1
00:00:01,380 --> 00:00:07,200
So let us look at the application of Newton's second law to solving kinetics problems by making use

2
00:00:07,200 --> 00:00:08,200
of an example.

3
00:00:08,850 --> 00:00:15,570
So what we have here is a crate of mass m 100 kilogram, which is being pulled across a surface by force

4
00:00:15,810 --> 00:00:19,790
F of one point five klayton as shown in the figure.

5
00:00:20,070 --> 00:00:26,260
So you can see that force is applied at a certain angle of three degrees with the horizontal.

6
00:00:26,940 --> 00:00:33,810
And what we need to do is that if the coefficient of friction mirror is zero point four, we need to

7
00:00:33,810 --> 00:00:35,460
determine the acceleration of this.

8
00:00:35,460 --> 00:00:35,850
Great.

9
00:00:36,790 --> 00:00:44,290
Now, Newton's second law states if is equal to M multiplied by A, and what are we going to do is we're

10
00:00:44,290 --> 00:00:47,860
going to take this crate and we're going to represent it as a free body.

11
00:00:47,890 --> 00:00:49,860
We're going to draw a free body diagram.

12
00:00:49,870 --> 00:00:55,990
So we're going to indicate all the forces that work onto the street and we're going to sum those forces

13
00:00:55,990 --> 00:00:57,490
in the different components.

14
00:00:58,060 --> 00:01:02,460
And we are going to equate that to M multiplied by eight.

15
00:01:02,630 --> 00:01:06,580
We have the mass of the crate and then we can solve for the acceleration.

16
00:01:06,610 --> 00:01:11,510
So this is an example of using Newton's second law to solve a kinetic problem.

17
00:01:12,220 --> 00:01:17,980
So what we have in the figure is the great negated by the Green Square with the forces working in on

18
00:01:17,980 --> 00:01:18,110
it.

19
00:01:18,850 --> 00:01:22,090
So working down, we have the force of gravity IFG.

20
00:01:23,370 --> 00:01:28,290
This great is moving to the right, so in the opposite direction, that is to the left, we have the

21
00:01:28,290 --> 00:01:38,640
force of friction if, if and then we have the applied force of one point five Klayton, five 1500 Newton.

22
00:01:39,030 --> 00:01:43,250
And we can break that force down into an X component and a Y component.

23
00:01:44,210 --> 00:01:49,990
This force is at an angle of 30 degrees with the x axis horizontal, so we can easily break it down.

24
00:01:50,040 --> 00:01:53,620
We have to break it down because we're going to send the forces in the different directions.

25
00:01:54,940 --> 00:02:02,530
So let us find these different forces first before we apply Newton's second law, first of all, we

26
00:02:02,530 --> 00:02:04,760
find the force due to gravity.

27
00:02:04,780 --> 00:02:09,400
Now, that is M times G where G is your gravitational acceleration.

28
00:02:10,060 --> 00:02:11,740
M is the mass of the great.

29
00:02:11,740 --> 00:02:17,740
We know that is 100 multiplied by nine point eight one and that gives us nine hundred and eighty one.

30
00:02:17,740 --> 00:02:21,850
Newton going downwards for the force of gravitation.

31
00:02:22,690 --> 00:02:29,000
Now we break the applied forced up into its X and Y component.

32
00:02:29,530 --> 00:02:40,060
So if X would be if cosine of the angle, so that would be 1500, cosine of 30, that gives us the component

33
00:02:40,060 --> 00:02:42,500
of the applied force of the X direction.

34
00:02:42,640 --> 00:02:45,300
That is one thousand two hundred and ninety nine Newton.

35
00:02:45,850 --> 00:02:52,290
And to get the component in the Y direction, we say the force multiplied by some 30.

36
00:02:52,900 --> 00:02:57,610
So that will be one thousand five hundred and thirty and that gives us 750 Newtons.

37
00:02:58,630 --> 00:03:03,010
Then we can calculate the frictional force, not the frictional forces.

38
00:03:03,010 --> 00:03:10,660
The friction coefficient multiplied by the normal force and the normal force is the force.

39
00:03:10,660 --> 00:03:14,560
That is the reaction to that gravitational force.

40
00:03:15,130 --> 00:03:20,920
Now we have the gravitational force going down and then we have the Y component of the applied force

41
00:03:20,920 --> 00:03:21,680
going up.

