Re: Two Weekend Puzzles

[John Mallinckrodt <ajm@CSUPOMONA.EDU>]

> For your weekend pleasure, here are two notes;
> from a hobby casting list, and a mech.eng
> newsgroup respectively.
> Enjoy!
>
> Brian.
> p.s. I don't know the answers...
>
> (1) From: "Kenneth Mayer" <mayerk@surfree.com>
> addressed to this list: <hobbicast@yahoogroups.com>
>
>   Using an 18" length of copper tubing capped at
> one end and pressure gauge in the other, filled
> with fuel oil, the heat from my hands was enough
> to generate over 200 psi in the pipe.
>
> [My question: estimate a plausible diameter
>     of the pipe]

I guess I don't see why the diameter should matter.  In any event
copper will surely have negligible expansion relative to the fuel oil
so all we should need to do is counter the thermal volume change with
sufficient pressure.

Doing a weighted average over the results of a brief web search (with
weighting coefficients proportional to the apparent authority of the
site), the bulk modulus of fuel oil seems to be something like 15,000
psi and its thermal
expansion is about .05% per degree F.  So if the hands warm the oil
by 30 degrees F, we'd get a fractional volume increase of about 1.5%
which could be countered with a pressure increase of 1.5% of 15,000
psi or 225 psi.  Close enough for me.

> (2) From Craig Joyce  naughty_sonny@yahoo.com
> addressed to sci.engr.mech   ng.
>
>   There is a prism (triangular solid) of mass m1 with
> base angle alpha.
> A body of mass m2 is released from rest on the incline
> of m1. We have to find the horizontal acceleration of m1.
>   All surfaces have negligible friction. This is a
> standard problem with the twist that m1 is not fixed.
>
> I am drawing a crude diagram to illustrate the problem
>   statement.
>   [I set this note to Courier New fixed pitch to
> preserve the ascii art below.]
>
>            /\ /|
>           /m / |
>          / 2/  |
>           \/   |
>           /    |
>          /     |
>         /  m1  |
>        /_______|

This is an old classic.

Conserve momentum in the x (rightward) direction:

     m2*a2x = m1*a1

Apply NII to m2 in the x and y (downward) directions:

 x:  N*sin(alpha) = m2*a2x

 y:  m2*g - N*cos(alpha) = m2*a2y

The acceleration of m2 *wrt the wedge* is the vector sum of the
acceleration of the m2 wrt the inertial frame and the acceleration of
the inertial frame wrt the wedge AND it must make an angle alpha with
the horizontal.  So

   tan(alpha) = a2y/(a1+a2x)

Four equations, four unknowns.  Eliminate a2x, a2y, and N and solve for a1:

        sin(alpha)cos(alpha)
   a1 = -------------------- * g
        m1/m2 + sin^2(alpha)

Does the right things when alpha = 0 and pi/2.  Might be right.

--
John Mallinckrodt        mailto:ajm@csupomona.edu
Cal Poly Pomona          http://www.csupomona.edu/~ajm

This posting is the position of the writer, not that of SUNY-BSC, NAU or the AAPT.