Physics (10th Edition)

$a=0.1m$ and $b=0.3m$
1) Case a): The lever arm between $F_1$ and the axis of rotation is $b$, and $F_1$ causes the square to rotate counterclockwise, so $\tau_1\gt0$ . Therefore, $\tau_1=+F_1b$ The lever arm between $F_2$ and the axis of rotation is $a$, and $F_2$ causes the square to rotate clockwise, so $\tau_2\lt0$ . Therefore, $\tau_2=-F_2a$ The net torque is zero, so $$F_1b-F_2a=0$$ $$F_1b=F_2a$$ $$\frac{b}{a}=\frac{F_2}{F_1}=3 (1)$$ 2) Case b): In case b), remember that the side of the square is $1m$. The lever arm between $F_1$ and the axis of rotation is $1-a$, and $F_1$ causes the square to rotate counterclockwise, so $\tau_1\gt0$ . Therefore, $\tau_1=+F_1(1-a)$ The lever arm between $F_2$ and the axis of rotation is $b$, and $F_2$ causes the square to rotate clockwise, so $\tau_2\lt0$ . Therefore, $\tau_2=-F_2b$ The net torque is zero, so $$F_1(1-a)-F_2b=0$$ $$F_1(1-a)=F_2b$$ $$\frac{1-a}{b}=\frac{F_2}{F_1}=3 (2)$$ Solving (1) and (2), we get $a=0.1m$ and $b=0.3m$