Sign of the sum of Virtual work (Direction)

Question

I have trouble understanding Virtual Work (in statics) as I understand what's happening but I do not understand the signs or directions.

This is what I know. If the force acts in the opposite direction to the displacement then it's taken as negative. But sometimes we don't do this and instead we take the derivative of an angle. Then, the derivative will account for the direction. (Do we also assume the direction of the derivative and plot it and that is why the following part happens?)

In this exercise, the work would be the work done by the spring in the positive y and x axis (standard) and the $W$, multiplied by their respective distances. In this case I took $d_y=-0.45\sin\theta\text{d}\theta$ and $d_x=-0.45\cos\theta\text{d}\theta$.

What I thought was substituting $d_y$ and $d_x$ into $\text{d}U= Wd_y + F\cos\alpha\text{d}x + F\sin\alpha\text{d}y$.

However in the solution it is said that instead is $-Wd_y - F\cos\alpha\text{d}x + F\sin\alpha\text{d}y$.

What am I misunderstanding? I hope it is clear what I'm trying to say.

Answer 1

I came up with yet another result.

Assuming $a$ is the angle of spring force with the horizontal, I end up with the following equilibrium around point B

Note that:

• $ds = 0.45\cdot d\theta$ , where $ds^2 = dx^2 + dy^2$ and the angle of $ds$ with the negative horizontal is $\theta$.

Although $ds$ doesn't come into the calculation it is important because it determines the direction of dx, dy. More specifically, for increasing $\theta$ the direction of $ds$ is pointing down and to the left.

• $R_{BC}$ is perpendicular to $ds$ (therefore produces no work).

So the virtual work (I'll denote it as $VW$ to avoid confusion with W) for each force is:

• $VW_{R_{BC}}=0$ perpendicular to $ds$ (therefore produces no work).
• $VW_W = W\cdot dy$
• $VW_F =-Fx \cdot dx- Fy \cdot dy$

The reason this is negative for weight, is because the displacement dx and dy is opposite to the direction of the force.

Therefore I end up with:

$$dU = W\cdot dy -Fx \cdot dx - Fy \cdot dy$$