Stability and equilibrium problems

[dtavob]

Homework Statement

An object is restricted to movement in 1-D. Its position is specified along the x-axis. The potential energy of the object as a function of its position is given by U(x)= a(x4 − 2b2x2), where a and b represent positive numbers. Determine the location(s) of any equilibrium point(s), and classify the equilibrium at each point as stable, unstable, or neutral. (Enter your answers from smallest to largest. Enter NONE in any unused answer blanks. Use any variable or symbol stated above as necessary.)

Homework Equations

F = - dU/dx

The Attempt at a Solution

What I did was find the second derivative of the given function, which is = -12axˆ2 + 4ab^2), then I tried using a=2 and b=3 and then use x=-1,0,1 to see if it's stable, unstable or neutral. But, the way the web-homework wants me to input the answers is where I'm lost.

I was able to get the correct answers for the classification of the 3 equilibrium points, but I don't know how to input the other part.

x =_________ ---stable
x =_________ ---unstable
x =_________ --- stable
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Homework Statement

A uniform rigid pole of length L and mass M is to be supported from a vertical wall in a horizontal position, as shown in the figure. The pole is not attached directly to the wall, so the coefficient of static friction, μs, between the wall and the pole provides the only vertical force on one end of the pole. The other end of the pole is supported by a light rope that is attached to the wall at a point a distance D directly above the point where the pole contacts the wall. Determine the minimum value of μs, as a function of L and D, that will keep the pole horizontal and not allow its end to slide down the wall. (Use any variable or symbol stated above as necessary.)

Homework Equations

FnetX = 0
FnetY = 0
TorqueNet = 0
Friction = μsFn

The Attempt at a Solution

What I did is:

FnetX= 0
Fn-Tcos(x) = 0
FnetY = 0

μsFn+Tsin(x)-mg = 0
TorqueNet = 0

TorqueRod - TorqueTension = 0

L/2mgsin90 - LTsin(x) = 0

If what I did is right then the answer would be: μs = [mg-Tsin(x)]/Tcos(x)
but the problem is asking me to use L and D in my final answer, that's where I'm lost...

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Thanks in advanced and I'm sorry for such a long post =/.

 

[anathema]

Hello,

For the first problem, you are on the right track. The equilibrium points are the values of x where the force is zero, so you have to set the force equal to zero and solve for x. In this case, the force is the negative derivative of the potential energy, so you would set -dU/dx = 0. Plugging in the given values for a and b, you would get -12ax^2 + 4ab^2 = 0. Since a and b are positive, you can divide both sides by 4ab^2 to get -3x^2 + 1 = 0. Solving for x, you get x = ±1/√3. These are the two equilibrium points. To classify them as stable, unstable, or neutral, you would need to look at the second derivative of the potential energy at each point. When the second derivative is positive, the equilibrium point is stable. When the second derivative is negative, the equilibrium point is unstable. And when the second derivative is zero, the equilibrium point is neutral. So, you would need to plug in the values of x = ±1/√3 into the second derivative, -12ax^2 + 4ab^2, to determine their stability.

For the second problem, you are also on the right track. The key is to realize that the force of friction is the only force keeping the pole from sliding down the wall, so you need to set the force of friction equal to the other forces (tension and weight) in order to find the minimum value of μs. So, you would set μsFn = Tsin(x) + mg and then solve for μs. Fn is just the normal force, which is equal to the component of the tension force that is perpendicular to the wall. You can find this using trigonometry, specifically the tangent function. The tangent of the angle x is equal to the opposite side (Tsin(x)) divided by the adjacent side (Tcos(x)). So, you can solve for Tcos(x) and then plug that into the equation for the normal force (Fn = Tcos(x)). Then, you can plug that into the equation for μs and solve for μs in terms of L and D. I hope that helps! Let me know if you have any further questions.