Did I Solve the D'Alembert/Virtual Work Monkey Problem Correctly?

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The discussion centers on solving the D'Alembert/Virtual Work problem involving a uniform plank and a climbing monkey. The user derived the equation a = (2bTtanΦ - Mgb)/mg, using static equilibrium principles and virtual work. A key point raised is the difference between treating variables as independent versus dependent, particularly in the context of tension in the string. The consensus is that the user approached the problem correctly, and the method of summing moments at the base of the wall is validated as a reliable technique.

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I have a question with process when using D'Alembert/Virtual Work.

I have the example question:
A uniform plank of mass M is leaning against a smooth wall and makes an angle α with the smooth floor. The lower end of the plank is connected to the base of the wall with a massless inextensible string. A monkey of mass m starts climbing on the plank. If the maximum tension force on the string before it breaks is T, up to what maximum height can the monkey climb to?

M =mass of plank
m = mass of monkey
g = force due to gravity
2b = length of plank
a = distance monkey is up plank
x = length of string, distance between wall and where plank meets floor
y = height of center of mass of plank
z = height of center of mass of monkey

Assuming a static case just before the string breaks we have:

x = 2bcosΦ
y = bsinΦ
z = asinΦ

and

δW = Tδx + Mgδy + mgδz

If I set δW to 0 and take δx, y and z to be the derivatives of x, y and z with respect to Φ I get

0 = MgbcosΦdΦ + mgacosΦdΦ - T2bsinΦdΦ
a = (2bTtanΦ-Mgb)/mg

Which is a similar process to what I would use in the more simple example without the monkey however I saw this http://imgur.com/FqjZv" which seems to take the partial derivatives of each of the independent variables then sum them to get the virtual displacements. That second example then goes on to set each displacement to zero and get equations of motion. In comparison in the first example I simply took the derivative with respect to Φ which solves simply.

In the monkey example z has a in it which I think could be seen as an independent variable so:
Have I done the monkey question right?
In the linked example is the person taking the partials of each θ then adding the results as I suspect?
How do I know when/where to use these seemingly different techniques?

Thanks.
 
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To answer your first question:
Sum the moments where the cord is attached to the base of the ladder and set it equal to zero. Solve for 'a' and see if you get the same result.
 
This is the same problem as one where you would ask for the tension in the cord if you specified how far up (a) the plank the monkey went. In this case you would not treat the tension as a variable.
 
LawrenceC said:
Sum the moments where the cord is attached to the base of the ladder and set it equal to zero. Solve for 'a' and see if you get the same result.

I don't know how to sum the moments so I can't do this.

LawrenceC said:
This is the same problem as one where you would ask for the tension in the cord if you specified how far up (a) the plank the monkey went. In this case you would not treat the tension as a variable.

I'm trying to understand what you mean here. Are you saying not to treat 'a' as a variable and therefore my approach or the monkey example would be correct?
 
It is interesting that you would be taught the method you are using before learning the method I cited. It was quite the opposite when I was an undergraduate. I believe you did the problem correctly because I got the same answer using the method of determining the reaction at the wall, summing moments (torques) and setting to zero since the plank is stationary. This relates distance 'a' to the tension.

Regarding the example you posted, I have not had a chance to give it more than a cursory look.
 

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