College Preparatory Program • Saudi Aramco
Newton’s Second Law
Newton’s Second Law
Firm knowledge of vector analysis and kinematics is essential to describe the dynamics of
physical systems chosen for analysis through Newton’s second law. Following problem
solving strategy will allow you to tackle the problems with greater ease.
Problem solving strategy:
1) Write the equation of motion Fnet = ma. Where, Fnet is the net force and a is the
acceleration vector and is equal to ax i + ay j.
2) Draw a free body diagram (FBD) which shows all forces acting on the object/ objects.
Do not draw the components of the force on the FBD.
3) Specify a coordinate system. Take x-axis in direction of the motion and y-axis
perpendicular to the motion.
4) Express every force in a vector form.
5) In order to get the resultant force, Fnet, add all force vectors.
6) From the x and y components of Fnet, construct your x and y equations of motion. If the
acceleration has a component in x direction, the x-component of the resultant vector is
equal to max, otherwise it is equal to zero. Also, if the acceleration has a component in y
direction, the y-component of the resultant vector is equal to may, otherwise it is equal to
zero.
7) Depending on the question, solve the equation or the set of equations simultaneously for
the unknown.
Example (1): Motion of a block on a frictionless incline plane
The figure below shows an object of mass m moving on a frictionless inclined plane which
makes an angle θ with the horizontal.
a) Find the acceleration of the block in terms of the constants g and θ.
b) If the block started from rest and travelled a distance d, find the final speed of the block
and the time it took to cover this distance.
m
θ
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Newton’s Second Law
SOLUTION:
a)
Our object is an accelerating block on a frictionless incline plane. Now, let’s follow step-bystep the problem solving strategy and solve for the acceleration.
STEP (1): Write the equation of motion Fnet = ma. Where, Fnet is the net force and a is the
acceleration vector and is equal to ax i + ay j.
STEP (2): Draw a free body diagram (FBD) which shows all forces acting on the object/
objects. Do not draw the components of the forces on the FBD.
Based on our problem, we have only one object (the block) and there are two forces acting
on the block. The forces are:
1) Fg : Force of gravity due to the earth is pulling the block downward
2) N: the normal force due to the surface is pushing the block perpendicular to the plane.
N
Fg
STEP (3): Specify a coordinate system. Take x- axis in direction of the motion and y- axis
perpendicular to the motion.
The Fig. below shows the motion is in direction of x- axis while y-axis is perpendicular to the
motion. The forces Fg and N acting on the block are also shown.
y-axis
N
-i
j
-j
i
a
Fg sinθ
θ
Fg cosθ
θ
x-axis
Fg
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Newton’s Second Law
STEP (4): Express every force in a vector form.
This can be done by resolving every force vector into its components.
Therefore, from the geometry of the above figure, vectors Fg and N can be written as
N=Nj
Fg = Fg sinθ i - Fg cos𝜃 j
(1)
(2)
STEP (5): In order to get the resultant force, Fnet, add all force vectors.
Based on our problem, equations (1) and (2) can be added, we get
Fnet = N + Fg = N j + Fg sinθ i - Fg cos𝜃 j
Since Fnet = ma
and
a = ax i + ay j, we can write
Fg sinθ i + (N - Fg cos𝜃) j = m (ax i + ay j)
Where,
(Fnet)x = Fg sinθ is the x-component of the resultant force
(3)
(Fnet)y = N - Fg cos𝜃 is the y-component of the resultant force
(4)
STEP (6): From the x and y component of Fnet, construct your x and y equations of motion.
If the acceleration has a component in x direction, the x-component of the resultant vector is
equal to max, otherwise it is equal to zero. Also, if the acceleration has a component in y
direction, the y-component of the resultant vector is equal to may, otherwise it is equal to
zero.
Based on our problem, the motion of the block is in direction of x while there is no
acceleration in y direction. Thus, from Eq. (3) and (4) the equations of motions are:
Fg sinθ = max
and
N - Fg cos𝜃 = 0
STEP (7): Depending on the question, solve the set of equations simultaneously for the
unknown.
As we see, we were able to construct our equations of motion for the block. The
acceleration can be found from the first equation, Fg sinθ = max . The second equation has no
College Preparatory Program • Saudi Aramco
Newton’s Second Law
use in this problem unless we are asked to find the magnitude of the normal force N. Thus,
with Fg = mg, we get
mg sinθ = max.
