Question

A nonuniform horizontal bar of mass 15 kg is supported by two massless wires against gravity. The left wire makes an angle 35° with the horizontal, and the right wire makes an angle 60°. The bar has a length of 8 m. The bar is in equilibrium. Find: (a) the tension of the left wire (in N); (b) the tension of the right wire (in N); (c) the position of the center of mass of the bar in m), measured from the bar's right end.

Answer 1

you Tesinde Tisinpa Sind Ti sinh T at Ti Coso, (T2 02 Tester Tisinde ars Tilesd, ² Tz les pe za bol. Tz loso2 ing Component of force along x-directions $, 235° Component of fore cibug y- Tisind, a Ty sind = ng - the value in equation and 1 we get TCos35 a 7 Costo T. Sin 35+ Ty Sin 60 1589.8 solving both equatia wegel, T 2120.875 N T. 73.780N @ Tension in left wire 'Tis 73.780N Tension in right wire T22 120.875 N Q ny on putting b T + 2 d2 di P re K

The rotational Equiliberium of block is, Ti sind,) (2-x) = (Tz Sin 2 XL) Also from T, caso = Tz Cos 02 (Jeran equation ) Tyz To Lost, los 02. Putting the value of Tz in equation (Ti sind) (L-~) Ty los cos x Sin ca ) xu L-N = (los o sinta) (les br) (sinti n 8-U n 2 Cos 35 x Sin 60° los 60° x din 35° n = 2.30m distance of Center of mans from right and > 230m (or) 2. Just simply prette all the value in equation (11) to get the value of n' (Ti sindi) (2-2) 2 (Tz sin P2) (h) (73.780 fin 35) (8-4) (120.875 X sin bo su n 22.30ml

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