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I'm sure your're sick of me by now, but heres my question

  1. Jan 17, 2006 #1
    i've included the text of the only one of the threads I accidentally created that contains my problem, the others were created by accident becuase my browser returned a database error, I had no idea they were actually posting (and blank messages to boot!)

    sorry for the repeats guys, my comptuers on the blink, is there a way to delete multiple posts? the other posts dont contain my question, so here goes:

    hi guys, this is my first time posting but here goes:

    I'm trying to understand the concept of inclined planes for my AP physics midterm exam, and for some reason its not clicking anymore, my teacher is, unfortuantely, less than useless (i went to ask him about it today after school and he told me that you solve the problems a certain way, becuase thats how "smart people" determined you do it)

    I'm looking at a basic picture of an inclined page in my textbook, with an angle theta of 30 degrees, i've drawn a picture of this plane in my notebook and done as the book says: I aligned my coordinate system to the top of the plane, mg (Fg) is pointing down at an angle in relation to the coordinate system, but is vertical otherwise. now my textbook and teacher tell me that the X component of gravity (Fgx) i simply use mg * sin (theta) and to find the Y component (Fgy) i use mg* cos (theta), but i dont understand this, sin (theta) should give me something like mg / (whatever the hypotenuse of the traingle is)

    I think its apparent that i'm bloody useless at this , so i'd appriciate any help! if it includes a primer on trig even better! but please someone explain to me why you solve inclined plane problems this way!

    thank you so much in advance everyone!

  2. jcsd
  3. Jan 17, 2006 #2
    Oh, where you are lost goes back to geometry. So, your plane is inclined at an angle Ø (theta), and if you use similar triangles then you know that the angle from the y-axis to the vector component of the force of gravity is that same angle Ø. Then we can use trigonometry to find the x-component of the force of gravity in relation to the inclined plane. You might have learned sinØ=y/r or opposite/hypotenus, and with the way our similar triangle is oriented it means that the x-component will really be the side opposite to the angle Ø. So now we know that the x-component is [tex]F_{g}sin(\theta)[/tex] or [tex]mg*sin(\theta}[/tex].

    Sorry I don't have a good diagram, but maybe someone else does.
  4. Jan 17, 2006 #3


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    Welcome Justin to the PF forums!! :smile: I think you will find this a very interesting and informative forum (check around the other topics too)..
    Yes, just go back in and edit any duplicate posts, you should see a delete option. Select that and delete unwanted posts.. You can only edit within 24hrs of your original post. If longer than that, you can pm (write a personal message) to a moderator and they will help you edit too.
    :smile: sounds like your teacher may not remember (or perhaps was not taught) how some of the elements were derived.

    I see Mindscrape has already started you off in the right direction, so I won't duplicate what was said. However I can point you to a diagram that makes the analysis clearer. In their diagram note that the angle of the inclined plane [itex]\theta [/itex] on left is equivalent to the [itex]\theta[/itex] on the right diagram. That you can prove to yourself, using geometry of similar triangles. So in the incline's frame of reference the axes may be noted x' and y' and so [itex]Fx' = mgsin\theta[/itex] and [itex]Fy'=mgcos\theta[/itex]
    Last edited: Jan 17, 2006
  5. Jan 17, 2006 #4
    thank you so much!

    Guys, you are collectively lifesavers, thank you so much for your help...i took a look at minscrapes post, while holding it next to a print out of the page that Ouabache linked me too and it all clicked, i dont know what i was missing, or what my teachers problem is, but i feel so relieved! thank you again!

    -- Justin
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