17 - Calculating Vector Components in Physics, Part 1 (Component form of a Vector)
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- Опубліковано 13 тра 2024
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In this lesson we begin the study of vector physics, which is the part of physics that deals with using vectors to solve problems. The first step in solving vector problems is usually that break up the vector into components.
Using trigonometry, we break up the vector into a horizontal component and a vector component, and we use the component form of a vector to solve the physics equations in each direction separately.
For example, we use the horizontal components of of the force and velocity vector to solve the motion in the horizontal direction (x direction). Separately, we use the vertical components of the same vectors to solve for the vertical motion..
![Solving Projectile Motion Problems in Physics - [1-4-7]](http://i.ytimg.com/vi/oKIaH2TftpA/mqdefault.jpg)
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What are the steps when you aren’t dealing with a right triangle? The resultant vector in the problem I’m working through isn’t oriented head to tail and the angle isn’t starting straight from the x-axis. I’m like two seconds away from withdrawing from my physics course.
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There's a mistake, if you say vector d = 10 m then that's wrong, a vector is not a number.
So vector d should be 10 m*r(hat) where r(hat) = i(hat)*cos(theta) + j(hat)*sin(theta).
SIr, where is the part two?
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Stupid question - if in 2D case alpha +beta = 90deg, how is in 3D case? is alpha+beta+gamma = 180deg always?
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the vector means a direction right but , why do we need two numbers to calculate a single direction ?
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Suppose a vector is given directly North or West, how do you break that vector into x and y components?
Same exact thing. Nothing changes. Suppose the vector is pointing straight north. That means it is pointing directly ALONG the y-axis. So the angle this vector makes with the x-axis is 90 degrees. So, the X-component of this vector is: (magnitude)*(Cos 90) = (magnitude)(0)=0. So the x-component is zero, which makes sense. Nothing is pointing in the x-direction. The y-component is: (magnitude)(sin 90) = (magnitude)(1) = magnitude. This means that y-component is the full magnitude of the vector. This makes sense because the ENTIRE vector is pointing in the y-direction. I'll leave it for you to work it out for a west pointing vector. Very similar results.
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Thank you for this video but at 25:11 you comment on how we should be thinking in our minds that SINE will chop for the vertical component and COSINE will chop for the horizontal component. I found this is not true for triangles like yours that are rotated 90 degrees and therefore the COSINE and SINE values are opposite.
Instead wouldn't it be best to associate it with the following below?
Your scenario will not always be true otherwise. Depending on the orientation of the triangle the Opposite Side and Adjacent Side values can either be Dₓ or Dᵧ.
|D|cos(