The problem is as follows:
Find the tension on wire labeled (1) as shown in the figure.
The alternatives given in my book are as follows:
$\begin{array}{ll} 1.&\frac{1}{2}mg\\ 2.&mg\\ 3.&\frac{3}{2}mg\\ 4.&2mg\\ 4.&\frac{5}{2}mg\\ \end{array}$
In this particular problem I'm lost at how should I proceed with the vectors. Can somebody help me on how is the interaction between the weight on the sphere which is labeled as 2m and the block which weight is m?.
Typically I would try to show some effort but here I'm stuck at the very beginning because I'm confused at how should I put the weight?.
Assuming the masses of the wires and the friction in the pulleys are negligible, the tension in the upper left wire ($\ T\ $, say) is the same along its whole length, as is the tension in the lower left wire. The only horizontal forces on the mass of $\ 2m\ $ at the bottom are the horizontal component, $\ T_h=T\cos\theta\ $, of the tension in the upper left wire, and the tension in the lower right wire, which must be in balance, telling you that the latter must be $\ T_h\ $.
The only forces on the mass of $\ m\ $ on the left are the tension, $\ T\ $, in the upper left wire, acting vertically upwards, the tension $\ T_h\ $ in the lower right wire, acting vertically downwards, and the force, $\ mg\ $, of gravity, also acting vertically downwards. Since these forces are in balance, what does this tell you about the relation between $\ T_h=T\cos\theta\ $ and $\ T\ $?
The only vertical forces acting on the mass of $\ 2m\ $ at the bottom are the vertical component $\ T_v=T\sin\theta\ $ of the tension in the upper left wire, acting upwards, and the force, $\ 2mg\ $, of gravity, acting downwards. Since these forces are in balance, what does that tell you about the value of $\ T_v\ $?
From these observations you should be able to get equations for $\ T\sin\theta\ $ and $\ T\cos\theta\ $ which you can square and add to get an equation for $\ T^2\ $ as polynomial in $\ T\ $ itself and multiples of $\ mg\ $. The quadratic terms in $\ T\ $ in this equation should cancel out, leaving you with a simple linear equation which you can solve for $\ T\ $. Remembering that it's $\ T_h\ $ that you're actually asked to find, you now have to back substitute for $\ T\ $ into your equation for the former quantity to find its value.