# Virtual Work to find equilibrium condition

1. Jan 18, 2016

### Ren

1. The problem statement, all variables and given/known data
Find the tension T needed to hold the cart shown (pic included) in equilibrium, no friction. Using virtual work, and force components.
(I don't care about signs, just looking for the magnitude of tension with quick reasoning)
(not homework, just studying virtual work)

2. Relevant equations
Variables:
Incline angel Theta = 30 degrees. Mass on incline has weight W.

3. The attempt at a solution
Using force components I get the solution given in the book. My work in included in 2nd pic. I identify the horizontal component of Wcos(theta) as the force that must be balanced by Tension. This horizontal component of W is W*cos(30)*sin(30) = √3W/4, correct.

Not my main question, but I'm wondering how this makes sense. To me, it seems wrong since this component, Wcos(30)sin(30), is already cancelled by the horizontal component of W*sin(30). Meaning the component I circled in my 2nd pic, labeled with T, is not really there since one could argue that the component of weight along incline, Wsin(30), has an equal and opposite horizontal component. What happens to this 2nd horizontal component, W*sin(30)cos(30) that would be pointing to the right?

My main question is how to solve this using virtual work.
My first thought was to equate the work done by the Weight as mass falls a vertical distance H, with the work down by tension as large cart moves horizontal distance X. Being 30 degree incline, X = √3 H. But setting these works equal : T*X = W*H gives me T = W/√3. To me, I really want to look at the work done by tension in moving the cart a horizontal distance, that seems natural, but I don't know what component of weight over what displacement I should take as the balancing work, if it's not what I have above. I could try component of Weight along incline, Wsin(30), dotted with vertical distance H, so I would need to multiply by cos of angle between them (60 deg), so: (W/2)*H*cos(60) = (W/4)H, set this equal to tension work, T*X, but this still gives (W/4)*H = T*√3 H, T = (W/4)(1/√3), this √3 is still in the denominator.

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2. Jan 18, 2016

### TSny

Hello, Ren. Welcome to PF!

For your first question, I believe you need to be careful in defining your system. You can take the incline alone and balance the external forces on the incline. Or you can take the incline and block together as "one object" and balance the external forces on that object.

For your second question, I don't think you have the correct relation between the horizontal virtual displacement of the incline (X) and the corresponding vertical component of the virtual displacement of the block (H). Once you get that right, your virtual work relation TX = WH should give the correct answer. It helps to note that due to the constraint of the rope, the block must move along a circle.