12.2 Examples of static equilibrium  (Page 4/9)

Strategy

Notice that in our frame of reference, contributions to the second equilibrium condition (for torques) come only from the y -components of the forces because the x -components of the forces are all parallel to their lever arms, so that for any of them we have $\text{sin}\theta =0$ in [link] . For the y -components we have $\theta =\pm 90\text{°}$ in [link] . Also notice that the torque of the force at the elbow is zero because this force is attached at the pivot. So the contribution to the net torque comes only from the torques of $T_y$ and of $w_y.$

Solution

and the y -component of the net force satisfies

[link] and [link] are two equations of the first equilibrium condition (for forces). Next, we read from the free-body diagram that the net torque along the axis of rotation is

[link] is the second equilibrium condition (for torques) for the forearm. The free-body diagram shows that the lever arms are $r_T=1.5\text{in}\text{.}$ and $r_w=13.0\text{in}\text{.}$ At this point, we do not need to convert inches into SI units, because as long as these units are consistent in [link] , they cancel out. Using the free-body diagram again, we find the magnitudes of the component forces:

We substitute these magnitudes into [link] , [link] , and [link] to obtain, respectively,

When we simplify these equations, we see that we are left with only two independent equations for the two unknown force magnitudes, F and T , because [link] for the x -component is equivalent to [link] for the y -component. In this way, we obtain the first equilibrium condition for forces

and the second equilibrium condition for torques