Answer
(a) The heaviest weight that can be put on the bar is a weight of 550 N.
(b) The weight should be placed a distance of 0.614 from cable A.
Work Step by Step
(a) The sum of the tension in the cables must equal the sum of the weights (bar and weight). Let $W_b$ be the weight of the bar. Let $W_w$ be the weight of the added weight.
$W_b+W_w = T_A+T_B$
$W_w = T_A+T_B-W_b$
$W_w = 500.0~N+400.0~N-350~N$
$W_w = 550~N$
The heaviest weight that can be put on the bar is a weight of 550 N.
(b) Let's consider an axis of rotation about the center of the bar. Since the magnitude of the torque from the 500 N tension is larger than the torque from the 400 N tension, in order for the net torque to equal zero, the weight's center of gravity must be closer to the 500-N cable. Let $r$ be the distance from the center of the bar that the weight's center of gravity is located. Note that the bar's weight exerts zero torque about this axis of rotation.
$\tau = 0$
$(400~N)(0.75~m) - (500~N)(0.75~m)+(550~N)~r = 0$
$(550~N)~r = (500~N)(0.75~m)-(400~N)(0.75~m)$
$r = \frac{(500~N)(0.75~m)-(400~N)(0.75~m)}{550~N}$
$r = 0.136~m$
The weight should be placed a distance of 0.136 from the center of the bar. We can find the distance from cable A.
$0.75~m-0.136~m = 0.614~m$
The weight should be placed a distance of 0.614 from cable A.
