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More on Newton's second law

Newton's second law can be used to solve two-dimensional motion problems. If any force vectors are acting at an angle, they can be broken into their horizontal and vertical components using trigonometry. Then, the net force in each dimension and the object's mass are entered into Newton's second law (F=ma) to determine the resulting acceleration in each dimension. Created by David SantoPietro.

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  • marcimus orange style avatar for user S Chung
    Do all these forces happen at the same time any any given moment?

    Also, do the same principles of acceleration apply here? For example, if the meteorite was going in the forward direction (positive) and the overall acceleration turned out to be negative, would this then become negative acceleration (i.e. slowing down)?

    Thanks in advance for any help given! :)
    (12 votes)
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    • male robot hal style avatar for user james
      Suppose there is a meteor traveling with a constant velocity of 1000 km/h and there is no force acted on it. As it enters the gravity field of Earth, earth will start pulling the meteor with a constant force of 100 N resulting in 10 km/h^2 constant acceleration. Now there is only one force acted on meteor.

      When meteor's velocity reaches 1500 km/h, we send a rocket to save the earth. The rocket starts pulling the meteor in the opposite direction with 50 N. Now there is two forces acting on meteor at the same time. The net force on meteor is 100-50=50 N towards the earth and it still accelerates but now its acceleration decreases to 5 km/h^2.

      When meteor's velocity reaches 1750, we send a second rocket pulling the meteor with the same 50 N. Now net force on meteor is 100-50*2= 0 N. Since net force is 0, meteor stops accelerating but it continues with a constant velocity of 1750 towards earth.

      Now we have to send a third rocket. Net force on the meteor becomes 100 - 3*50= - 50 in the opposite direction from earth, and the meteor; first starts to slow down from 1750, then stop completely, and eventually start moving in the opposite direction.
      From first rocket onwards there are allways 2 or more forces acted on the meteor. Even when the meteor moves away from earth after third rocket, the force of earth's gravity continues to act on it.
      (46 votes)
  • leaf green style avatar for user Ali Mohsin
    READ BEFORE CONTINUING: I think i solved this but I thought I'd share it cause if i had to go through the trouble of going through this you should too. plus it's probably wrong so check it out if you want to

    I have a question, when I find the acceleration for the forces at , Do I use the acceleration I get for x and y forces in the pythagoras theorem? Or do I use the total net of the x and y forces for the pythagorean theorem?

    Heres my work if it sounds confusing:

    Horizontal forces:
    1) +50N - 30N + 25N - 40N = +5N
    2) +5N / 10kg = 0.5 m/s^2 (force/mass = acceleration)
    3) acceleration of horizontal forces: 0.5 m/s^2

    Vertical forces:
    1) +48N - 28N = +20N
    2) +20N / 10kg = 2 m/s^2 (F/m = a)
    3) acceleration of vertical forces: 2 m/s^2

    So do I now use the acceleration of the vertical and horizontal forces in the pythagorean theorem, like this?

    a^2 + b^2 = c^2
    0.5^2 + 2^2 = c^2
    and then this would be my total acceleration?


    Do I use the total net force of the horizontal and vertical forces in the pythagorean theorem:

    Horizontal force total: +5N
    vertical force total: +20N

    a^2 + b^2 = c^2
    +5N^2 + 20N^2 = c^2

    Cause if I use the total acceleration of the horizontal and vertical forces in the pythagorean theorem, how do I find the one total net force of the asteroid?

    SOLVED: I realised that either method of first figuring out the total acceleration, or figuring out the total net force, would both result in the same answer. For some reason I never got far enough to realise in the end, as long as I have two of three factors in "F = m * a" I'll always find the third factor.

    The total net force would be the square root of 425, or sqrt425. To make sure the answers were the same to that solving the total net force w/ acceleration first, I first figured out what the square root of 425.

