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Newton‘s 2nd Law: Example, 2 Dimensions Part 2

by Jared Rovny
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    00:01 So the question now is if we found the velocity, or sorry the accelerations in both the x and the y direction, we can also ask what is the speed as well as what is the velocity after one hour which we can get from the equations of motion.

    00:14 So first, let's find the velocity, which again will include not only the magnitude, but the directions since velocity is a vector.

    00:20 First of all, we will again do a vector sort of equation with the velocity being v plus acceleration times time and then we'll treat this in the x and y direction separately so the initial velocity of our object was zero so we just get rid of the v nought, the v zero, and then we'll just look at the x direction and the y direction of your final velocity after it has been accelerated for a certain period of time.

    00:42 Since we've already solved for the acceleration, which I've kept below here so you can see them, you use multiplied by the time, however long those acceleration are acting for, and you can find the new velocities and so you can see that our acceleration equation don't change much since the initial velocity was zero and went away, all we have to do is multiply the accelerations by the time.

    01:01 So these equations will look very similar but with a new unit of time in them and now we have the velocity in the x direction and the final velocity in the y direction as well.

    01:10 If we actually plug the numbers if you'd like to check these and practice on your own, You can actually plug in all the different variables that we've given in this problem, the forces, the masses, and the angles, and you should be able to get numbers like these if you're practicing this yourself and now we actually have numbers, the numerical values for both the velocity in the x direction and the velocity in the y direction.

    01:33 Now the question is, if we know the velocity, meaning we know both components, since it's a vector.

    01:38 We know the x and the y, can we find the speed which is just a number telling you what the total actual speed of the object is regardless of what direction it's acting.

    01:47 To do this, what we're going to do is look at a triangle one more time.

    01:51 We have a velocity in the x direction and a velocity in the y direction so we have an a and we have a b, so the total speed of the object will just be the length of the hypotenuse in this triangle.

    02:02 In other words, it's what we would call the vector sum of a and b.

    02:07 So looking at this triangle, c would be sort of considered a vector sum of a and b, as vectors.

    02:12 And so if you want to find the magnitude of c, the value of c which in this problem would be the speed, we'll just need to find c as we would normally do for a right triangle and for that we will use Pythagoras theorem, which is c squared equals a squared plus b squared.

    02:28 In our problem we already have the velocities, as we've solved and those will be represented as the legs of this triangle, we have the horizontal velocity and the vertical velocity as well.

    02:38 So to find the magnitude of velocity, in total, which will be the speed of the object, all we have to do is apply the Pythagorean Theorem which is to take the square root of the sum of squares of the two legs of your triangle, which is the velocity in the x direction and then the y direction.

    02:56 So we plug in these velocities, you should be able to get a total velocity of something like 10.460 meters per second and that again because is just the magnitude of the velocity, is in fact the speed since I haven't specified the direction.

    03:08 This is a good review of the angles.

    03:12 We've just gone over a pretty complicated problem and few of them using different angles, so this table which I won't go over in detail right now but there is something you have as reference will be a good table to be brushed up on, you'll be able to recall these angles really quickly.

    03:25 This will help you get through problems much more efficiently to know what these angles are.

    03:30 Just so you're aware of what this is saying, in the first column you have an angle written in degrees, and then the second row, sorry you have the angle again written but in radians this time.

    03:39 So that first row and the second row are the same.

    03:41 They are both the angle but they are just rewriting the angle in either angle as degrees or as a radians and then we have sine and cosine for some of these very special and very often used angles as we've already seen 30 and 45 be used.


    About the Lecture

    The lecture Newton‘s 2nd Law: Example, 2 Dimensions Part 2 by Jared Rovny is from the course Force.


    Included Quiz Questions

    1. Fx = 69.28N, Fy =40 N.
    2. Fx = 6.928N, Fy =40 N.
    3. Fx = 6.9N, Fy =40.1 N.
    4. Fx = 40N, Fy = 69.28 N.
    5. Fx = 6N, Fy =4 N.
    1. 0.698 m/s2 (we will take Fx=69.28, because direction of acceleration is in the direction of motion).
    2. 0.4 m/s2 (we will take Fy = 40 because direction of acceleration is in the direction of Fy).
    3. 6.98m/s2.
    4. 4 m/s2.
    5. 69.8m/s2.
    1. vx= 13.22 m/s, vy = 1.53 m/s.
    2. vx = 1.53, vy = 13.22 m/s.
    3. vx = 15.3 m/s, vy= 13.22 m/s.
    4. vx = 11.53 m/s, vy= 13.22 m/s.
    5. vx = 1.53 m/s, vy = 1.322 m/s.

    Author of lecture Newton‘s 2nd Law: Example, 2 Dimensions Part 2

     Jared Rovny

    Jared Rovny


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