Physics Tutorial: Newton's Second Law of Motion

In this Physics tutorial, you will learn:

• What is the relationship between force and acceleration in a motion?
• What is the relationship between mass and acceleration in a motion?
• What does the Newton's Second Law of Motion says?
• What is the procedure used to study the motion of objects when forces acting on it are considered?
• What advantages offers the use of a force diagram when dealing with Newton's Second Law of Motion?
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4.6Newton's Second Law of Motion

Introduction

Suppose you are trying to make a heavy object move from rest. Is this action easy for you or not? Can you make this object accelerate as a steady rate for a long time? Why?

What about a heavy object when it is moving? Is it easy for you to stop it at instant? Explain.

The Relationship between Force, Mass and Acceleration. Newton's Second Law of Motion

In the previous tutorial, "Newton's First Law of Motion. The Meaning of Inertia", we stated that heavy objects are more inert than light ones as they have a greater mass. It was explained that Inertia is a physical concept related to objects' tendency to preserve their previous state of motion. As a result, the objects move at constant velocity when the forces acting on them are balanced.

But what happens when forces acting at the same object are not balanced? Does the object still move at constant velocity?

The answer is NO. When forces are not balanced, the resultant force at the object is different from zero. As a result, the object will move in the direction of the resultant force as shown in the figure below.

The vector equation of resultant force in this case is:

FR = F1 + F2

However, since F1 and F2 have opposite directions, the scalar equation of the resultant force will be

FR = F1 - F2 > 0

As a result, the object will move in the direction of the resultant force (here, in the direction of the largest force, i.e. due right).

If the object was at rest, its initial velocity v0 is zero. When it starts moving, the magnitude of velocity v is not zero anymore. As this process (making an object move) takes some time t, there will be an acceleration that is not zero, based on the equation (and definition) of acceleration. It is known that the equation of acceleration caused by this non-zero resultant force (as stated in the Physics tutorial "The Meaning of Acceleration. Constant and non-Constant Acceleration"), is

a = v - v0/t ≠ 0

Greater the magnitude of the resultant force, greater the acceleration it causes at the object in which the forces act. Therefore, acceleration is (directly) proportional to the resultant of forces acting at the same object. Mathematically, we write this conclusion as

Equation 1

a ~ FR

On the other hand, it is obvious that there is a relationship between the acceleration and mass. Thus, greater the object's mass (more inert an object is), more difficult is to make the object move. This means the most massive objects experience smaller acceleration as they cannot gain much velocity when a force (or a number of forces) acts on them. Therefore, acceleration is inversely proportional to the mass of object. Mathematically, this inverse proportionality is written as:

Equation 2

a ~ 1/m

Combining the relations (1) and (2), we obtain the mathematical expression of the Newton's Second Law of Motion:

Equation 3

a = FR/m

In words, the Newton's Second Law of Motion states that:

"The acceleration an object gains due to the action of a force (or the resultant of some forces) is directly proportional to the force itself and inversely proportional to the mass of the object."

Example 1

Five forces are acting at a 40 kg object as shown in the figure below.

The magnitudes of these acting forces are: F1 = 120 N, F2 = 70 N, F3 = 40 N, F4 = 100 N and F5 = 100 N. Calculate:

1. The acceleration of the object and its direction
2. The velocity of the object after 12 seconds if it starts moving from rest
3. The displacement of the object during this interval

Solution 1

a. We have to calculate the resultant force acting at the object first, in order to find the acceleration later. Thus, from the figure we can see that the forces F4 and F5 cancel each other, as they are equal in magnitude and opposite in direction. Therefore, no resultant force exists in the up-down direction.

As for the left-right direction, we have:

FR = F1 + F2 + F3

The scalar representation of the above equation considering the forces direction is

FR = F1 - (F2 + F3)
= F1 - F2 - F3

Substituting the values, we obtain:

FR = 120N - 70N - 40N
= 10N

The positive result for the resultant force means its direction is due right as it complies with the direction of the largest force.

Therefore, based on the Newton's Second Law, we have:

a = FR/m = 10N/40kg = 0.25 m/s2

The acceleration is in the direction of the resultant force as when a vector (here the resultant force) is divided by a positive scalar (here the mass), the result is a new vector (the acceleration) whose direction is the same as the original vector. Therefore, the direction of acceleration is due right as well (this means the object speeds up due right).

b. Based on the results obtained at the part (a), we have:

v = v0 + a × t

Substituting the known values, we obtain

v = 0 + 0.25 m/s2 × 12s
= 3 m/s

The displacement ∆x is calculated through the known kinematic equation

∆x = v0 × t + a × t2/2
= 0 × 12 + 0.25 × 122/2
= 18m

As you see, there is a strong relationship between Kinematics and Dynamics when studying the motion of objects.

