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The reason you can’t treat up as positive for both of them is that then the acceleration for the two objects would not be the same. Since they are moving together, it’s important to set it up so that either both are moving in the positive direction or both are moving in the negative direction so that they have the same acceleration.

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Watch through the video the first time to get the basic ideas. Make sure that you’ve watched the videos on Newton laws before the current one to get the background you need. Then go through the video a second time pausing it each time I’m about to solve a problem. Make yourself try to work out the solution. If you get stuck, then watch the video just up to the point where you got stuck, then pause the video again and keep going. Learning physics is like learning a sport. You have to actually practice the process of doing it. You can imagine if I told you all about how to play a sport like golf but you never actually got out there and swung the club that you wouldn’t be able to do it. So that’s the key.

It's ok to have vector signs on the vectors when drawing the free body diagram. However, once we are plugging into Fnetx=max and Fnety=may there should not be vector signs since we are dealing with the components of the vectors.

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You are correct that if the brick wall exerts a force on the car, the car exerts an equal and opposite force on the wall (Newton's 3rd Law). However, when we are looking at the motion of the car, we only care about the forces acting ON the car. For example, the car is also pulling up on the earth through gravity (reaction to Earth's gravity pulling down) and pushing down on the surface of the Earth (reaction to Normal Force). If we were looking at the brick wall, then we would care about the force the car is exerting on it.

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The box is not moving at all in the y direction, so it’s definitely not accelerating in the y direction. That’s how we know that may =0. In other words, all of the forces in the y direction have to add to exactly 0. You’ll notice that’s how we then are able to solve for the normal force. Since the applied force is partly in the upward direction, notice that the normal force is less than mg. The key point here is that if an object is moving only in the x direction then you know the acceleration in the y direction has to be equal to zero

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That would only work for the simplest problems. I’m teaching you how to do problems that are more difficult

Only in the most simple problems- not generally

I understand that it's given visually right to the question. But, what if there's no visuual representation? (I'm not hating, I'm just genuinely asking. I'm cramming this, I have a test in few hours HAHAHA)

If the angle weren't given, then there would not be enough information to solve the problem. However, I could have given you the acceleration and then asked you to find the angle. The analysis would be the same, but you would plug in acceleration to solve for theta. That's really the key - no matter what you are given, the analysis works the same way, then just solve for the variable you need based on the variables given.

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Yes, the sign represents the direction. So, for example, the magnitude of the force of Friction is Fk. If it is directed to the left, when we put it in Newton’s second law we would write -Fk.

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College level calculus based course for science and engineering majors

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I used the 0.85m to calculate the acceleration. Go to: ruclips.net/video/PgrohbgwxxU/видео.html to see where the 0.85m gets plugged in to the kinematic equation to find the acceleration. Thanks

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@@smithjomiddlesexedu Got a 76 😞… However I am ready to get my test back, identify my mistakes and weak points, and then do better on the retake. Managed to get my grade from a B to a B+ at least.

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For M2 which is on the incline, we rotate the axes so the X direction is along the incline, and the Y direction is perpendicular to the incline. This is what we always do for objects on incline. So M2 is only moving in the x direction. Since there’s no friction in this problem, we don’t need to figure out the normal force, so we don’t need to do anything in the Y direction for M2.

There is no applied force acting on the object. The block has an initial velocity up the incline (but no force in that direction). Another way to think about it - it took a force to get the object moving initially, but now it has a velocity, and that force is no longer acting on the object, so it is not part of the problem. A common mistake students make is to assume there has to be a force in the direction of motion. Just remember that velocity does not equal force!

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@@smithjomiddlesexedu 😊 thank u sir u really teach phy concepts well than my offline teachers.

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They incorrectly think of the two masses, pulling down on the rope as opposed to realizing that the rope is simply transmitting the force from one mass to the other. That’s why you always have to draw the free body diagram for a particular object and really pay attention to the forces acting just on that object.

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@@smithjomiddlesexedu I did very well ! 19/20 Thank you professor

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