Southern Illinois University

Lecture 3

In this lecture, we will START applying Newton's equations to investigate the dynamics of a particle. Let's start with two simple examples, a block sliding down a frictionless plane and a plane with friction. Notice that, both these problems can be solved using Kinematics equations. In the Homework # 1, you will do both these problems using the Kinematics equations.

In addition to applying the Newtonian mechanics for the motion of a particle, you will alsolearn about the static and kinetic frictional coefficient.

There is an inetrative simulation embedded (from phET). In this lecture, you will ONLY see if you can open this simulation and play with it. (Just check the compatibility of these simulations in your computers and let me know)

End of the class, we also discuss the other types of retarding force. We will do a problem with the velocity dependent ratarding force.

3.1 Example 1: A Block sliding down a frictionless inclined plane
3.2 Frictional Forces

- Static Frictional Coefficient

- Kinetic Frictional Coefficient

3.3 Example 2: A Block sliding down an inclined plane with a friction:
3.4 Other types of retarding forces
3.5 Retarding Forces in Horizontal Direction

3.1 EXAMPLE 1: A Block sliding down a frictionless inclined plane:

A block is sliding down a frictionless plane, that is inclined by an angle theta . What is the accelleration of the block? What is the velocity of the block after it moves from rest a distance x0 down the plane?

Last lecture, we discussed about the choice of our coordinate system (as shown in the figure) Now Mark the forces using the chosen coordinate system.

- The Normal force N is acting in the ŷ direction.
- The gravitational force Fg can be decomposed in to x and y directions.

Knowing the force acting on the system, we can apply the Newton's second law of motion: Both in x and y directions.

In y direction: There 's no accelleration in y direction.

In x direction: There is an accelleration in the x direction. The only force in the x direction is coming from the gravitational force.

Where ẍ is the acceleration along the x direction, (That is what we are looking for):

Now let's use the fact that the gravitational force equal mg, that

Now let's put the equation in the form:

Now by integrating these equations and applying the boundary conditions, we can find out the equations of motion for the particle. That is how the velocity depend on time and how the position depend on time.

By integrating the above equation, we get:

where C1 is a constant.

We use the initial conditions (boundary conditions) to find the unknown integration constant C1.

At v(t=0)=0 and x(t=0)=0. The block starts with the zero velocity and we set the coordinate system such that at time t=0, the position coordinate of the block equals zero.

This conditions give C1=0: Now

------------------- (1)

Let's write this equation as,

By integrating again:

By considering the initial condition, at t=0 x(t)=0 : Therefore C2=0.

Now we have the position as a function of time:

-------------------- (2)

Now if we focus on the question, we want to find out the velocity of the block after it moves a x0 distance on the plane. So we need to write down the velocity as a function of the position (instead of as a function of time).

We can find t from equation (2) and substitute in the equation (1).

From (1):

Substituting that back in the equation (1):

Finally, we get,

We can now find the velocity of the block at any position on the plane.

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2.2 Force of Friction

Frictional forces are occured when one surface is in touch with another surface. Frictional forces keep things from moving. These forces adjust itself to counter the force exerted to move the object. There is a point where the frictional force cannot excedd a particular maximum value. The maximum frictional force s obtained by the experimentally obtained formula:

where µs is the static frictional coefficient. In another words, the static frictional force gives the force needs to keep the object from moving.

When the object starts moving, there's still a frictional force that oppse the motion. This force is called the force of kinetic friction. Kinetic frictional force is usually less than the maximum static frictional force.

3.3 Example2: A Block sliding down an inclined plane with a FRICTION:

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Let's work on the example.

If the coefficient of friction between the block and the plane in the previous example is µs=0.4, at what angle will the block start sliding if it initially is at rest.

The frictional force is in the -x direction, as shown in the figure. So the equation of motion in the y direction will remain the same.

Now the equation of motion in the x-direction will be changed.

The static friction force keeps the block at rest. As you increase the angle, the Frictional force increase. There is a particular inclined anlge, for which the frictional force cannot hold the block from moving.

Now by using the form of the frictional force:

We get:

Given the fact that the static frictional coefficient equal 0.4, we can calculate the inclined angle of the plane at which the block start to slide. (The angle for this problem equals to 22 degrees.

Once the object start moving, we need to use the coefficient of Kinetic friction to calculate the frictional force.

Let's visualize this problem in an interactive simulation.

------ PhET Simulation -------

Click to Run

The above Mathematica CDF player was downloaded from wolfram website, and I have modified it to add the frictional force.

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As the Block starts moving:

After the block in the previous example begin to slide, that means, the gravitational force is larger than the frictional force, the coefficient of kinetic froctions needs to be taken in to account. Let's find the accelleration of the block at an angle 30 degrees. (We earlier found out that at 22 degrees of inclination, the block start to slide.)

Again, similar to the previous case, we only need to consider the motion in the x direction (because y direction direction equation of motion stays the same as in non-frictional case.

Frictional Force:

Now the equation of motion in the x direction becomes:

By plugging the numerical values, we can get a numerical value for the accelleration of the Box at angle 30 degrees.

3.4 Other types of retarding forces

In the previous problem, we have done a frictional force, a type of a resistive force throughout the motion.

Will all resistive forces be constant through out the motion? Are there time dependent (or position or velocity dependent) resistive forces? That is the question for the next section.

Here we are talking about a type of a resistive force, which is a function of a instantaneous velocity of the object. This means, the force keeps changing as the object moves.

For example, if an object falls in a constant gravitational field, there is the gravitational force mg acting on the particle. Additionally, there can be an additional retarding force: Fr:

Usually, the force is proportional to some power of the speed. It can also come in complicated velocity dependance. But in this context, we only do the retarded froces with a simple velocity dependance.

In the power law approximation:

where k is a positive constant, which specifies the strength of the retarding force. The last term in the above equation (along with the negative sign) tells that the rearding force is in the direction opposite to the direction of the velocity.

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3.5 Retarding Forces in Horizontal Direction:

Find the dispalcement and velocity of horizontal motion in a medium in which the retarding force is proprtional to the velocity.

Let's apply the Newton's equation along the horizontal direction:

-------------- (A)

Now, in order to get the integration constant C1, we use the initial conditions. Let's sat at time t=0, the object started with a velocity v0.

Then,

That is:

---------------- (B)

This situation can be visually represented as:

The velocity exponentially decreases because of the retarding force as shown in the majenta curve.

Let's integratie the velocity equation (eq.1) again, to find out the displacement x as a function of time:

Now, we can use the initial conditions to find the integration constant C2. At t=0, x=0 (That's the choice of the coordinate system)

So C2=v0/k; Therefore, the position coordinate can be written as:

That tells, the position coordinateexponentially approaches the value v0/k. The result can be graphically represented as follows.

Now we have solved for the velocity as a function of time and the position as a function of time. Now we are going to find the velocity as a function of the position, such that we can find the velocity at any given displacement.

Now from that, we get,

(Here we have used the equation (A) above. By integrating the last equation,

And finally we get,

That tells, the velocity decreases linearly with the displacement.

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THUSHARI Jayasekera - ASSistant Professor

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