42
00:03:22,120 --> 00:03:30,100
So the difference between those two forces will be the normal force that is applied onto the great by

43
00:03:30,130 --> 00:03:31,900
the surface on which it is pulled.

44
00:03:32,530 --> 00:03:37,090
So to get the normal force, we need to take nineteen eighty one.

45
00:03:37,090 --> 00:03:39,510
That's the gravitational force going downwards.

46
00:03:39,910 --> 00:03:48,520
We need to subtract if y the y component of the applied force that is seven hundred and fifty and then

47
00:03:48,910 --> 00:03:52,660
we have the normal force on the body.

48
00:03:53,820 --> 00:03:58,830
And then we multiply that by the coefficient of friction, which is zero point four, and then we get

49
00:03:58,830 --> 00:04:01,830
the frictional force of ninety two point four meters.

50
00:04:03,240 --> 00:04:09,390
So we can write these down on the free body diagram like we've done here, we've got the frictional

51
00:04:09,390 --> 00:04:15,390
force, we've got the gravitational force and we have two components of the applied force.

52
00:04:15,660 --> 00:04:18,380
Now we are ready to apply Newton's second law.

53
00:04:18,900 --> 00:04:26,130
Newton's second law says F is equal to multiplied by eight and we some the forces in the X direction

54
00:04:26,130 --> 00:04:29,000
because we want to get the acceleration in the X direction.

55
00:04:29,490 --> 00:04:36,340
So you can see that we say to the right, which is our positive direction.

56
00:04:36,370 --> 00:04:42,750
We've got a force of one thousand two hundred and ninety nine Newton and to the negative X direction,

57
00:04:42,750 --> 00:04:51,510
we have a force of ninety two point four Newton, so we subtract ninety two point four from one to nine

58
00:04:51,810 --> 00:04:58,470
and we say that is our some of the forces in the X direction that is equal to M multiplied by eight.

59
00:04:58,740 --> 00:05:05,220
And we know M is 100 and we can solve four and we get 12 meters per second squared.

60
00:05:05,880 --> 00:05:11,190
So what is the acceleration in the Y direction while that is zero, because remember this great is on

61
00:05:11,190 --> 00:05:15,120
a surface, so it doesn't move in the direction of the forces in the Y direction.

62
00:05:15,960 --> 00:05:18,060
Cancel out the R balanced.

63
00:05:18,450 --> 00:05:22,290
We have a gravitational force going down on 981 Newton.

64
00:05:22,890 --> 00:05:30,090
We've got the Y component of the Applied Force going up of 750 Newton and then we have the normal force

65
00:05:30,090 --> 00:05:33,030
also going up of 231 Newton.

66
00:05:33,810 --> 00:05:39,600
And if you add the normal force, it goes up and the Y component of the blunt force, you'll see that

67
00:05:39,600 --> 00:05:42,650
is exactly the same as the gravitational force going down.

68
00:05:42,930 --> 00:05:45,750
So there is no acceleration in the wind direction.

69
00:05:46,140 --> 00:05:51,750
If you have a body with a mass moving in the X and Y direction, you can still solve it with Newton.

70
00:05:51,750 --> 00:05:56,310
Do you just some of the forces in that direction.

71
00:05:56,640 --> 00:06:01,710
And you said that is equal to the mass, multiplied by the acceleration in the X direction.

72
00:06:02,070 --> 00:06:04,890
Then you some the forces in the Y direction.

73
00:06:04,890 --> 00:06:09,570
You said that is equal to the mass, multiplied by the acceleration in the Y direction.

74
00:06:09,810 --> 00:06:11,760
And in both cases you solved in four.

75
00:06:12,270 --> 00:06:14,910
But in this case we only have movement in that direction.

76
00:06:15,240 --> 00:06:18,080
So eight is 12 metres per second.

77
00:06:18,530 --> 00:06:26,220
Now, this is just an example of how you would use Newton's second law to solve kinetics problems.

78
00:06:26,490 --> 00:06:28,140
And it is quite straightforward.

79
00:06:28,950 --> 00:06:35,760
As long as you remember, you can only some forces that are in the same direction, positive or negative.

80
00:06:35,760 --> 00:06:43,380
So let's say X direction or Y direction, and you then equate it to the mass multiplied by the acceleration

81
00:06:43,380 --> 00:06:46,020
in that direction, all those forces.