Discussion
A bar of uniform cross section and weighing 100 N is held horizontally using two massless and inextensible strings S1 and S2 as shown in the figure.
The tensions in the strings are
Answer: B
The tensions are T1 = 0 N and T2 = 100 N.
The bar is uniform, so its 100 N weight acts at its midpoint, and string S2 is attached exactly there. Take moments about the midpoint: T2 has no moment arm, the weight has none either, and T1 acts L/2 away. For rotational equilibrium T1 × L/2 = 0, so T1 = 0, and vertical balance then gives T2 = 100 N.
Splitting the weight 75/25 or 25/75 would be right only if the strings were placed elsewhere, and T1 = 100 N with T2 = 0 would leave the weight with a moment about S1, so the bar would rotate.
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