Fisika kelas X - Dinamika Gerak Partikel part 1 - Penerapan Hukum Newton 1

Hello brothers and sisters, welcome to the BIG course channel with Cobain, on this channel Cobain will help you in learning mathematics, physics and chemistry in this video Cobain will discuss the dynamics of particle motion, yes, brothers and sisters, in the dynamics of particle motion, we will discuss Newton's law, Newton's law 123, this is Koko talking about Newton's law first, yes,

Newton's law first talks about sigma F, so the term is sigma F, which is zero, the number of gays, so the number of gays is zero, what does the word number of gays is zero, this is the first thing we can call it as understanding that it is silent

silent, not moving the second one can be called a straight line of order straight line of order or the movement is GLB the value in the movement is GLB if the straight line of order if the straight line of order means that the acceleration is zero

The acceleration is zero, A. The next one, the straight line of the rule shows that the acceleration is constant. The big value of the acceleration is constant. So Newton's law speaks of the state of the object is silent, it has a GLB, has a zero acceleration with a constant speed.

In Newton's law, for example, there is a burden, an object that is hung

with a rope, a system of ropes, so the object that is hung with a rope arrangement like this, this object cannot move, why is it held in the sky, for example, this roof, yes, the roof is held, it cannot move, so this is called silent or balanced, silent, balanced, not moving, it's the same, so it doesn't move, yes, this is what we call

Newton's first law principle because there is no movement if there is a string arrangement like this, then the children will draw the tension of the string so for example there is a string, there are three strings, we assume that there are three strings 123 later here it is called string 1 why is this string 1 in the direction of

This is against each other, so the system of the rope style always comes out of the point of the system, so for example, he is tied to the sky, this means

roof, roof of the house, it means that the tension of the rope will come out like this, the direction for the arrangement later here, there is a rope tie, it means that the tension of the rope also comes out, so the tension of the rope comes out of the system. The object, yes, for this one, it's the same, for example, you think this is the second rope, it means that here there is a arrangement of two ropes, so the directions are also opposite, this is T2, we consider it T2,

So this is the first rope, the second rope, and this is also the same. Here is the third rope.

So why is there something coming out of this object too? Well, this is an object system, so this string tension also comes out of the object. So it's like that, so this is the third Newton's law reaction action system. So this is what you learned in Newton's law 3, there is a reaction action, so it's interesting to attract each other, so there will be a reaction action later.

Newton's law 3 talks about reaction action Newton's law 3 is the same as the reaction pattern

so the values ​​are the same, the style is just different minus, so it means different directions so if we push something, what we push is definitely trying to fight it doesn't want to, so it's like this, it's trying to hold it, so it's like this one wants to go to the right, this one wants to go to the left, that's reaction action, the value of the style is the same, that's Newton's law 3, similar to this, this is T2, this is reaction action,

for the object system, the object must have weight the weight value w always goes down we draw it from the point of the center of the object so if we draw it like a box, this is the diagonal center so the image w starts from the middle

So the heavy image is the symbol of the heavy style, it's in the middle of the object, yes, so if we have a style image, sometimes there are teachers who have to be really put where it has to fit, yes, this is the placement of W in the middle, okay, for this weight, the value of the formula is m times g, yes,

times the acceleration of gravity. Well, later here, I assume the acceleration of gravity is 10, not 9.8. Maybe it's easier to calculate. So it's considered 10 m/s². So the small g is

10 m/s2. If the question is, determine the value of T1, T2, and T3, the tension of the rope 123. The question is, count the tension of rope 1, tension of rope 2, and tension of rope 3.

this is the question of the rope tension pattern for this system of things that are hanging here later there are some systems there the first is the system of the rope the system of the rope that is tied the second is the weight of the rope, do you need this one?

this one doesn't need to be why? the sky's surface we can't calculate the time value so we can't calculate it so later we only work on the system of the tied rope this is like two ropes tied together and then one like this hanging on this burden so later there are two jobs this is the first work you do first this is the first work you do the first is the object system

the thing that is hung. Well, this one, the first system, this has a "W", here is the T3 rope,

we just repeat the picture so that it doesn't look confused, so it's like being taken to another place, the system is repeated, yes, later how it works because this system is not moving, yes, it means the style that works on this object system, the total style, the definition is zero,

We usually have the top of the top is plus, the bottom is minus. Because this is zero. But later if there is a movement that is in the direction of the plus movement versus the direction of the minus movement, it is Newton's law 2. I just assume that the top is plus. So later T3 minus W is zero.

Why do you draw the T3 only upwards, why not downwards in the system of objects in this hanging? Because what works on this object is the T3 upwards that comes out of the object. So which pattern is used? The pattern that comes out of the object, that is, the one that concerns the object.

if this one is not, the one down here is related to this binding system so it is not used later this is for the next work the binding system later we will calculate first T3 is W W is the formula M times G, the mass times G, this is what I gave earlier, the mass is 2 kg, this weight is 2 kg, so later T3 is

M times G means it will produce 2 times 10, this is 20 Newton. This is for the first system. For the second system, we work on the binding, this binding. So for the second system, this is a replica, this is the binding system, this is T1, this is T2, this is T3.