    (sqrt will mean square root here)

    sqrt425 = 20.61552813
    Then, you divide by the mass to find acceleration.
    sqrt425/10kg = 10kg/10kg * acceleration
    this leaves us with 2.061552813 = acceleration

    2.061552813 is an important number because if we now solve this problem using the total acceleration first

    (meaning we now find "F" or force instead of finding "a" or acceleration like we just did previously),

    we'll see that at the end of our pythagorean theorem process of finding the total acceleration (which looks like this: 0.5^2 + 2^2 = sqrt4.25) will get us the square root of 4.25, which is also equal to 2.061552813. And the square root of 4.25 when multiplied by the mass, 10kg, will also equal the square root of 425 (sqrt425 = 20.61552813)

    Here's what I mean if I didn't express my thoughts properly:

    TOTAL NET FORCE: sqrt425
    sqrt425 = 20.61552813
    TOTAL acceleration: sqrt4.25
    sqrt4.25 = 2.061552813

    NET FORCE TO acceleration:
    sqrt425/10kg = 2.061552813 (which is also sqrt4.25)

    acceleration TO NET FORCE:
    sqrt4.25 * 10kg = 20.61552813 (which is also sqrt425)

    So the point of me showing this is that my thoughts are that if the answers I get are identical to each other and identical to the answers I get from the pythagorean theorem process to find either total net force or total acceleration, then that means these must be the right calculations for acceleration and net force in the end. It's like checking to see if your answer is right by solving it the other way around. is this right
    (22 votes)
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  • blobby green style avatar for user Bond
    When a body is floating on water, what are the net forces acting on it?
    (13 votes)
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    • aqualine ultimate style avatar for user Sarthak Agrawal
      While a body is in water or in any other fluid(regardless of whether it is floating or sinking), 2 forces act on the body, First is it's weight acting downwards, and second is the upward force exerted by water or fluid on the body in the upward direction, this force is known as upthrust. Now, for a body to float, the upthrust must be equal to the weight of the body. Hence, the net force would be zero.

      You can learn more by reading Law Of Flotation and Archimedes Principle.
      (21 votes)
  • piceratops seed style avatar for user Maci Biser
    What is Newton's second law
    (11 votes)
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  • leafers seed style avatar for user victoria martin
    When a body is floating on water, what are the net forces acting on it?
    (8 votes)
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  • duskpin ultimate style avatar for user Aditya1997
    What is pythagoreas theoreom?
    (5 votes)
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    • leaf grey style avatar for user pornhub premium
      In mathematics, the Pythagorean theorem, also known as Pythagoras's theorem, is a relation in Euclidean geometry among the three sides of a right triangle. It states that the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. The theorem can be written as an equation relating the lengths of the sides a, b and c, often called the "Pythagorean equation":

      a^2 + b^2 = c^2 ,
      where c represents the length of the hypotenuse and a and b the lengths of the triangle's other two sides.

      Although it is often argued that knowledge of the theorem predates him, the theorem is named after the ancient Greek mathematician Pythagoras (c. 570 – c. 495 BC) as it is he who, by tradition, is credited with its first recorded proof. There is some evidence that Babylonian mathematicians understood the formula, although little of it indicates an application within a mathematical framework. Mesopotamian, Indian and Chinese mathematicians all discovered the theorem independently and, in some cases, provided proofs for special cases.

      The theorem has been given numerous proofs – possibly the most for any mathematical theorem. They are very diverse, including both geometric proofs and algebraic proofs, with some dating back thousands of years. The theorem can be generalized in various ways, including higher-dimensional spaces, to spaces that are not Euclidean, to objects that are not right triangles, and indeed, to objects that are not triangles at all, but n-dimensional solids. The Pythagorean theorem has attracted interest outside mathematics as a symbol of mathematical abstruseness, mystique, or intellectual power; popular references in literature, plays, musicals, songs, stamps and cartoons abound.
      (13 votes)
  • leafers seed style avatar for user victoria martin
    How we know the angle if we're only know the magnitude for horizontal and vertical direction?
    (4 votes)
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    • starky ultimate style avatar for user KLaudano
      angle = arctangent(magnitude of vertical force/magnitude of horizontal force)

      Note: You may have to change the sign and/or add Pi radians (180 degrees) to get the correct angle depending on the signs of the force vectors. If the vertical force is negative, multiply the angle by -1. If the horizontal force is negative, multiply the angle by -1 and add Pi radians.
      (4 votes)
  • duskpin ultimate style avatar for user Skrive.25o8
    My textbook states that Newton’s II law as “ The rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of force.” Here it is different. So which one is correct?
    (3 votes)
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    • hopper cool style avatar for user obiwan kenobi
      Both are correct, here is how it works:

      F = ma
      F = m( delta v/delta t) (acceleration is velocity divided by time)
      F = (delta p)/(delta t) (mass times velocity is momentum, which is denoted by p).

