Problem of statics reposted for the benefit of answerers who missed to answer?

The answer that I was about to post in the previous question and had to be done over again is as follows:

Let (x, x/√3) be the coordinates of the contact point on the right, and (z, -z√3) that of the left one, the bottom of the trough being at (0,0). We find that in order that the distance between the two is 4, z must be as follows:

z = – (1/√3)(√(12 – x²))

Thus, the height of the mass is as follows:

√(12 – x²) + (1/4)(x/√3 – √(12 – x²))

which forms a nice arc (see link for graph). The maximum occurs when x = √(3/7) (using a bit of calculus), so that the coordinates of the left and right points are:

(-3√(3/7), 9/√7)

(√(3/7), 1/√7)

The slope then works out to 2/√3, so that the angle is ArcTan(2/√3) = 49.1066

It’s easy to jump into the conclusion that as x moves from 0, the mass drops in elevation monotonically until the rod can’t go any further. But in fact, it rises very slightly first before dropping. And that maximum is not a stable point, in that if the mass is deflected in any way, it will drop.

Source(s): http://i254.photobucket.com/albums/hh120/Scythian1…

Great problem:

The mass will follow an ellipse with, in the reference frame of the inclined plane, will have an equation

(Here is a wolfram plot of the tilted ellipse. Note that the coordinate system is not equal: http://www.wolframalpha.com/input/?i=%281%2F2+y+%2… )

x’² + y’²/9 = 1

or

x’ = 1/3 √ (9 – y’²)

Now converting back to standard coordinate

y = y’ cos 60 + x’ sin 30

y = √3/2 y’ + 1/2 x’

y = √3/2 y’ + 1/6 √ (9 – y’²) )

max h will be at dy/dy’ = 0

dy/dy’ = 0 = √3 /2 – 1/6 y’ / √(9- y’²)

1/6 y’ = √3 /2 √(9- y’²)

1/36 y’² = 3/4 (9 – y’²)

28/36 y’² = 27/4

y’ = 9 √ (3/28)

x’ = 1/3 √ (9 – y’sup2;)

x’ = √ (1 – 27/28)

x’ = √ (1/28)

Converting to standard coordinates

y = h = 9 √ (3/28) * √3 /2 + 1/2 √(1/28)

y = 14 √(1/28)

x = -1/2 * 9√ (3/28) + √3 /2 √(1/28)

x = -4 √ (3/28)

x = -2 √ (3/7)

I haven’t computed the angle, but it looks close to your result.

(have to go, I’ll come back and rework it in a could of hours)

**************************

(I’m back)

Conclusion

The easy way to figure this out is to go back to the reference coordinates of the plane. There

x’ = √ (1/28)

Since we know this is perpendicular to the y’ axis and we know the length (hypotenuse) to y’ axis is 1, we can get the angle in the angle to the y’ axis

θ’ = asin (x’/1)

θ’ = 10.8933°

Now we convert to a horizontal frame of reference

θ = θ’ – 60

θ = -49.1066°

And all is right with the world.

*Note: This is an unstable equilibrium.

……………………………………..

Kudos to Scythian and to MD for their different approaches to this interesting problem, and TU to C22 who was supportive of my approach.

@Madhukar Daftary

I see you worked out your answer from mechanical equilibrium. Wouldn’t it be easier to minimize potential? All that is needed then is to write an equation for the possible positions of the mass, when the angle is given. FGR seems to be on to it.

Just a thought.