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Inclined planes question?

Can anyone explain this?

When dealing with an inclined plane with an object on it, in order to find the forces along the x-axis, why do you use sin(theta) multiplied by the weight of the object. Also, with the y-axis, you use cos(theta). In the problem I am looking at, the Cartesian coordinate has been tilted so that the x axis is parallel with the incline plane. In math, I always thought that cos(theta) related to the x direction and sin related to y.

If this doesn't make sense, comment and I can clarify with an example from my lecture.

Thanks

Update:

@ izelkay

Isn't there only one component for weight pointing in the y direction?

Update 2:

@ izelkay

Forget the question I asked above, the example made it click in my head

Thanks, I'll give you BA as soon as I can

3 Answers

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  • 8 years ago
    Favorite Answer

    Ignore what you think you know and just look at your free body diagram. When you tilt the axis so that the y axis is perpendicular to the inclined plane and the x axis is parallel to the inclined plane, draw the components of the weight force. When you draw the y component, you should notice that it will be cos(theta), and the x component will be sin(theta).

    Here's an example:

    http://dunningrb.files.wordpress.com/2007/10/300px...

    I, personally, would have draw the mgsin(theta) so that it connects the mgcos(theta) and the mg. It's easier to see why it's mgsin(theta) that way.

  • 8 years ago

    You need to draw a free body diagram

    draw an arrow straight down where the object meets the plane, this is the weight or mg, from the tip of draw an arrow going up to where the object meets the plane so that the arrowhead meets the tail of your first arrow, this arrow should be perpendicular to the the plane, this is your normal force, the force of the plane pushing up on your object, now draw a third line from somewhere on your second line, that it meets the tip of the first (weight) arrow, this third arrow should be parallel to the plane, Now you have a little triangle of three arrows inside the triangle that makes up your plane. A little geometry will show that the angle between arrows 1 and two is the same as the angle between your inclined plane and the ground, you can work that out, or just trust me on this. Now, using the angle between arrows one and two, see that arrows two and three ( the normal force arrow and the arrow in the direction of the plane) are perpendicular, so you have a right triangle, the arrow in the direction of the plane is the vector component of the weight that is in the direction of the inclined plane, and it is OPPOSITE your angle, so its value is mgsin a (a is your angle) The normal force arrow is ADJACENT to your angle so its value is mgcos a

  • ?
    Lv 4
    5 years ago

    A wedge, a chisel, a pry bar, a knife, an awl. evaluate how a knife works, you have an exceedingly sharp part (like the backside of the susceptible plane) however the blade gets thicker as you pass faraway from the sting. This makes the item being decrease unfold aside, so as that the reducing is far less puzzling.

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