0:00
Hi there.
00:01
So for this problem, we are told that this force can either push the block upward at a constant velocity or allow to slide that war at a constant velocity.
00:13
Now, we are told that the magnitude of the force is different in the two cases, while the directional angle, we're going to call this angle theta, is the same.
00:31
Now, kinetic friction exits between the block and the wall, and the coefficient of kinetic friction is given, and that has a value of 0 .25.
00:47
Now, the weight of the block is also given, and that weight is equal to 53 newtons, and the directional angle of 4th of 4th, that angle theta, is equal to 42 degrees.
01:02
Now we need to determine the magnitude of the force f for two cases.
01:13
The case a, the first one that we are going to do in here, is for when the block slides up the wall.
01:33
To solve this problem is convenient first to draw the diagram of forces.
01:39
So let me just put this one, this in here.
01:44
Let's just draw is the block, and then we are going to have the normal force that is always perpendicular to the surface of contact.
02:00
We have its weight downward and also we have two components for the force f.
02:12
So we have one force that is to the right.
02:17
This is the x component of the force and since it is it is, it is, lights up so that the force, the white component of the force should be up war.
02:35
Now because we know that the block is moving upward, we know that there is going to be a frictional force that opposes the motion, and that frictional force then should be done where.
02:53
Now, with that said, we can apply newton's seconds law to assume all of these force.
02:58
So for this block going up, we sum the forces in the x component.
03:07
So we will have that those are.
03:09
We have the x component of the force.
03:12
We set it that is positive because it is pointing to the right.
03:16
And this minus the normal force.
03:20
And then, and this should be equal to the mass times the acceleration in that direction.
03:27
However, we know that the block is not moving.
03:30
In the x direction.
03:31
So this should be equal to zero.
03:35
Now, with the angle defined as it is in the picture, the horizontal or x component is opposite of the angle, and therefore goes with the sign instead as the usual cosine.
03:50
So as you can see in here, so you can see the angle is given with respect to the vertical, so that the x component in this case, that we can solve from here is equal to the magnitude of that force, sine of theta, and from this, if we solve for the normal force, we obtain that this is equal to the normal force.
04:20
Now we do something similar for the sum of forces in the y component.
04:27
So that will be.
04:30
The y component of the force and this minus the weight, because it is pointing downward, and this minus the frictional force, which is also downward, and this should be equal to the mass times the acceleration in that direction.
04:52
However, we are told that the block is troubling up the wall with a constant speed, and that means that if it is troubling with a constant speed, then its acceleration should be equal to zero, so this should be equal to zero.
05:11
Now with that said, we can write that the white component of this force is its magnitude cosine of the angle theta.
05:20
And then solving from this expression, we will have that this should be equal to the weight plus the frictional force.
05:28
And we know that the frictional force is defined as the normal force times the coefficient of friction.
05:35
And we can use the fat from before that the normal force is equal to the force sign of theta.
05:41
So that's what we are going to substitute in here the force, sign of theta.
05:53
Now, with this, we can start solving for the force, the magnitude of the force, and from this, we immediately obtain that the magnitude of the force should be equal to, we will have, in the denominator, we will have cosine of theta minus the coefficient, sine of theta, and in the numerator we will have the weight.
06:20
So now we just need to simply substitute all of this numerical value, so we will have cosine of the angle theta, and the angle theta is given, 42 degrees, this value right here.
06:32
Now we will have minus the coefficient that is 0 .25, sine of 42 degrees, and in the numerator, we will have the weight, and the weight is equal to 53 newtoms...