You draw all the forces and in each case you write down the following two equations:
Total force up=Weight
Total
torque counterclockwise about center of mass=0
(you have to pay attention to the *signs* since the forces are taken "up" as positive and the torque "counterclockwise" about the center as positive [you can also chose clockwise if you like, but at least all torques in the *same* direction about the center of mass])
The torque is (distance from center of mass) times (perpendicular force in counterclockwise direction). For example if the force on string P is called P and string P is 2m from the center of mass, then the torque is -2*P (negative since its not counterclockwise as we agreed). Likewise Q is 1m from the center so the torque is 1*Q.
a)
forces and torque equations:
P+Q=60
-2*P+1*Q=0
Solve this set of equations and get P=20N and Q=40N
b)
again forces and torque this time with a new force A at point A (and distance 2.5 from center):
P+Q-A=60
2.5*A-2*P+1*Q=0
and P and Q should be equal so
P=Q
Solving these equations (use P=Q first to get rid of Q and continue) you get
A=15N and P=Q=37.5N
c)
this time the extra force is B at B:
P+Q-B=60
-2*P+1*Q-2.5*B=0
for the bar to stay horizontal you want string P to pull and not push (which a string cant do). so
P>0
canceling Q in the first two equations you get an equation for P and pluging this in into the last you get
20-0.5*B>0
This corresponds to
B<40N
which is the maximum weight you can apply at B.
