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Friction on an inclined plane (freebody diag. included)

  1. Jul 23, 2008 #1
    1. The problem statement, all variables and given/known data
    what is the max angle the plane can be at so that the crate stays at rest.

    static friction coefficient: 0.60
    mass of crate: 10kg
    max angle of plane: ?

    my free body diagram http://img.photobucket.com/albums/v217/tylerc13/physprob.jpg

    my approach:

    I set it up so that the forces of +mg and -mgUk would = 0. My train of thought was that if they canceled each other out there would be no movement... guess not.

    edit: i just realized im not incorporating the frictional coefficient into this, and i might have it a bit backwards. would my positive (downward) force be mg*sin theta and the my negative (upward resistance from friction) force be N*Uk ?
    Last edited: Jul 23, 2008
  2. jcsd
  3. Jul 23, 2008 #2
    my *NEW* free body diagram http://img.photobucket.com/albums/v217/tylerc13/physprob-1.jpg

    my *NEW* approach:

    I reworked my free body diagram and i think i have it setup right, the upward force of friction is the weight in newtons multiplied by the frictional coefficient. the downward force on the crate is the weight in newtons multiplied by the sine of the angle of the plane.

    my thoughts are that if the angle must be such that it puts less force on the crate than force of friction the crate has at rest.

    so i calculated the force of friction on the crate as being N*Uk which is (mg)Uk or (10*9.8)*0.60 = 58.8N

    i figured then that the downward force exerted by gravity and the slope must not exceed this value so i used 58.7N as my target downward force to max the angle without going over the frictional force, thus keeping the crate in place.

    so i set mg*sin theta = 58.7 getting 98*sin theta = 58.7 then dividing both sides by 98 i get; sin theta= 58.7/98 or

    sin theta = .59898 then inverse sin .59898 to get an angle of 36.8 degrees

    but this answer is getting marked wrong... where and I going wrong?
  4. Jul 23, 2008 #3
    the upslope friction force is NOT the weight times the coefficient of friction. It is the c.o.f times the normal force which you marked as N. What is N equal to?

    Also, whichever value you get for the friction force, use it in your calculations (aka don't use 58.7 instead of 58.8) because when the forces are equal, the mass still doesn't move. it is when the downslope force becomes greater then the max friction force that the block starts accelerating down.
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