CHAPTER 12

Static equilibrium

Practically the whole of Chapter 12 can be summarized by

a body will be in complete equilibrium (i.e. no accelerations) when both the net force and the net torque are zero

The first condition guarantees the absence of linear, i.e. translational, accelerations, while the second implies no angular, i.e. rotational, accelerations.

A legitimate question: we know that the condition $\Sigma \vec{F} = 0$ is independent of the choice of reference axes. But what about the torques ? Torques are defined with respect to a given point, so even if the torques add up to zero with respect to a given point, couldn't their resultant be non-zero with respect to another point?

Fact : if the vector sum of a set of forces is zero, then the vector sum of their torques is the same regardless of the reference point

note that I am saying that the sum of the torques is the same, not that it is necessarily zero. One could obviously have a situation where the forces add up to zero, while the torques do not. But if the torques are 0 with respect to a given point, they will be zero with respect to any point, and the body will be in both translational and rotational equilibrium, i.e it will be moving with constant (possibly 0) linear velocity and constant (possibly 0) angular velocity.

Proof : assume $\Sigma_{i} \vec{F}_{i} = 0$. With respect to an arbitrary point O1, the torque sum is

$\Sigma_{i} \vec{\tau}_{1i} = \Sigma_{i} \vec{r}_{1i}\times \vec{F}_{i}$

If we now choose a point O2, we have

$\Sigma_{i} \vec{\tau}_{2i} = \Sigma_{i} \vec{r}_{2i}\times \vec{F}_{i}$

but, if $\vec{r}_{12}$ is the position of O2 with respect to O1, $\vec{r}_{2i} = \vec{r}_{1i}-\vec{r}_{12}$,therefore

$\Sigma_{i} \vec{\tau}_{2i} = \Sigma_{i} (\vec{r}_{1i}-\vec{r}_{12})\times \vec{... ...{F}_{i} - \vec{r}_{12}\times\Sigma_{i} \vec{F}_{i} = \Sigma_{i} \vec{\tau}_{1i}$

due to the fact that $\Sigma_{i} \vec{F}_{i} = 0$

Special case : what is the condition of equilibrium for a body supported at one point and under the sole influx of gravity?

Since one can assume that the force of gravity is applied at the body's center of mass, for the torque to be zero then the moment arm with respect to the support point has to be zero, i.e. gravity's line of action has to go through the point of support. Therefore the body will be in equilibrium when its center of mass is exactly above (or below) the support point.

The conditions $\Sigma_{i} \vec{F}_{i} = 0$ and $\Sigma_{i} \vec{\tau}_{i} = 0$ can in general lead to a fairly complex situation (the two vector equations correspond to six equations among the components). In many cases the problem can be reduced to a two-dimensional one : if all the applied forces are coplanar, i.e. they belong to the same plane (let's say the xy plane), then the conditions to be satisfied for equilibrium are:

$\Sigma_{i} F_{xi} = 0, ~~~ \Sigma_{i} F_{yi} = 0, ~~~ \Sigma_{i} \tau_{zi} = 0$

Examples : Problems 12.16,12.41

Section 12.4 : read and digest. I will not discuss it, nor include it in the tests, but it is useful information