      F = ma and F = (delta p)/(delta t) are two different ways of stating the exact same thing. Hope this helps!
      (3 votes)
  • blobby green style avatar for user EdgarEat
    Sin30x45N gives me a negative answer but the force is horizontal, up and positive. What do I do in this case? Thanks in advance.
    (2 votes)
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  • blobby green style avatar for user kaycie.ebert
    When calculating the net force, why are things like normal force and weight included?
    (3 votes)
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Video transcript

- [Instructor] If you're face to face with a sophisticated Newton's Second Law problem, you're gonna need a sophisticated understanding of Newton's Second Law. That's what I'm gonna try to provide you with here, so that no matter what scenario you're faced with you can apply this law in a correct way. Most people know Newton's Second Law is F equals MA, which is fine, it's a simple way to understand it, and it's fine for simple problems, if I had an asteroid for instance, of mass m, out in outer space so there's no air resistance or friction, and there was only one force on it, a force F, and that force pointed to the right, let's say that force was 50 newtons, well I could plug the 50 newtons into the force, I could plug the mass of the asteroid, let's just say it's 10 kilograms, into the mass, and I'd find the acceleration of the asteroid, in this case, 50 over 10 would give me five meters per second squared. But, what if we had extra forces on this asteroid? What if there was another force that pointed to the left, that was 30 newtons? So let's call, let's name these now, let's call this F1, this 50 newtons, let's say that's the magnitude of that force, let's say F2 was the magnitude of the 30 newton force, it points to the left, yes, that's the negative direction, but let's just say these forces here are just giving the magnitude of it and then the direction is specified by the direction of the arrow. Now what would I do? Well, to handle this we need to understand that the left hand side here isn't just force, it's the net force. Or, you can call it the sum of the forces. So to denote the net force, we often write this Greek letter sigma, and sigma is a mathematical symbol that represents the sum of whatever comes after it. So this is the sum of the forces. Because F comes afterward. If I had G it would be the sum of the G's, and if I had H it would be the sum of the H's. And this is a little confusing already. People hear sum of, phonetically, and they think oh, sometimes they're like oh, so, some of? Like a few of? No no no, we mean all of, all of the forces. That's what this means. You add up all of the forces, that will equal the mass times the acceleration. So in this case, we'd take this 50 newtons, I can take 50 newtons because it goes to the right, and, I mean we can call leftward positive if we really wanted to, if there was a good reason, but unless otherwise specified, we're gonna just choose rightward as positive and upward as positive, so this 50 has to be positive. And I can't now, by sum of, I add up the forces, but I have to add them up like vectors. This force here is a vector. Forces are vectors, and so I have to add them up as vectors. This is a vector equation. I can't just take 50 plus 30 to get the answer, because vectors that point to the left we're gonna consider negative, and vectors that point to the right we'll consider positive, and so I'll take 50 newtons minus 30 newtons. That's what's gonna be equal to the mass times the acceleration, so I could plug in 10 kilograms if I wanted to, multiply by A, and in this case I'd get 20 over 10 is two, meters per second squared. So you have to add these up like vectors, and if I had more forces it'd be just as easy to deal with, I can just add them up as vectors, so if I had another force here that was maybe, that's maybe 25 newtons, we'll call that F3. And let's say there's another force that points to the left, this one's gonna be 40 newtons, so we'll have 40 newtons to the left, we'll call this one F4, well, I can just keep including these. I can just add them up as vectors, the 40 newtons points to the left, so that's gotta be a negative, I'll put negative 40 newtons, and then this 25 points to the right, I'll make that positive, so positive 25 newtons, and I can find my total force, my net force, my sum of the forces, and that would allow me to figure out the acceleration. So, there's one problem here though. A lot of times physicists don't like writing this anymore, at least physicists that are interested in education don't like this form of Newton's Second Law as much, a lot of them don't, and the reason is, there's a misconception students have. They think that as they're drawing forces here, the M times A is also a force. They want to draw an extra force on this asteroid, maybe it points to the right, that is mass times acceleration. But mass times acceleration is not a force. Mass times acceleration is just what the net force happens to equal. So if you add up all the net forces, or sorry, if you add up the net force on an object, which is adding up as vectors all the forces on the object, that will just equal MA, it happens to equal MA. But it is, MA is not a force in and of itself, so you cannot draw this as a force, don't draw any of that, that is not a force, that's just what the sum of the forces happen to equal. So upon realizing this, physicists were like oh, okay, this is causing confusion, so let's just write an equally