00:01
Hi there.
00:01
So for this problem, we have first the situation that is shown for problem five.
00:08
Okay, so we have a surface, and on this we have two crates.
00:19
One of this is g, and the other is h.
00:23
Okay, so we're applying some force to g.
00:29
That force is f.
00:31
We know that there are forces at the nanjid or its weight and were.
00:39
The normal force.
00:43
Then also the force of comte in here, this will be the force of comtat or the normal force between g and h to g.
01:04
Well, for h we have its normal force.
01:08
Also we have that force that we are applying to this the frictional force oh sorry if we got the frictional force to g which is to the opposite direction frictional or g while for h as well in this direction and um the force of comtat then in this case it will be g h okay, so then with that said, let's start with problem a of this problem.
01:43
The question is, what is the force that bobs h at source on bob's g? now, with this information, what we can do is to apply newton's second's law to determine this force.
02:03
First of all, we can calculate the total mass of this system.
02:08
So that total mass is the sum of the mass of g plus the mass.
02:12
Of h and that will give us a value of 4 .20 plus 6 .30 so that will give us a value of 10 .5 kilograms.
02:25
Okay.
02:26
Now the next step is to calculate the net force acting on the system.
02:33
So the applied force is 50 and a total frictional opting on both boxes that total frictional force is 15 newtoms, okay? does the net force acting on the system is just the force that we're applying to this minus the total frictional force? so that will be 15 newtoms minus 15 newtoms and that will give us a value of 35 newtons.
03:02
We're doing all of this so we can obtain the acceleration of the system because the acceleration of the system is the net force divided by the total mass.
03:10
So that will be in this case 35 newtons.
03:13
This is, you divided by the mass, the total mass that will calculate that is 10 .15 kilograms, okay? so that will give us about 3 .33 meters per second square.
03:25
So the next step is to calculate the force that bobs h at surce on g.
03:32
So bob's g is being pushed by bob's h.
03:38
The force that boffs age exerts on boffs g is equal to the net force acting on bob's g.
03:45
The net force on bops g, we know that that will be calculated as just the mass of g times the acceleration that we just determined.
03:55
So that will be 4 .20 kilograms, this times 3 .33 meters per second square.
04:03
So that will give us a value of 13 .99 newtoms.
04:08
So that will be the answer for part a of this problem.
04:11
Now for part b, the question is about what is the force that pops g at source on h? now, according to newton's third law, the force that pops g at source and bobs h is equally magnetic but opposite interruption.
04:27
So then in this case, that force, so the first one should be negative because it is in the opposite direction.
04:40
So this should be positive.
04:44
Be in the same magnet that is 13 .99 newtoms...