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Horizontal component of gravity on inclined plane

  1. Mar 6, 2010 #1
    I'm working on a problem in which a block sits on an inclined plane of angle @ above the horizontal. Assuming there is no friction, the block will slide down the plane with a force of mg*sin(@). The trouble I'm having is figuring out the horizontal and vertical components of that force.

    Intuitively, I feel like the horizontal force should be mg*sin(@)*cos(@). My reasoning is that the mg*sin(@) force is just any other force at that point, so I can multiply is by cos(@) to get the component that is parallel to a horizontal line. But from what I've been able to find, the actual horizontal component should be mg*tan(@).

    Could someone give me an explanation for this, or point me in the right direction? I'm having trouble searching for an answer because I'm not really sure what to call this horizontal force (thus the somewhat confused title of this post.)

    Thanks.
     
  2. jcsd
  3. Mar 6, 2010 #2

    rock.freak667

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    the weight is mg,when you split that into its components that are perpendicular and parallel to the plane, you will see that parallel to the plane is mgsinθ and perpendicular is mgcosθ. Does it make more sense now?

    Did you mean horizontal and vertical with respect the incline or with respect to the horizontal that the angle is measured from?
     
  4. Mar 6, 2010 #3
    Horizontal with respect to the horiztonal base. So, the weight is directed vertically downward. Some of that weight (wgcosθ) is normal to the plane, while the rest (wgsinθ) is parallel to it. Now, some of that parallel component is directed vertically, and some is directed horizontally. I'm trying to figure out what the horizontal component is.

    In other words, what portion of mg is perpendicular to the original weight vector?
     
  5. Mar 6, 2010 #4

    rock.freak667

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    I see, in your diagram (if you have one) if you draw a line parallel to the horizontal through the mass you will see that the angle made by the plane and the drawn line is also θ.

    So the mgcosθ, is split into vertical and horizontal components as well. You should now be able to find the components now.


    It looks like your initial thinking is correct.
     
  6. Feb 28, 2011 #5
    I always think of it like this: if the object pulled or push is slanted on an angle, then the gravitational force and the normal force will both be mgcostheta but just opposite in direction, and if the object pushed or pulled is on a horizontal surface for example if one of the fat dudes at like home depot are pulling stuff and pushing stuf at an angle to move it, then the gravitational force and the normal force will be mgsintheta but in opposite dircetions. make sense?
     
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