good form of Newton's Second Law, in terms of algebra, but that also makes it so that people aren't so susceptible to falling into this misconception, and this alternative version of Newton's Second Law looks like this. The acceleration equals the net force divided by the mass. And you might be like, so what? We just divided both sides by mass, who really cares? Well here's why it's better. Because people are much less like to think that acceleration itself is a force. They're much less likely to say oh, acceleration is a force over here, I mean people might do that but they're less likely. So acceleration is not a force, it's a vector, but it's not a force vector. And the other reason this equation is nice, it shows us the relational dependence of acceleration, it shows us that the net force is what will give us acceleration, and the more net force we have, the bigger the acceleration. So it shows us that the acceleration is proportional to the net force, or the sum of the forces. And it shows us that the acceleration is inversely proportional to the mass. So the bigger the mass, the less acceleration you have. So it's another reason this equation's nice, shows us what acceleration actually depends on in terms of net force and mass. So there's one other problem here. So this is better, this is already better, now we know it's net force, now we can write it this way and not fall into the misconception that MA is a force. There's a problem though, what if I introduced another force over here? Let's say I introduced a force that points downward. Let's say this force was 28 newtons, and this is F4, well we did F4, this is F5. So we have F5, 28 newtons downward, you might think, oh, I know how to deal with this now, can't I just sneak this 28 in over here and write it as a negative 28 because it goes down, can't I just put negative 28 right there and it turns out you cannot do that. That is not allowed, and the reason is, just like we couldn't take 50 plus 30, because we're adding these up as vectors, and leftward meant negative and rightward meant positive, we can't take a vertical force and just add that magnitude, or subtract it, from the magnitude of the horizontal force. That's not allowed, a horizontal force and a vertical force added up will not equal the sum of, or the difference of, the magnitudes. What I'm saying is this. Think about it this way, if you had a certain amount of force to the right, and a certain amount of force upward, to add these up it would not equal this value plus this value, you'd have to add them up with the Pythagorean theorem. To add vectors this way, you'd get this vector right here. That would be your total vector. So you'd have to do a squared plus b squared equals the total force squared over there between those two vectors and you might be thinking oh great, I don't want to have to do trigonometry here. And it turns out you don't have to, not yet at least. If these are the only forces we have, we don't have to do it this way, I'm just trying to show you that you cannot simply naively add up 50 minus 28 and expect to get the total answer right. But here's what you could do. You could take only horizontal forces, deal with those in the horizontal direction first. And only vertical forces, and deal with those in a vertical direction, so it's the same trick we always play as physicists, we say alright, we're gonna divide and conquer, we're gonna take all horizontal forces, and put those into their own equation because the horizontal forces should only affect the horizontal acceleration. So if I just want horizontal acceleration, I can take only horizontal forces, add those up, and get the horizontal acceleration. Or, I could take only vertical forces. Add those up, and I'd get the vertical acceleration. So if I take this, I'll make an equation for each direction independently, and I know I can find each component of the acceleration by just using the forces in that particular direction. So this is a nice trick, we're gonna deal with each direction independently. And then if we really wanted, say, say these were acceleration vectors, say we wanted the total acceleration, we were talking about forces before, but all vectors add up the same way. You can use the Pythagorean theorem. If you figure out A in the X direction, the total acceleration in the X direction and you figure out the total acceleration in the Y direction, you could figure out the total acceleration, the magnitude of it, by again using the Pythagorean theorem, and so this is just a way, this is a handy way of dealing with these forces that point in multiple directions. Let me stick one more force in here, just because we should have one that points up and then we have all possible directions here. Alright so if this force, F, what are we on, six, is gonna be, maybe that's about 42, that's a good number, newtons upward, how do we deal with this? Well we already dealt with the X direction. This was the X direction, this gives us all the forces in the X direction, that equals mass times acceleration in the X, it's not written in this form, it's just multiplied with the 10 on the right hand side instead of divided here, but this is the formula you could use to relate forces in the X direction to the acceleration in the X direction. We're essentially just taking all of these forces, plugging them into the net force in the X, dividing by the mass, and we'd find the acceleration in the X direction. For the Y direction, we could say that acceleration in the Y direction would be the net force in the Y direction, so how