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Please help: Normal force for box on an inclined plane = mg/cos(theta)?

  • Thread starter igul222
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  • #1
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Homework Statement



A box sits on an inclined plane. Given its mass and the angle of the incline, what is the magnitude of the normal force?

Homework Equations



I don't know- my understanding is N=mgcos(theta), but my physics textbook says otherwise...

The Attempt at a Solution



My attempt at two different solutions (different reference frames, I think). I don't know which one is correct, or why the other one is wrong.

[PLAIN]http://dl.dropbox.com/u/413086/Screen%20shot%202011-01-31%20at%2012.26.50%20AM.png [Broken]

UPDATE: I found my problem- the box is accelerating in one direction. Helpful link: https://www.physicsforums.com/showthread.php?t=157790
 
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Answers and Replies

  • #2
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Should there not also be a force due to friction pointing back up the slope? When you put that into your free body diagram have a look at all the individual x and y components of each force, F_n, f_s and F_g.
 
  • #3
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If you'll notice in either coordinate system way of looking at it (left side is x-oriented along incline, right side is x-y conventional), you have an unbalanced force in the +x-direction (+x down the ramp or to the right, respectively).

You appear to be using Newton's Laws, but you've neglected friction (the only thing keeping the box from sliding down the ramp). After you include that (remember, Friction is FUN, i.e., f_s=mu_s*F_n), you can appropriately solve for the normal force.

The really good news is that you drew in your angles correctly. That's probably one of the stickiest points for many intro physics students.
 
  • #4
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Yes, that's it- the box is accelerating down the plane. Thanks everyone!!
 
  • #5
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My only problem--and I might be wrong because I'm new to physics-- is that your lines of force above the box (your right triangle) might be drawn incorrectly. I think the line of force representing your force of gravity should be drawn as the hypothenus (sp.?) of the triangle, and the line of force connecting your force of gravity and your normal force should be drawn parallel to the slant of the inclined plane. That way, cosine of theta = normal / mg, and then the normal force equals mg * cosine (theta). The way you have drawn it cosine of theta = mg / normal which gives normal = mg / cos theta, which I think is wrong. Or am I wrong?
 

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