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A uniform rod of mass m and length L is suspended horizontally by two vertical ideal strings attached to its two ends. If T1 is the tension in the left string and T2 is the tension in the right string, which of the following is correct when the system is in static equilibrium?
- T1 + T2 = mg and T1 = T2 = mg/2
- T1 > mg/2 and T2 < mg/2
- T1 = mg and T2 = 0
- T1 + T2 = 2mg
Correct answer: T1 + T2 = mg and T1 = T2 = mg/2
Solution
For a horizontal uniform rod of mass m in equilibrium under two vertical strings at its ends: (1) Force balance gives T1 + T2 = mg. (2) Torque balance about the centre gives T1*(L/2) = T2*(L/2), so T1 = T2. Therefore T1 = T2 = mg/2. Both tensions equal half the weight.
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