would we deal with this, only vertical forces are gonna be affecting the vertical acceleration, so I can take this F6 which is 42 newtons, it points up, we're gonna treat that as positive because we don't have a good reason to treat down as positive, 42 up and that's the convention we usually pick, 42 minus 28, 28 points downward, we typically choose that as the negative direction, now we can divide by the mass. We'll divide by 10 kilograms, this gives us our acceleration in the vertical direction. And now that we have both of these, we could if we wanted to do AX squared plus AY squared equals the total A squared to find the magnitude of the total acceleration. Alright, let's step it up one more notch and see what happens, let's make it one step harder. We're gonna move this out of the way, I'm gonna make a little room. So let's say this 40 newtons is still applied, but I'm just gonna put it here so it's out of the way. Let's say there was one more vector involved, one more vector that points this way, and let's say the size of that vector, we'll call this one F7, let's say F7 is 45 newtons. Okay. 45 newtons applied at an angle, let's just say from that point of 30 degrees. How do you deal with this? This is an even more sophisticated Newton's Second Law problem. Here's where it starts to frighten people, they don't know what to do, a lot of times they try to just throw this 45 newtons into one of the equations, they think the 45 maybe should go in the X equation because it points to the left, that's horizontal, that's X, but it also points vertical, sometimes they throw the whole 45 into the vertical equation, maybe they do plus 45 over here, but that's wrong, you can't do that, and you can't do that because only vertical forces and components of forces can go in this vertical equation, and only horizontal forces and horizontal components of forces can go into this horizontal equation. So what we have to do at this point, I think you know what we have to do, we have to break this up. So we have to take this 45 newtons, that points up and to the left, we have to break this up into how much of this force points left, how much of this force points upward. So we're gonna have to figure out what is this component of this force that way, in the X direction, what is this component in the vertical direction, I'll call this F7 in the Y direction, and this component here would be F7 in the X direction, it's getting a little cluttered there, sorry about that. We have to figure out what these components are, and once we figure out what those are, I can associate the F7 X into the X equation, and the F7 Y into the Y equation, but I can't put the whole 45 into either equation because all 45 newtons is not directed vertically or horizontally, only part of it is. So we have to figure out with more trigonometry, we're gonna use this same rule or the same idea over here, but instead of the Pythagorean theorem, we're gonna take this 45 newtons and break it up into components, and the way we do that is with the definition of sine and cosine. So if this is, I'm just gonna make it bigger over here, 45 newtons, if this side is 45 newtons, then this side would be the adjacent to this 30 degrees, this F7 in the X direction, and this side would be the opposite to that 30 degrees, so that's F7 in the Y direction. Now we use sine and cosine, let's use the definition of cosine. The definition of cosine, theta, is gonna be the adjacent over the hypotenuse. So the adjacent to this 30 degrees is the side touching that angle, which is F7 X. So F7 in the X direction, over the hypotenuse is the total magnitude of this force vector which is 45 newtons. So I can solve now for F7 in the X direction, F7 in the X direction is gonna be 45 newtons times cosine of 30 degrees. Now I can take this F7 in the X direction, I can take this, whatever it is, this is just a number, you can calculate it if you want, and I'm gonna take this and I'm gonna plug this straight into the X direction, I'm gonna put it right into here, as, let's see, should it be positive or negative? Here's where it's tricky. The F7 Y points up, the F7 X points left, this is the X component, so it's the leftward that we care about, the fact that it points up doesn't matter in terms of X, but this component points left so we have to include it as a negative 45 newtons times cosine of 30. And now for the Y direction we can use the definition of sine of theta. Sine of theta is F7 in the Y direction, which is the opposite, because sine is opposite over hypotenuse. So in this case the opposite is F7 in the Y direction, over the hypotenuse, the hypotenuse is the total vector which is 45 newtons, the magnitude of the total vector, so, we get that F7 in the Y direction is gonna be 45 newtons times sine of 30 degrees. So I can take this, this is the Y component, I can plug this over into, how am I gonna get there without crossing a line? I'm gonna take this and plug it, over here. Right into this Y directed equation. Should it be positive or negative? The Y component points up, so it's gonna be plus 45 newtons times sine of 30 degrees, whew. Okay, so, Newton's Second Law, now you're equipped, you know how to use it, these forces might not be asteroid forces, maybe they're tension or gravity or normal forces or frictions, maybe there's forces in all different directions, but these rules still apply no matter what the force, whether it's up, down, left, right, or diagonal, now you know how you can figure out how to use Newton's Second Law no matter what direction the force is pointed in.