 0.1 80 9.8 = (80)𝑎
𝑇2 − 𝑇1 − 78.4 = 80𝑎 ……(2)
Considering the 25 kg
object
𝑊3 = 𝑚3 𝑔
𝑇2
Direction of
movement
Applying Newton’s second law
𝐹 = 𝑚3 𝑎
𝑚3 𝑔 − 𝑇2 = 𝑚3 𝑎
25 9.8 − 𝑇2 = 25 𝑎
245.0 − 𝑇2 = 25𝑎…(3)

15. We are left with mathematics, we have three equations and three unknowns
The equations
𝑇1 − 98 = 10𝑎 ….(1)
𝑇2 − 𝑇1 − 78.4 = 80𝑎 ……(2)
245.0 − 𝑇2 = 25𝑎…(3)
(1)+(2)+(3): 𝑇1 − 98 + 𝑻 𝟐 − 𝑻 𝟏 − 𝟕𝟖. 𝟒 + 𝟐𝟒𝟓. 𝟎 − 𝑻 𝟐 = 10𝑎 +𝟖𝟎𝒂 +25𝑎
68.6 = 115𝑎
𝑎 = 0. 6 𝑚𝑠−2
Substituting a in (1) we get 𝑇1 − 98 = 10(0.6) 𝑇1 − 98 = 6 𝑇1 = 104 𝑁
Substituting a in (3) we get 245.0 − 𝑇2 = 25(0.6) 245.0 − 𝑇2 = 15
𝑇2 = 230.4 𝑁

16. Problem 116 Chapter 4:
As the diagram shows, two blocks are connected by a rope that pass over a set of pulleys. One block
has a weight of 412 N, and the other has a weight of 908 N. The rope and the pulleys are massless
and there is no friction.
(a) What is the acceleration of the lighter block?
(b) Suppose that the heavier block is removed, and a downward force of 908 N is provided by
someone pulling on the rope as part b of the drawing shows. Find the acceleration of the
remaining block.
(c) Explain why the answer in (a) and (b) are different.

17. Problem on inclined plane
• Incline plane problems are more complex than other force problems
because the forces acting on the object on the incline plane are not
perpendicular. Therefore, components must be used.
θ
𝑊 = 𝑚𝑔
𝑚𝑔𝑐𝑜𝑠𝜃
𝐹𝐴
𝐹 𝑁
First lets identify forces acting on the object if the red object is stationary
𝑊 = 𝑚𝑔
𝑚𝑔𝑐𝑜𝑠𝜃
𝐹 𝑁
𝐹𝐴
If the object is stationary hence:𝐹𝐴= 𝑚𝑔𝑠𝑖𝑛𝜃
𝐹 𝑁= 𝑚𝑔𝑐𝑜𝑠𝜃
If the object is moving up the plane
Ignoring friction and up taken
as positive hence:
𝐹 = 𝐹𝐴 − 𝑚𝑔𝑠𝑖𝑛𝜃 = 𝑚𝑎

18. Problem on inclined plane
• Incline plane problems are more complex than other force problems
because the forces acting on the object on the incline plane are not
perpendicular. Therefore, components must be used.
θ
𝑚𝑔𝑐𝑜𝑠𝜃
𝐹𝐴
𝐹 𝑁
𝑊 = 𝑚𝑔
If the object is moving up the
plane
with friction and up taken
as positive hence:
𝐹 = 𝐹𝐴 − 𝑚𝑔𝑠𝑖𝑛𝜃 − 𝑓 = 𝑚𝑎
𝐹 = 𝐹𝐴 − 𝑚𝑔𝑠𝑖𝑛𝜃 − 𝝁𝒎𝒈𝒄𝒐𝒔𝜽 = 𝑚𝑎
Similarly If the object is moving down the
plane
with friction and down taken
as positive hence
𝐹 = 𝑚𝑔𝑠𝑖𝑛𝜃 − 𝐹𝐴 −𝒇 = 𝑚𝑎
𝐹 = 𝑚𝑔𝑠𝑖𝑛𝜃 − 𝐹𝐴 −𝝁𝒎𝒈𝒄𝒐𝒔𝜽 = 𝑚𝑎
𝑓
θ
𝑚𝑔𝑐𝑜𝑠𝜃
𝐹𝐴
𝐹 𝑁
𝑊 = 𝑚𝑔
𝑓

19. 1. Two forces act on a 4.5-kg block
resting on a frictionless surface as
shown. What is the magnitude of the
horizontal acceleration of the block?
The answer is A
A) 1.8 m/s2
B) 1.2 m/s2
C) 0.82 m/s2
D) 3.2 m/s2
E) 8.9 m/s2

20. 2. A 10-kg block is connected to a 40-kg
block as shown in the figure. The
surface on that the blocks slide is
frictionless. A force of 50 N pulls the
blocks to the right.
What is the magnitude of the acceleration of the 40-kg block?
A) 0.5 m/s2
B) 1 m/s2
C) 2 m/s2
D) 4 m/s2
E) 5 m/s2
The answer is B

21. 3. A 10-kg block is connected to a 40-kg
block as shown in the figure. The
surface on that the blocks slide is
frictionless. A force of 50 N pulls the
blocks to the right.
What is the magnitude of the tension T in the rope that connects the
two blocks ?
The answer is B
A) 0 N
B) 10 N
C) 20 N
D) 40 N
E) 50 N

22. 4. A block is at rest on a rough inclined plane and is connected to an
object with the same mass as shown. The rope may be considered
massless; and the pulley may be considered frictionless. The
coefficient of static friction between the block and the plane is µs; and
the coefficient of kinetic friction is µk.
What is the magnitude of the static frictional force acting on the block?
A) mg sin q
B) mg cos q
C) mg(1 – sin q)
D) mg(1 – cos q)
E) mg
The answer is C

23. 4. A block is at rest on a rough inclined plane and is connected to an
object with the same mass as shown. The rope may be considered
massless; and the pulley may be considered frictionless. The
coefficient of static friction between the block and the plane is µs; and
the coefficient of kinetic friction is µk.
If the rope were cut between the block and the pulley, what would be
the magnitude of the acceleration of the block down the plane?
The answer is E
A) g
B) g – k sin q
C) g – k cos q
D) g(tan q – k sin q)
E) g(sin q – k cos q)

24. 5. A block is at rest on a rough inclined plane and is connected to an
object with the same mass as shown. The rope may be considered
massless; and the pulley may be considered frictionless. The
coefficient of static friction between the block and the plane is µs; and
the coefficient of kinetic friction is µk.
If the mass of the suspended object is doubled, what will be the
acceleration of the block up the plane?
The answer is D
A) g(2 – k sin q)
B) 2g(k sin q – cos q)
C) g(2tan q – k sin q)
D) g(2 – sin q – k cos q)
E) g(2cos q – k sin q)

25. Additional Question
One 3.2-kg paint bucket is hanging by a
massless cord from another 3.2 kg paint bucket,
also hanging by a massless cord.
a. If the buckets are at rest, what is the tension
in each cord?
b. If the two buckets are pulled upward with an
acceleration of 1.60 m/s2 by the upper cord,
calculate the tension in each cord.