How Does Beam Center of Gravity Affect Wire Tension and Pulse Timing?

[azila]

Homework Statement

A 1740-N irregular beam is hanging horizontally by its ends from the ceiling by two vertical wires (A and B), each 1.30 m long and weighing 2.80 N. The center of gravity of this beam is one-third of the way along the beam from the end where wire A is attached.
If you pluck both strings at the same time at the beam, what is the time delay between the arrival of the two pulses at the ceiling? (what is delta t)

Homework Equations

ok,
wave speed = T/mass per length
Summation of Torque= -rbwb + rsTb = 0
Tb = rcomWb/(length of string)

The Attempt at a Solution

So, first of all what is the radius of the center of mass, like how would I find it? After I get the Tension in string b, then how would I solve the problem? Any guidance would be helpful..Thanks..

 

[Astronuc]

So, one needs to find the tensions in both wires.

Let the weight act a distance L/3 from A and 2L/3 from B.

We have T_A + T_B = 1740 N. Sum of forces = 0.

Then pick one of the ends and use the fact that the sum of the moments must also = 0.

Take the weight 1740 N at L/3 which must be equal and opposite the moment formed by T_B at L.

 

[azila]

ok, i don't understand...could you explain it in other words...please

 

[Astronuc]

In a static problem like this the sum of the forces must equal zero, so the sum of the tensions in the cables must equal the weight (force of gravity) of the beam.

If the weight of the beam was equally distributed, then the weight would act at the midpoint, and the tensions in the wire would be equal. But the weight distribution is non-uniform, so it acts at L/3 from one wire (I took A) and is 2L/3 from the other wire (B). So wire A will bear more weight (and have a greater tension) than B. Using the sum of the moments about pivot point A will allow one to solve for the tension in B.

A moment is just the product of the length of the moment arm and the normal force acting at that moment arm.