A Bar Suspended by Two Vertical Strings

[laxmanic33]

Homework Statement

A rigid uniform horizontal bar of mass m1 = 85.00 kg and length L = 5.300 m is supported by two vertical massless strings. String A is attached at a distance d = 1.500 m from the left end of the bar and is connected to the top plate. String B is attached to the left end of the bar and is connected to the floor. An object of mass m2 = 2000 kg is supported by the crane at a distance x = 5.100 m from the left end of the bar.

Throughout this problem, positive torque is counterclockwise. Use 9.81 m/s^2 for the magnitude of the acceleration due to gravity.

(1) Find T(A) the tension in string A.
(2) Find T(B) the magnitude of the tension in string B.

Homework Equations

Ta = (Torque due to weight of bar + Torque due to weight of object) / (l/2)
Tb = Ta*d - m1g(L/2) - m2gx

The Attempt at a Solution

I got Ta = 68111.63, It says that I am close, rounding error, and it has to be in 4 sig figs.

I got Tb = 34056.45, and no dice with this one.

 

[Doc Al]

Ta = (Torque due to weight of bar + Torque due to weight of object) / (l/2)
Tb = Ta*d - m1g(L/2) - m2gx

I don't understand how you arrived at these equations.

To find Ta, set up a torque equation about the point where string B connects to the bar; To find Tb, set up a torque equation about the point where string A connects to the bar.

 

[TVP45]

Are you using T for both torque and tension? Some units would be a big help to you.