Question 1201559
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Let T1 be the tension of the rope 30°;  T2 be the tension of the rope 45°.


T1 and T2 are the <U>magnitudes</U> of the forces; their dimensions are newtons.


Force F1 has horizontal component T1*cos(30°) = {{{T1*(sqrt(3)/2)}}}.

Force F2 has horizontal component T2*cos(45°) = {{{T2*(sqrt(2)/2)}}} in opposite direction.


Equilibrium condition in horizontal direction gives you this equation

    {{{T1*(sqrt(3)/2)}}}} = {{{T2*(sqrt(2)/2)}}},   or

    {{{T1*sqrt(3)}}} = {{{T2*sqrt(2)}}}.        (1)


Force F1 has vertical component T1*sin(30°) = {{{T1*(1/2)}}}.

Force F2 has vertical component T2*sin(45°) = {{{T2*(sqrt(2)/2)}}}.


Equilibrium condition in vertical direction gives you this equation

    {{{T1*(1/2)}}} + {{{T2*(sqrt(2)/2)}}}} = 20*g,   or

    {{{T1}}} + {{{T2*sqrt(2)}}} = 40g.     (2)


Thus you have a system of two equations (1) and (2),  and now your task is to solve it.


For it, replace in equation (2) the term  {{{T2*sqrt(2)}}}}  by {{{T1*sqrt(3)}}}, based on equation (1).

You will get then

    T1 + {{{T1*sqrt(3)}}} = 40g,   or

    {{{T1*(1 + sqrt(3))}}} = 40g,

    T1 = {{{(40g)/(1+sqrt(3))}}}.


From equation (1), you get then

    T2 = {{{T1*(sqrt(3)/sqrt(2))}}} = {{{((40g)/(1+sqrt(3)))*(sqrt(3)/sqrt(2))}}}.


At this point, the problem is just solved.


If you want to get the numerical values for magnitudes / (tensions)  T1  and  T2,  use your calculator.
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Solved.