00:01
Hi there, so for this problem, we have a block with a mass that is given and that mass is equal to 10 .8 kilograms and rest on an inclined plane with a slope, with a slot, slope of 37 degrees.
00:30
Yeah, 37 degrees.
00:34
And the coefficient of a static friction between the block and the incline plane is also given, and that is equal to 0 .21.
00:47
Now with that set for par b of this problem, we are asked about to determine the minimum force f required to push the block up the incline.
00:59
Okay, so let me just draw this situation that we have in here.
01:04
We have an incline, an angle in here, an angle tita.
01:12
We have a block on top of this.
01:17
We have a force that is being applied along the incline.
01:22
This is the force that we're going to call the force f.
01:27
The other forces that are acting on this are, of course, the weight of this, which is just the mass times acceleration due to the gravity.
01:41
We have the normal force that is perpendicular to the surface of contact.
01:47
And because this block is moving upward, or we are applying a force in order to move this upward, there is a frictional force to the, we're going to call it just simply aft, to the left.
02:08
So we can set the a axis in here.
02:11
This is going to be the y -axis, the x -axis, okay.
02:21
And then in here, this angle makes an angle of theta with respect to the vertical.
02:29
We're gonna set that the forces pointing towards the left are negative and towards the right are positive.
02:38
And now we just need to apply new time second slu, to the forces, for example, let's start with the forces acting on the, on the x direction.
03:05
So we have, as you can see from the figure, we have the force f positive because it is pointing towards the right.
03:21
This minus the frictional force and this minus.
03:27
Minus the x component of the weight, which is the mass times the acceleration data gravity times the sign of the angle theta.
03:36
Then we set this, we want this condition at just the minimum force, so we assume that this is equal to zero.
03:47
Then summing the forces, adding the forces, adding on the y direction, we have the normal force minus the y component of the weight, which is the mass times the acceleration, the gravity times the cosine of theta, then this is equal to zero.
04:03
So from this last expression, we obtained that the normal force is the weight cosine of theta.
04:11
And now we know that the condition of the frictional force is that is going to be greater or well, in this case, because we want the minimum force that will be equal to the frictional force is equal to the coefficient of the coefficient of static friction times the normal force...