Why don't we follow the T1 and T2? Because it touches the roof of the house. So we don't use it. We use the T1, T2, T3 on this system. This is included in the binding system. So if there is a string binding, it is included as another system. Now,

we make this first, yes, the vector direction in a BAP vector lesson, we make it like this, if there is a D53, this is a tool for the Y-Solid, the X-Solid in this knot, this 53 will be put here later, yes, this 53, 37, this 37 later, yes, the angle of contact, for example, so if you want to repeat the picture here,

this is the T1 that is less inclined this is the T2 that is more upright this is T1 this means 37 degrees this is 53 degrees so if the children have 53 degrees here this angle is 53 degrees 37 degrees this is 37 degrees I draw here the angle of the opposite, the opposite means now this

for us in the bar factor, this is like directed to the Y-shape, which is horizontal, which is horizontal, T1 and T2, we project it, this is the projection here and here, this is for T1, yes, the term is made into a box, a string of ropes, the term is so later here is T1 cos, why cos? Because it's close to the angle, the one close to the angle is cos,

the one here, this one, later this means T1 sin, yes, so the one to the left is T1 cos, the one above is T1 sin, because the one near the cos angle is far from the 2 sin angle, yes, understanding from the projection vector, for this one, T2, we also made this, we also made this, yes,

this one is near the angle 53, this one means that the angle is touched by the angle, the sum is so later this is T2 cos 53, the one above this can be called T2 sin 53 so the point is that the one near the angle is times cos, the one far from the angle times sin so the understanding of the vector is like that if the projection is made is true

Why is the T3 not needed in the projection? No need, because it is already in the Y-shape. The Y-shape is straight, the Y-shape is flat. Later in the string system, it means that there are two Y-shape, X-shape and Y-shape. So what's the point? If the children get the style is a lot, there are three in one system, it makes the projection of the Y-shape and Y-shape. Later we work twice.

Sigma gaya for example, sigma gaya at the x-axis is zero. Why zero? Because it doesn't move the system, especially it's a bond, this bond doesn't have time, so it's zero.

then Sigma fx means the flat direction, yes, this is the one to the right, this is the T2 cos 53, there is one to the left, T1 cos 37, later we will work, the one to the right we consider plus, yes, T2 cos 53 is reduced, T1 cos 37 is zero, so we will do this work directly, yes, cos 53 is 3/5, yes,

this times 3/5, this min is just an change, yes, it's the same as T1, cos 37 is 4/5, now the 5 is subtracted, in other words, we get this T2 is 4/3 of T1, that's it, for the next one, this is a sigma fY work, yes, so what is the number of values ​​in the sum of y

the number of gaias on the source Y, the gaias on the source Y are the straight ones, there are two top pairs, namely T2 sin 53 and T1 sin 37, we assume the top plus, like the previous one, it means that later this T2 sin 53 plus T1 sin 37 minus T3

= 0, now later, the children have already obtained the equation T2 is 4/3 T1, this is a substitution, just enter it here, T2 is replaced, so later T2 is 4/3 T1 sin 53 is 4/5, plus T1 sin 37 is 3/5, minus T3, I've got it, this 20,

-20 = 0, now let's calculate this first, this means 16/15 T1 + 3/5 T1 = 20, now this is the same as the denominator first, 15 is 9, 9 + 16 = 25/15 T1 = 20, so T1 is

300 divided by 25, now this is 12 N, if the T1 is 12 N, put it here, you get T2, 4/3 times T1, 4/3 times 12, it means you get 16 N, this is how the system works, the object is hung with a rope,

which does not move This includes the application of Newton's law 1 the silent object system This is for children who may still be confused with the explanation of the law regarding the projection of the sum of x and y maybe you can see it in the video of the vector for the link is in the description below

If there is a question, there are two types of patterns. The first is the pattern to the right 40, the second pattern 30 to the left.

This is a slippery area, so it's not rough. Determine the style and direction needed so that the object is still, not moving. If the children are left like this, it's slippery, it means that this object must move. Why? Because the right direction is bigger, 40, this is 30, so it must move to the right.

If the question is to keep things quiet, so that they don't move, there must be another style. Another style means that the value must be balanced. So, for example, to the right 40, to the left 30, it means that the other style must be balanced.

to the left by 10 why? because to balance right 40 means the left should also be 40 means how many styles are needed? means the third style is 10 Newton to the left this is also one of the applications of Newton's law first so that the object is silent I have an example of the following question

there is a weight of 2 kg pulled by the right axis the field is rough, the static shift coefficient is 0.4 if the field is rough, in the question there is a shift coefficient, static shift coefficient, there is also kinetic if the kinetic is specific for objects that are moving, it is kinetic

For the static shift coefficient, it is usually used to check whether the object is still or moving. The static shift coefficient is mu s, so the symbol is like M but it is curved, so mu s. The static shift coefficient is s.

