Work done by a force in a deformable body

Question

Consider a deformable body that is acted upon by several forces at different points of application.

Under these forces, the body remains at rest but deforms and the particles have some displacement. Consider the force $F_1$. This force initially had its point of application (POA) at A which then moved to B, resulting in a displacement of $s$. This movement of POA was caused by all the forces not just $F_1$.

Assuming $F_1$ was a constant force, will the work done by the force $F_1$ be equal to $$W = F_1s$$ ?considering that s was caused by all the forces not just $F_1$.

This question aims at achieving the conceptual clarity of whether the displacement in the definition of work should necessarily be caused by the force for which work is evaluated or whether we just care about the displacement of the point of application irrespective of which force caused it.

Answer 1

Work is a strange but useful mathematical concept. It is very handy because it is conservative in inertial frames. It also has an intuitive meaning most of the time, but not always. This is what gets people into trouble. Engineers have to accept the formal definition, not the intuitive definition, and this is what is being tested here.

The question as posed is very unrealistic. So the intuitive sense of what is going on is misleading. But it can be handled by simply using the definitions at hand.

The reaction forces are indeed necessary, but they do no work in this reference frame because they aren't associated with any displacement. Is this realist? No - this will not happen in the real world. But it is how we are told to treat this problem .

So all of the other forces evaluate to zero work. The only work is the constant force (again, unrealistic) acting on an deformable body over distance s. So the integral of $\mathbf{F} \cdot ds$ over the interval s is just $\mathbf{F} \cdot s$.

From a comment:

The problem statement doesn't indicate that all other forces are associated with zero displacements.

Okay, but the question is about the work "done by the force $F1$". Some mechanism produces $F1$, and the amount of work it does is $F1 \cdot s_1$.

You can ignore the details of the free body and the other forces. In the real world, there would only be one way to parameterize the other forces in terms of $F1$ so that $F1$ is constant and some body reference point remains stationary throughout.

We can loosen that condition a bit and only require that the net acceleration of the entire body is zero (final ketetic energy equals initial kinetic energy). Now there are multiple ways to parameterize the other forces in terms of $F1$, and $s_1$ is dependent on which parameterization you choose, as are all the other displacements $s_2 - s_n$. The total work done on the body by all forces is constant if the shape is the same. The relationship between the possible displacements is that they are linear transforms. But the work done by $F1$ is still $F1 \cdot s_1$ whatever $s_1$ turns out to be.

Now we can remove all restrictions, let the free body accelerate as it will. Let it deform as it will. There are infinite parameterizations of the other forces in terms of $F1$. It still doesn't matter. The work done by $F1$ is still $F1 \cdot s_1$. That is why we really like to work with work. Problems can be separated into tractable pieces if you can parameterize the pieces.

Answer 2

Individual work done on the body by a single force equals $F*s$ as shown on the sketch to the right. The work done by multiple forces equals the vector sum of the forces ($R$) and the absolute/net displacement ($d$).

Answer 3

The work of a force is by definition vector multiplication of the force by displacement. $$W =F*x$$

Let's use two examples to clarify. In a pendulum, the work done on the part of the descending of the mass is:

$$W= F*h=mg*h$$

Regardless of the fact that towards the bottom of the movement the pendulum doesn't need any force to move, it moves by inertia!

Or Consider a bar on a frictionless table rest.

If you apply force F to the middle of the bar it obviously accelerates.

$$F=m*a$$

But the same bar if we apply the force, not to its center it both accelerates and turns. Same force F now does more work. We just add all the components of the work and set it as the work done by force F.

Edit

Here is another example.

Consider a vertical thin rod supporting a mass of 1kg placed on a flat plate on top of the rod penetrating the deformable material.So F= 9.8N.

We measure the penetration in two cases. once without any other forces applied to the material the other time with some other forces and measure the two penetrations. ds1 and ds2.

in both cases the work is

$$W= F_1*S_1 \ or \ W = F_1*S_2$$ The amount of deformat/ penetration varies, and also the work done by F1 varies, but the quantity of work is difined by W= F ds.