# Virtual work isn't working

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1. Jun 4, 2015

### John1951

1. The problem statement, all variables and /known data
Help! I'm stuck on an exceedingly simple statics problem, number 2.24 in the New Millennium edition of exercises for the Feynman lectures.

The problem consists of an inclined plane (inclination angle 30 degrees) on wheels with a frictionless weight W on the slope and held there by a string parallel to the slope and fixed to a wall to the right. In the absence of a tension on the inclined plane/cart, the cart would move to the right under the force of W.

The problem is to find the leftward tension required to keep the plane/cart from moving, using a) forces and b) virtual work.

With forces, it's very easy. The downward force of W is just W. It's component along the slope is Wsin(30) and the x-component of that is Wsin(30)cos(30) or W(square root of 3)/4, which is the answer given in the book.

The virtual work method is also very easy (or so I thought). If the required restraining force T shifts the plane/cart to the left a small distance x, the weight W slides along the plane a vertical distance y and the work done is Wy. Since y/x is tan(30), the work done is Wtan(30)x. As the work Tx done in moving the cart to the left must equal the work Wtan(30)x done in raising weight W vertically, T=Wtan(30)=W/square root of 3, which is wrong.

Where did I go wrong?

Thanks!

2. Relevant equations

3. The attempt at a solution

2. Jun 4, 2015

### BiGyElLoWhAt

You've equated Tension (a force) to work (an energy) * a ratio (unitless number).

3. Jun 4, 2015

### John1951

Thanks for you reply. What I equated, though, were two "works": 1). The work done by the force (tension) T in (reversibly) moving the inclined plane to the left a small distance x, and 2) the work done by the inclined plane as it moves the weight W vertically a small distance y. The two must be the same, unless I'm missing something.

4. Jun 4, 2015

### BiGyElLoWhAt

In your final statement, I see T=Wtan(30). Not T*x = ...

5. Jun 4, 2015

### haruspex

Doesn't your analysis assume the actual system is static? It won't be, because the weight will slide. If you imagine instead that the weight is constrained to move in a vertical line, so the system is static, then the virtual work analysis will be exactly as you have above, but the tension is now different.

6. Jun 5, 2015

### John1951

Thank you. I'm not sure I follow, though. Yes, the system is static, but the small shift of the inclined plane leftward and the resultant sliding of the weight rightward and upward along the inclined plane are imaginary (virtual). I don't see how I can imagine that the weight is constrained to move vertically as the problem itself specifies the constraints: 1) that the weight can slide without friction along the slope of the plane and 2) that the weight is attached to a vertical wall to the right of the plane by a string. Since any movement of the weight at the end of the presumably fixed-length string would be perpendicular to the string, this constraint would do no work on the weight.

7. Jun 5, 2015

### SammyS

Staff Emeritus
Sure!

The result of using Algebra after equating two expressions for work.

8. Jun 5, 2015

### haruspex

I may have misinterpreted the question.
As I read it now, there are two cables, one tying the weight to a wall on one side and the other attached to the plane. We want the tension in the second required to achieve stasis.
Since there are no other horizontal forces on the weight+plane system, the two tensions are equal, T. If the normal force on the weight is N then $W=N\cos(\theta)$, $T=N\sin(\theta)$, so $T=W\tan(\theta)$.
When I first read the problem I thought the plane was free to slide. That reduces the tension in the cable attached to the weight to $W\sin(\theta)\cos(\theta)$.

Edit: ignore all that. When I wrote the above I forgot the cable attached to the wall was parallel to the slope, not horizontal.

Last edited: Jun 5, 2015
9. Jun 5, 2015

### John1951

Thank you! I'm sure you have answered my question, but I don't quite see it yet. The correct answer (given in the book) is Wsin(theta)cos(theta) Which for theta = 30 degrees gives (sq root of 3)/4. I get this result using the forces method.

The problem arises when I use the virtual work method, which yields Wtan(theta).

I'm sure you've put your finger on where I've gone wrong, but where is it, exactly? You mention the inclined plane being able to slide. It does slide without friction from left to right on wheels. How does this affect the outcome?

Thanks again for your help with a problem si simple but which has me tearing my hair out.

10. Jun 5, 2015

### TSny

The work done by the weight of the block would be -Wy (i.e., negative).
Careful, here. Check again whether or not the vertical rise of the block will be y = x tan(30). Does the block move only vertically when the incline is shifted a small amount to the left? Hint: Does the string attached to the block remain extended as the incline is shifted?

11. Jun 6, 2015

### John1951

Thank you. Yes, the cable from W to the wall (parallel to the slope of the inclined plane) remains extended. Since the slope is constant, each Leftward translation x of the plane along its wheels results in W sliding upward and to the right along the plane.

12. Jun 6, 2015

### TSny

OK. Not sure if you got it to work out now. The key is to see the direction that the block moves relative to the earth frame of reference.

13. Jun 9, 2015

### John1951

You are quite right and now I see your point! Since the weight is at the end of a thread, as the inclined plane shifts horizontally beneath it, the weight moves not vertically, but normal to the slope of the plane (tangent to the circle described by the weight moving at the end of the thread). Once I realized this, it was easy to get the correct answer. Thanks!