Remark! If the resultant force is negative, it mean that resistive forces are greater than the moving force. As a result, the acceleration will be negative as well. This causes a slowing down in the object's motion. As an example in this regard, we can consider a rolling ball along a horizontal surface. After throwing it, the moving force becomes zero as no force is pushing the ball anymore. Frictional force now is the only force acting on it. We know that frictional force has an opposite direction to the motion, so the resultant force is negative. This causes the ball to slow down and eventually stop.

The Advantages of Drawing a Force Diagram When Dealing with Forces

When several forces are acting at an object (especially when forces are at different directions), it is better to start with drawing a force diagram. This helps us understand better the situation and estimate the possible direction of motion prior of making any calculation.

A force diagram includes all forces acting at an object despite some of them maybe are not involved in the motion process. The figure below shows a simple force diagram for an object moving at constant acceleration along a rough horizontal surface.

Weight and Normal Force are equal and opposite, and this fact is outlined through the equality in length of the vectors representing these forces. On the other hand, since the object is accelerating, the length of the moving force vector must be greater than that of frictional force vector and also their direction must be opposite because frictional force is a resistive force that acts in the opposite direction of motion (which in this example is due right).

If we use the standard notation for the directions of forces (x for the horizontal and y for the vertical), we have two horizontal forces (moving force F and frictional force f) and two vertical (weight W and normal force N) ones. We do not write the gravitational force Fg in this diagram as it is substituted by the weight.

When one or more forces do not lie only according one of the basic directions, we can use the components to represent them as shown in the figure below.

Since the acting force is neither completely horizontal, nor vertical, we split it into components. As a result, the weight (and normal force as well) will be smaller in magnitude than the gravitational force as its value is reduced because of the upward effect of the vertical component Fy of the moving force. The vector equation of weight is

W = -N
= Fg - Fy
= m × g - |F| × sin θ

This change also affects the value of frictional force f as

|f| = μ × |N|
= μ × (m × g - |F| × sin θ)

If the acting force in the figure above was sloped down, the object's weight would be greater than the gravitational force as the vertical component of the acting force Fy is directed downwards. As a result, the equation of weight W (as opposite to the normal force N), is

W = -N
= Fg + Fy
= m × g + |F| × sin θ

Look at the figure:

Example 2

:

A 14 kg object is pulled along a horizontal plane by a 40N force, as shown in the figure below.

If the force acts at 37° sloped down direction, what is the displacement of the object in 20 s if it starts moving from rest? The friction coefficient between the object and the horizontal plane is 0.2. Take cos 37° = 0.8, sin 37° = 0.6 and g = 10 m/s2.

Solution 2

The first thing to do is to draw a force diagram as explained in the theory. The diagram is shown in the figure below.

The Newton's Second Law of Motion for this case is

Fx - f = m × a

We have

Fx = |F | × cos θ = 40N × 0.8 = 32 N

and since the vertical component of the force F is directed upwards, we also have:

|f |=μ × |N|
= μ × (m × g - |F| × sin θ )
= 0.2 × (14 × 10 - 40 × 0.6)N
= 0.2 × (140N - 24N)
= 23.2N

Now we can substitute these values in the formula of Newton's Second Law of Motion to obtain the acceleration. We have

32N - 23.2N = 14 kg × a

Therefore,

a = 32N - 23.2N/14kg
= 8.8N/14kg
= 0.63 m/s2

Now, let's use the kinematic formula

∆x = v0 × t + a × t2/2

to find the displacement ∆x, where the initial velocity v0 = 0 as the object starts moving from rest. Hence, we obtain after substitutions,

∆x = 0 × 20 + 0.63 × 202/2
=126 m

When the object lies on a slope, such as an inclined plane shown in the figure below, the reader has two possible choices:

1. To split in components according the horizontal and vertical direction the Normal force, frictional force and any possible moving force (only gravitational force is vertical) and then making the calculations accordingly
2. To rotate the axes in such a direction that one axis fits the plane (any possible motion occurs according this direction) and the other axis is perpendicular to the plane (it fits to the normal force). In this case, only gravitational force needs to split in directions. Look at the figure:

Even if no moving force is acting on the object, it tends to go downslope due to the x-component of the gravitational force Fg(x). Therefore, a frictional force f appears in the opposite direction of this moving tendency.