So, mu_s is used to make sure that the object is still or moving. We want to know if the object is still or moving. If this field is rough, it is seen from mu_s, it is seen from the static motion. But if we put mu_k, the kinetic motion coefficient,

means the condition of the object must be moving, yes, this is what I am told is static because the potential is later how the object condition is silent or moving if given a style of 5 and 10, that's why I'm only told that the

What if there is mu s and mu k? To check the condition of the object, the children use mu s and mu k as a function of the function of the function of mu k. mu k is usually used when it is known that it is moving. Later it is asked how fast it is, later it is included in Newton's law 2. I still explain Newton's law 1.

Mu S is for the motion of the object, later here the students will talk about FS, which Koko mentioned earlier, static shift pattern, if Mu S is the coefficient, the coefficient is the level of roughness of the field, that is the shift coefficient, static shift pattern

this static gesture has a formula mu s times n what is n? n is the normal mode or the pressure of an object against its field so here is the term n this is the normal mode or what we call the pressure of an object against its field or the floor if the floor is the floor if it presses the board against the board

The pressure of the object against the weight is the normal style, straight and straight. That is the normal style. So, what I mean is, we start to answer, this question is to determine the condition of the object if given the 5 style, given the 10 style, how is it? Is it still or moving? First, if there is a burden given by mass, we have to draw the weight first, the heavy style. This is weight.

which is the formula m times g earlier, then this object accumulates in the field, it means it has a normal pattern, so there is n, n is always drawn from the floor from the field, yes, this is actually centered, yes, I mean this is actually a line, yes, this is a little far away so that the picture is clear, later here there is n

straight, so if the plane is flat, it means that the n is straight upwards, so if there is a slope, it means that the n is a bit slanted, because it has to be straight, that's not normal,

Now if there is a rough area, it means that the children have a moving style, the moving style is against the direction of movement So for example, this is a child pulling the style to the right, it means that this should automatically be a moving object to the right, so the moving style is to the left

The style of the gesture is the image when on the floor between the boundary of the object and the floor, the image is like this So it's like a row of floors when it's close to the floor, the floor is called FS

Why FS? Because I want to check and make sure the condition of the object. This is FS. If you calculate the kinetic gas coefficient, how? It means that the children directly write FK. Because the object immediately moves, is pushed, is pulled. For the next question, I will answer it. I will calculate the value of the normal pattern first. This normal pattern,

we have a good picture, the pattern is finished, there is no other pattern, there is already a pattern F, a pattern C, there is already a pattern N, there is a pattern W, so the children always draw the pattern system first in Newton's law of motion dynamics, it must be drawn first, then we start to answer, to find N

This object has a flat movement, so in other words, this is the source Y, this is the source X, the source Y has a value of 0, why 0? Because the system is silent, it must be silent, why? It's impossible for the object to be like this, the object moves like this, there is no up and down like this, it's impossible, the field is flat like this. In other words, to find the normal force, we will use the term

Sigma f = 0 to find the normal way. Why is n 0? Because it's silent, because the system of the object moves to the right and left. n-v = 0, so the normal way will get the value of w. w is m times g,

Well, the mass is 2 kg, so later the n is 2 times 10, so the normal force gets 20 Newton, the unit is Newton, for the force, the unit is Newton.

this is normal, so to calculate the normal pattern we use the Y-sigma system, Fy-sigma to determine the condition of the object moving or not, silent or moving, we check it through the X-sigma we see between the F pattern and the FS

which one is bigger? So to make sure the condition of the object, it is compared to the F and Fs. For this F, for question A, we are told that the value is 5. So we answer question A first. This is 5 newtons. For Fs, the formula is mu s times n. Mu s times n, I'll calculate first. The mu s is told to be 0.4 times the normal formula, we get 20. That means

get the value of 8 N. In other words, the style that was given 5 N, the FS gets 8 N. It turns out to be bigger than the FS is bigger. If the FS is bigger, it means that the condition of this object is still silent.

What does it mean? It means that the style we give, we pull it, it can't be. Why? Because the area is rougher. So if there is a large cabinet, if the lower area is rough, with the status, the coefficient, how much we pull, if we are not strong, it means that the area is very rough. That's it, that's called static shift coefficient.

For the question B, we compare again, F and Fs. B is given a force of 10 N. If given a force of 10 N and the Fs is US times N, it gets an 8. This is impossible to change. This is a static force, depending on the roughness of the floor. But if the force can be changed, how much we pull, how big the force.

it turns out that this is larger than F if the pull style is larger, then the logic is automatically that this object is moving why? we are strong to pull it means that this object will move

If the condition of this object is moving, then the children will continue to Newton's law. The question is how fast does the object move? We will appear in part 2, I will discuss Newton's law 2.

later in Newton's law 2, the children will know the term kinetic coefficient, so later there is mu k, the kinetic one, later the children will use mu k if it moves, why is this mu s? this is still making sure the object is moving or not

using mu s but if the children are directly asked to be told the kinetic kinetic coefficient 0.4 oh that must be the object moving why? because mu k has made sure the object is moving if the children are told 2 mu s and mu k means the children must first make sure the condition by looking for between f and fs how big if it's bigger than f means it's moving well, that's what I'll discuss later

in part 2 thank you for watching this video I hope this is useful for you all at school please support me by liking, subscribing and sharing to your friends thank you