So let us look at the application of Newton's second law to solving kinetics problems by making use
of an example.
So what we have here is a crate of mass m 100 kilogram, which is being pulled across a surface by force
F of one point five klayton as shown in the figure.
So you can see that force is applied at a certain angle of three degrees with the horizontal.
And what we need to do is that if the coefficient of friction mirror is zero point four, we need to
determine the acceleration of this.
Great.
Now, Newton's second law states if is equal to M multiplied by A, and what are we going to do is we're
going to take this crate and we're going to represent it as a free body.
We're going to draw a free body diagram.
So we're going to indicate all the forces that work onto the street and we're going to sum those forces
in the different components.
And we are going to equate that to M multiplied by eight.
We have the mass of the crate and then we can solve for the acceleration.
So this is an example of using Newton's second law to solve a kinetic problem.
So what we have in the figure is the great negated by the Green Square with the forces working in on
it.
So working down, we have the force of gravity IFG.
This great is moving to the right, so in the opposite direction, that is to the left, we have the
force of friction if, if and then we have the applied force of one point five Klayton, five 1500 Newton.
And we can break that force down into an X component and a Y component.
This force is at an angle of 30 degrees with the x axis horizontal, so we can easily break it down.
We have to break it down because we're going to send the forces in the different directions.
So let us find these different forces first before we apply Newton's second law, first of all, we
find the force due to gravity.
Now, that is M times G where G is your gravitational acceleration.
M is the mass of the great.
We know that is 100 multiplied by nine point eight one and that gives us nine hundred and eighty one.
Newton going downwards for the force of gravitation.
Now we break the applied forced up into its X and Y component.
So if X would be if cosine of the angle, so that would be 1500, cosine of 30, that gives us the component
of the applied force of the X direction.
That is one thousand two hundred and ninety nine Newton.
And to get the component in the Y direction, we say the force multiplied by some 30.
So that will be one thousand five hundred and thirty and that gives us 750 Newtons.
Then we can calculate the frictional force, not the frictional forces.
The friction coefficient multiplied by the normal force and the normal force is the force.
That is the reaction to that gravitational force.
Now we have the gravitational force going down and then we have the Y component of the applied force
going up.
So the difference between those two forces will be the normal force that is applied onto the great by
the surface on which it is pulled.
So to get the normal force, we need to take nineteen eighty one.
That's the gravitational force going downwards.
We need to subtract if y the y component of the applied force that is seven hundred and fifty and then
we have the normal force on the body.
And then we multiply that by the coefficient of friction, which is zero point four, and then we get
the frictional force of ninety two point four meters.
So we can write these down on the free body diagram like we've done here, we've got the frictional
force, we've got the gravitational force and we have two components of the applied force.
Now we are ready to apply Newton's second law.
Newton's second law says F is equal to multiplied by eight and we some the forces in the X direction
because we want to get the acceleration in the X direction.
So you can see that we say to the right, which is our positive direction.
We've got a force of one thousand two hundred and ninety nine Newton and to the negative X direction,
we have a force of ninety two point four Newton, so we subtract ninety two point four from one to nine
and we say that is our some of the forces in the X direction that is equal to M multiplied by eight.
And we know M is 100 and we can solve four and we get 12 meters per second squared.
So what is the acceleration in the Y direction while that is zero, because remember this great is on
a surface, so it doesn't move in the direction of the forces in the Y direction.
Cancel out the R balanced.
We have a gravitational force going down on 981 Newton.
We've got the Y component of the Applied Force going up of 750 Newton and then we have the normal force
also going up of 231 Newton.
And if you add the normal force, it goes up and the Y component of the blunt force, you'll see that
is exactly the same as the gravitational force going down.
So there is no acceleration in the wind direction.
If you have a body with a mass moving in the X and Y direction, you can still solve it with Newton.
Do you just some of the forces in that direction.
And you said that is equal to the mass, multiplied by the acceleration in the X direction.
Then you some the forces in the Y direction.
You said that is equal to the mass, multiplied by the acceleration in the Y direction.
And in both cases you solved in four.
But in this case we only have movement in that direction.
So eight is 12 metres per second.
Now, this is just an example of how you would use Newton's second law to solve kinetics problems.
And it is quite straightforward.
As long as you remember, you can only some forces that are in the same direction, positive or negative.
So let's say X direction or Y direction, and you then equate it to the mass multiplied by the acceleration
in that direction, all those forces.