On the other hand, Weight is not equal to the Gravitational Force as they have different directions. However, Weight is equal to one of the Gravitational Force components, the y-component. As a result, the Normal Force will be numerically equal and in the opposite direction to this y-component of Fg as well.

Example 3

What is the acceleration of the object sliding downslope the plane shown in the figure above if the mass of the object is 4 kg, the plane is inclined at 25° to the horizontal direction? The friction coefficient between the object and the plane surface is 0.12. (Use the approximate values, cos 25° = 0.9, sin 25° = 0.4, g = 10 m/s2).

Solution 3

The first thing to do is drawing a force diagram where all forces including their components are shown when necessary. We have the diagram already drawn so there is no need to draw it again.

It is obvious that if there is any motion (and therefore any acceleration), it exists only in the downslope direction. Therefore, the Newton's Second Law of Motion (in scalar mode) for this direction (x-direction) is:

Fg(x) - f = m × a

We have:

Fg(x) = Fg × sin θ
= m × g × sin θ
= 4 × 10 × 0.4
= 16 N

and

f = μ × N
= μ × m × g × cos θ
= 0.12 × 4 × 10 × 0.9
= 4.32N

Therefore, we obtain after the substitutions

16 - 4.32 = 4 × a
11.68 = 4 × a

Thus, the acceleration is

a = 11.68/4 = 2.92 m/s2

The Standard Procedure Used for Solving Exercises on the Newton's Second Law of Motion

Since the situations involving Newton's Second Law of Motion are a bit complicated, it is advisable to use the standard procedure shown below to solve them easier. This procedure consist in 6 steps:

1. Plotting a force diagram in which all forces acting at the object (or the system of objects) are shown,
2. Choosing a suitable pair of axis to show the direction of forces,
3. Drawing the components of forces where necessary,
4. Writing Newton's Second law of Motion according to both directions for all objects involved. The equation(s) of the perpendicular direction to the (possible) motion are used as auxiliary equation that helps to find any missing quantity that appears in the equation(s) according the direction of motion (in our previous example, we used the component Fg(y) of gravitational force to find N and then the frictional force f through it),
5. Substituting the known values in the active equation(s), i.e. in the equations that contain the forces according the direction(s) of motion, and
6. Making the calculations and finding the result.

Example 4

A system is composed by two objects connected through a non-extendable string that passes through a fixed pulley, as shown in the figure below.

The mass of the large object is M = 5 kg and that of the small one is m = 2 kg.

1. What is the tension T in the string?
2. What is the minimum friction coefficient μ so that the system remain at rest?

The pulley causes no friction to the string. Take g = 10 N/kg.

Solution 4

Let's apply the abovementioned procedure to solve this problem.

1. The forces diagram acting at the two objects is shown below:
1. The directions here are the standard ones, i.e. x for horizontal and y for the vertical.
2. Here we don't need to split any force vector in components as all, forces lie according one of the two basic directions.
3. Normally, for a system composed by two objects we would have four different equations: two for the horizontal and two for the vertical direction. However, it is not necessary to write any equation for the horizontal direction of the object m as it moves only vertically. Therefore, we write only three equations for this system:

i. For the object M according x:

T1 - f = M × a1x

ii. For the object M according y:

N - M × g = M × a1y

iii. For the object m according y:

m × g - T2 = m × a2y

The active equations are (i) and (iii). The second equation helps to determine the normal force N, which on the other hand is used to calculate the frictional force f.

Also, giving that T is the tension caused by the stretching of the string and since there is the same string passing through the pulley, we have T1 = T2 = T.

Another thing to mention here is that acceleration is zero because the problem says the system is not moving. Therefore, a1 = a2 = a.

1. The system composed by the two active equations, therefore is
T - μ × M × g = M × a = 0/m × g - T = m × a = 0
1. From the second equation of the system, we have

m × g - T = 0
T = m × g
= 2kg × N/kg
= 20N

6b

T - μ × M × g = 0
20 - μ × 5 × 10 = 0
20 = 50 × μ

Thus,

μmin = 20 N/50 N
=0.4

Whats next?

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