00:01
In this exercise, we're going to need to use the concept of inclined plane.
00:04
So let's give a brief review of how to deal with these problems.
00:09
So consider that we have an inclined plane that makes an angle of theta with a horizontal and a block of mass m on top of the inclined plane.
00:19
Let's add up a coordinate system such that the y -axis points upwards and the x -axis points downwards.
00:29
The dxx is actually parallel to the plane and the y axis perpendicular.
00:38
And let's draw a free body diagram on the block.
00:46
So on the block, the forces that act are the normal force that acts perpendicularly to the plane.
01:00
Then we have the gravitational force that acts downwards.
01:04
And also we have the friction force that acts control.
01:10
Contrary to the direction of movement of the particle.
01:15
And we can decompose the gravitational force into two components.
01:20
So we have, this is the total gravitational force.
01:24
We can decompose it in a component that is along the x -axis and a component that is along the y -axis.
01:35
So actually, it would be better represented like this to the component.
01:42
That is along the x -axis, i'm giving the name fx, and to the component that is along the y -axis, i'm giving the name f -y.
01:54
And f -x is equal to m -g sine theta, and f -y is equal to m g cosine theta.
02:05
So this is everything we need to know in order to solve an inclined plane exercise.
02:10
So in our case, particularly, we have this system that's shown here.
02:17
And in this figure, we can see that there is a block on top of an inclined plane.
02:25
The block a has a mass of 100 pounds, and the inclined plane is such, according to the book, that it's similar to a triangle that has size four, three, and five.
02:46
Okay, so from this we can find what is the angle of the inclined plane, block a has an initial speed of 5 feet per second, and the coefficient of kinetic friction between the plane and block a is equal to 0 .2.
03:06
The mass of block b is equal to 50 pounds.
03:13
And we have these three pulleys that are represented in blue, a, b, and c.
03:22
And our goal is to find what is the acceleration of a and what is the distance, the acceleration of block a, and what is the distance the block a travels before stopping? okay, so we need to draw the free body diagram for both blocks.
03:46
So for block a, this here is block a, we have the gravitational force, the x component of the gravitational force that acts parallel to the plane.
04:11
Then we have the normal force, the y component of the gravitational force, and we also have both the friction force and the tension, which i'm going to call ta.
04:28
This is tension in the first rope, this one here.
04:34
And remember that the tension along the same rope is the same, so the tension is the same along the row, basically.
04:44
And let's write the equations.
04:47
So notice that block a must be on top of the plane, meaning that it does not move in the y direction.
04:54
Remember that the y direction is this one, the x direction is this one.
05:01
So we have that n is equal to fy, so that the block doesn't move in the y direction.
05:07
Fy, as we saw, is mg cosine theta.
05:12
This is in the y direction.
05:15
Now in the x direction, we have fx minus ta minus ff is equal to the mass of block a, and actually here it should have written ma times the acceleration.
05:34
Now fx is ma times g times sine theta.
05:40
T .a is just the tension.
05:43
We don't know its value.
05:44
The friction force is the coefficient of kinetic friction times the normal force, which is mg cosine theta, and this is equal to m .a times a.
06:03
And here i'm going to index the acceleration with an a, just to show that this is the acceleration of block a.
06:11
Okay.
06:13
So this is the equation for block a.
06:15
Let's find the equation for block b.
06:21
In block...