Therefore,
ax = g sinθ
Or in vector form
a = g sinθ i
Note, maximum acceleration is when θ = 90o. This corresponds to a free falling object with
ax = g = 9.8 m/s2. Also, the acceleration does not depend on the mass of the object.
b)
Since the acceleration is constant, the final speed of the block can be found by using the
2
2
kinematic equation, 𝑣𝑓𝑥
= 𝑣𝑖𝑥
+ 2𝑎𝑥 𝑑. With vi = 0 and ax = g sinθ, we get
2
𝑣𝑓𝑥
= 2𝑔𝑑 sinθ
Therefore,
𝑣𝑓𝑥 =
2𝑔𝑑 𝑠𝑖𝑛 𝜃
The time it took the block to cover a distance d can be calculated using the kinematic
equation,
vfx = vix + ax t
With vix = 0 , 𝑣𝑓𝑥 =
2𝑔𝑑 𝑠𝑖𝑛 𝜃 and
2𝑔𝑑 𝑠𝑖𝑛 𝜃 = (g sinθ) t
t=
2𝑔𝑑 𝑠𝑖𝑛 𝜃
𝑔 𝑠𝑖𝑛𝜃
t=
2𝑑
𝑔 𝑠𝑖𝑛 𝜃
ax = g sinθ, we get
College Preparatory Program • Saudi Aramco
Newton’s Second Law
Example (2): Motion of a block on a rough incline plane
The figure below shows an object of mass m moving on a rough inclined plane which makes
an angle θ with the horizontal. If the coefficient of kinetic friction between the surface and
the block is 𝜇𝑘 , find the acceleration of the block in terms of the constants g, θ and 𝜇𝑘 .
m
θ
SOLUTION:
Problem solving strategy:
STEP (1): Fnet = ma. Where, Fnet is the net force and a is the acceleration vector and is equal to
ax i + ay j.
STEP (2): Draw a free body diagram (FBD) which shows all forces acting on the object
N
Ff
Fg
STEP (3) Specify a coordinate system and resolve the forces into x and y components. Take x-
axis in direction of the motion and y-axis perpendicular to the motion.
y-axis
N
Ff
-i
a
θ
Fg cosθ
θ
Fg
-j
Fg sinθ
x-axis
j
i
College Preparatory Program • Saudi Aramco
Newton’s Second Law
STEP (4): Express every force in a vector form.
There are three forces acting on the block: N, Fg and Ff. Therefore, from the geometry of the
above figure, the three force vectors can be written as
N=Nj
(1)
Fg = Fg sinθ i - Fg cos𝜃 j
(2)
Ff = – Ff i
(3)
STEP (5):
In order to get the resultant force, Fnet, add all force vectors, Eq. 1, 2 and 3,
Fnet = N + Fg + Ff = N j + (Fg sinθ i – Fg cos𝜃 j ) – Ff i
Fnet = (Fg sinθ – Ff ) i + ( N – Fg cos𝜃) j
(4)
STEP (6):
Construct your x and y equations of motion. The x-component of the resultant vector is equal
to max and the y-component of the resultant vector is equal to may. Thus from Eq. (4), the
equations of motion are:
Fg sinθ – Ff = max
N – Fg cos𝜃 = may = 0, since the acceleration in y direction is zero.
STEP (7):
Now, let’s solve for the acceleration.
Since, Ff = 𝜇𝑘 N = 𝜇𝑘 Fg cos𝜃, where we used Eq. (6) for the value of N. Thus, Eq. (5)
becomes,
Fg sinθ – 𝜇𝑘 Fg cos𝜃 = max
With Fg = mg, the magnitude of the acceleration is equal to
ax = g (sinθ – 𝜇𝑘 cos𝜃)
Or in vector form
a = g (sinθ – 𝜇𝑘 cos𝜃) i
(5)
(6)
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Newton’s Second Law
NOTE: If the surface is frictionless, i.e. 𝜇𝑘 = 0, the magnitude of the acceleration will be
reduced to g sinθ as discussed in example number (1). Also note, the acceleration of the block
on a rough plan is less than its acceleration on a smooth plane by the amount 𝜇𝑘 g cos𝜃.