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A 50 N boy hangs from it, as shown inFig. 2 (c ) .35 (a). Find the senstion in the two parts of the rope. |
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Answer» Solution :Resolving tension `T_(1) and T_(2)` into tworectangular components we have `T_(1) cos 15^(0)` acts horizontally and `T_(1) sin 30^(0)` vertically upwards `T_(1) cos 15^(0)` acts horizontally and `T_(2) sin 30^(0)` vertically upwards. As the system is in EQUILIBRIUM, so `T_91) cos `30^(0) =T_(2) cos 15^(0) or T_(1) XX 0.8660 =T_(20 xx 0.9659` or `T_(1) =(0.9659)/(0.8660) T_(2) =1.11T_(2)` and `T_(1) sin 30^(0) _T_(2) sin 15^(0) =W or 1.11 T_(2) xx 1/2 + T_(20 xx 0.2588 =50` or `0.55 T_(2) +0. 25 88 T_92) =50` or `0.8 T_(2) =50` or `T_(2) =50//0 .8 =62 .50 N` `T_(1) =1.11 xx 62 .50 =69 .37 N`  . Alternative solution `In Fig. 2(c ). 35 (b0 , `alpha= 90^(0) +15 ^(0) =105^(0)` `beta =90^(0) +30^(0) =120^(0)` and `gamma=180^(0)- (30^(0) +15^(0)) =135^(0)`. Using Lami`s Theorem, we have T_(1)/(sin alpha) T_(2)/(sinbrta) =W/(sin gamma)` :. T_(1) =W xx (sin alpha)/(sin gamma) =50 xx (sin 105^(0))/(sin 135^(0)) =50 xx (sin 75^(0))/(sin 45^(0)) =(50 xx 0. 9659)/(0.7071) =68.3 N` T_(2) =(W sin beta )/(sin gamma) =(50 sin 120^(0))/(sin 135^(0)( =(50 xx sin 60^(0))/(sin 45^(0)) =(50 xx 8660)/(0.7071) =61.24 N`. 
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