Re: [Phys-l] Ball on an inclined plane

[pschoch@nac.net]

Thank you both!  I rotated the axis to be along the plane and did the
solution -- I never thought to not do so.  Too many block on an incline
plane problems.

Thanks again,
Peter


> I agree with the analysis, and note that mg is approximately 10N.  Why
> would
> we (or the problem's creator) want to "divide out mg"?  The "solution"
> provided by the book seems to be unitless?!  Was the mass supposed to be
> 0.1Kg?
>
> ----- Original Message -----
> From: "Josh Gates-fac" <Josh_Gates-fac@nobles.edu>
> To: "Forum for Physics Educators" <phys-l@carnot.physics.buffalo.edu>
> Sent: Wednesday, August 02, 2006 6:37 PM
> Subject: Re: [Phys-l] Ball on an inclined plane
>
>
> > here goes:
> > What I'm getting from the problem description is that there's a vertical
> > wall (WLOG, put this on the left) and an inclined plane going from lower
> > left to upper right, theta above horizontal.  The forces on the ball
> > are:
> > gravity - mg down
> > force of wall - directed horiz. to right
> > force of inclined plane - directed normal to plane, towards upper left
> > of
> > diagram.
> >
> > net force on ball vanishes in all directions.
> > In particular, choosing right and up as positive,
> > Fnet, x = Fwall - Fplane sintheta = 0
> > Fnet, y = Fplane costheta - mg = 0
> >
> > from this,
> >
> > Fplane = mg/cos theta
> > and
> > Fwall = mg tantheta
> >
> > dividing out mg gives Feynman's solution.
> > jg
> >
> >
> >
> > Forum for Physics Educators <phys-l@carnot.physics.buffalo.edu> on
> > Wednesday, August 02, 2006 at 4:49 PM -0500 wrote:
> > > I have just finished reading "Feynman's Tips on Physics" and found it
> > > very
> > > enjoyable.  Then I got to the problems at the end of the book and the
> > > first one intrigued me.
> > >
> > > A Ball of Mass = 1 kg and radius of 3 cm is resting on an inclined plane
> > > against a wall, the incline is at an angle alpha.  The question is to
> > > find
> > > the force from the incline and the wall on the ball assuming negligable
> > > friction.
> > >
> > > Hubris is a terrible thing.  Thinking this was trivial, I plunged into
> > > the
> > > problem and got nowhere near the solution in the text:
> > >
> > > F(plane)= 1/cos(alpha)
> > > F(wall) = tan(alpha).
> > >
> > > I get (leaving mass unknown and g in the equations):
> > >
> > > F(plane) = Mg cos(alpha)
> > > F(wall) = 0.5Mg sin(2*alpha)
> > >
> > > Any help in where I went awry would be appreciated.
> > >
> > > Peter Schoch
> > > Sussex CC
> > >
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> >
> >
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