## Physics (10th Edition)

1) The translational forces: We have $\sum F=ma$ In A, $\sum F=4F$, so $a\ne0$ In B, $\sum F=2F-F-F=0$, so $a=0$ In C, $\sum F=F+F-2F=0$, so $a=0$ 2) The rotational torques: The axis of rotation is at the center, so in A and B, the force $\vec{2F}$ does not create any torques because it passes the center of the sheets. Take the lever arm for each force to be $L$, since $\sum \tau=I\alpha$; we have In A, $\sum\tau=FL-FL=0$, so $\alpha=0$ In B, $\sum\tau=FL-FL=0$, so $\alpha=0$ In C, $\sum\tau=FL-FL-2FL=-2FL$, so $\alpha\ne0$ Therefore, in A, $a\ne0$ but $\alpha=0$ -> (b) In B, $a=\alpha=0$ -> (c) In C, $a=0$ but $\alpha\ne0$ -> (a)