Question

can anyone help me with this A non-uniform rod of length 1.00 m is hung horizontally, supported by strings on both ends. If the center of gravity of the rod is 0.200 m from the left end, what is the magnitude of the tension applied by the string on the right end of the rod? 50% of the weight of the rod Cannot be determined unless the weight of the rod is given 80% of the weight of the rod 20% of the weight of the rod

Answer 1

The tension in the right string is 20 % the weight of the rod

(the vertical component of the tension)

.2 Fl = .8 Fr balancing torques

Fl / Fr = 4

Fl + Fr = W balancing vertical forces

4 Fr + Fr = W or Fr = W / 5

One can go farther and use H as the height of suspension above the rod

tan θ = .2 / H and tan φ = .8 / H

tan Φ / tan θ = 4

Use this result to find actual tension in string and one will find

Tl / Tr = 4 where T represents actual tension in string