A non-uniform rod has length 2m. The rod is placed on a support at its midpoint. It is in equilibrium when a particle of weight 50N is placed at one end and a particle of weight 30N on the other end. The particles are then removed and the rod is then suspended by two light inextensible strings attached to its ends. When the rod is horizontal, the tension in one of the strings is 120Nn. Find the two possible values for the weight of the rod. Trying to use moments hasn't been useful to me, because it introduces another unknown, namely, the distance between the centre of mass and the midpoint. I don't think this is what I need to do, but I may be wrong.
I have also tried to use simultaneous equations to find one value. M = mass of rod. R = reaction force
9.8M = 170 + R
9.8M = R – 80
This gives me an answer of 125N, but according to the mark scheme, 220N and 260N are the answers, and I have no clue how to get these values.
If you could offer any help, thank you very much.
Best Answer
Let x represent the center of mass from distance from the midpoint and W weight of the rod. By the equation of torque, we get xW+30-50=0, xW=20. There are two scenarios. Scenario 1 is the 120N string is attached to the left end of the rod.
xW+120-T1=0
T1=140
So the weight is, by using a not-torque equation, W-T1-120=0, W=260 N.
Scenario 2 is with 120 on the opposite side.
Torque equation gives xW+T2-120=0, T2=100. So by the standard force diagram, W-120-100=0, W=220 N.