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I am a new physics student and learning Newton's law at the moment.

I remember from the previous lessons that in order to find the horizontal component of force, the equation is $F \cos θ$ and for the vertical component it's $F \sin θ$.

My question on this example is: why the professor chose this direction to present the $x$-direction and not the opposite?

enter image description here

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  • $\begingroup$ Keep in mind that cosine of an angle is defined as the length of the adjacent side divided by the length of the hypotenuse ... it is NOT the length of the horizontal side divided by the length of the hypotenuse. Sine of an angle is defined as the length of the opposite side divided by the length of the hypotenuse. Whether these sides are horizontal or vertical depends on which angle of the triangle is involved. And note - this tends to be a common mistake among inexperienced physics students. $\endgroup$ Jan 23 at 17:09
  • $\begingroup$ (Terminology nitpick: ‘𝐹𝑐𝑜𝑠θ’ is an expression. An equation is something that equates two things — makes them equal — usually by putting an ‘=’ between them!) $\endgroup$
    – gidds
    Jan 23 at 22:04
  • $\begingroup$ I believe PSE is not the right place to answer this kind of questions. You could easily ask you tutor or consult a text book. $\endgroup$
    – Markoul11
    Jan 24 at 7:00

4 Answers 4

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If you are resolving a force into components in directions A and B at right angles to each other, the rule is

Component in direction A = Magnitude of force multiplied by cosine of angle between direction of force and direction A.

Component in direction B = Magnitude of force multiplied by cosine of angle between direction of force and direction B.

I used A and B only to try and make myself clear. There is no need to use any sort of label on the direction when you are doing the resolving, that is multiplying the magnitude of the force by the relevant cosine. So, usually, calling the directions '𝑥' and '𝑦' doesn't help. The exception, perhaps, is when the directions are horizontal and vertical, either literally or on a plan.

Your teacher has chosen to resolve the pull of gravity on the car into components in the direction of the 'downward' normal to the slope and the direction down the slope (for which we note that $\cos (90°-\theta) = \sin \theta$). [It's convenient to choose these directions rather than their opposites, because it avoids the components having to have negative signs in them, when expressed in terms of the slope angle, $\theta$.]

Have you checked that the magnitudes of the components marked on your diagram accord with the rule given above ?

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  • $\begingroup$ Thank you for your reply, So there is any constant rule to know what vector represents the cos component and which is the y component? or I should always look for the perpendicular vector and the parallel vector according to that slope? $\endgroup$
    – Ryan
    Jan 23 at 13:11
  • $\begingroup$ "So there is any constant rule to know what vector represents the cos component and which is the y component?". I'd say "no". My advice is not to think about $x$ and $y$ (except when you're dealing with horizontal and vertical). Simply use the rule about the cosine of the angle between the force direction and the direction in which you're resolving that I gave early in my answer. It's easy to remember and always works (as long as the directions in which you are resolving are at right angles to each other). $\endgroup$ Jan 23 at 13:21
  • $\begingroup$ In addition to what @PhilipWood already said: you can select whichever coordinate system you want. You could solve this problem in polar coordinates if you wanted to. The physics and the outcome of the problem will be the same. Typically we choose the coordinate system that makes the math easy. In the example in this question, the choice is clever, because nothing happens in $y$ for the cart, only in $x$ $\endgroup$
    – Bernhard
    Jan 23 at 14:51
  • $\begingroup$ @Ryan I've expanded my answer a little, in an attempt to make it as clear as possible. $\endgroup$ Jan 23 at 18:46
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and welcome to the world of physics.

The answer is rather simple, you take the cos component of the weight in order to make it perpendicular, recall that in the absence of other forces with vertical components, the magnitude of the weight is equal to the normal, and recall that the normal is a contact force that is always perpendicular to the surface of contact, then taking the cos component of the weight is necessary.

Similarly, as we were able to take a cos component of the gravitational force, we have a sin component that we must not neglect so that our answer is as neat as possible.

Hope this helps.

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  • $\begingroup$ Thank you for your answer, as I understand from your answer, no matter how the triangle will look like I mean no matter how to slope will look like, as long as the gravity force is drawn down, the perpendicular vector will be always represented as the cos component and the parallel vector will be the sin component? $\endgroup$
    – Ryan
    Jan 23 at 13:09
  • $\begingroup$ Yes, and you have to take the respective angle to statist those conditions (be extra careful with the angles) $\endgroup$
    – Doe Pual
    Jan 23 at 13:43
  • $\begingroup$ Satisfy* pardon my typo $\endgroup$
    – Doe Pual
    Jan 23 at 16:53
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On the diagram provided, I don't see any indication of an (x) direction. If I were solving that problem, I would choose an (x) axis which goes up the incline and then wraps around the pulley to go down on the right. Then both masses would have the same (x) acceleration.

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You have to take two perpendicular axis, Like you have taken previously.

We can take those two perpendicular axis at any where in any directions. And take component in those direction [ bcz as components are balanced, they (components) will be balanced even if we change ( roatate) two perpendicular axis ]

Now in this case it is easy to get solution if we choose axis so. Generally we chose one axis where movement is possible and obviously other axis perpendicular to that)

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    $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Markoul11
    Jan 24 at 7:09

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