Answer
(a) The distance between the forces should be 0.80 meters.
(b) The torque is clockwise.
(c) The distance between the forces should be 0.80 meters.
The torque is clockwise.
Work Step by Step
(a) Since $F_2$ is farther from the pivot at the left end of the rod, $F_2$ exerts a larger magnitude of torque.
$\sum \tau = 6.40~N~m$
$F_2~(3.00~m+L) - F_1~(3.00~m) = 6.40~N~m$
$(8.00~N)~(3.00~m+L) - (8.00~N)~(3.00~m) = 6.40~N~m$
$(8.00~N)~L = 6.40~N~m$
$L = \frac{6.40~N~m}{8.00~N}$
$L = 0.80~m$
The distance between the forces should be 0.80 meters.
(b) Since $F_2$ is farther from the pivot at the left end of the rod, $F_2$ exerts a larger magnitude of torque. Since $F_2$ exerts a larger magnitude of torque, the torque is clockwise.
(c) Since the pivot is at the location of $F_2$, the force $F_2$ exerts zero torque.
$F_1~L = 6.40~N~m$
$L = \frac{6.40~N~m}{F_1}$
$L = \frac{6.40~N~m}{8.00~N}$
$L = 0.80~m$
The distance between the forces should be 0.80 meters.
Since only the force $F_1$ exerts a torque about the pivot at $F_2$, the torque is clockwise.
