A uniform rod of length 4L and mass M is suspended from a horizontal
root by two light strings of length L and 2L as shown. Then the tension
and 2L as shown. Then the tension
in the left string of length Lis
LLLLLLLLL
(1) 2
(1) Mg
2L
(3) Mg
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Length of the rod = 4L (Given)
Mass of the road = M (Given)
Length from horizontal = L and 2L (Given)
The free body diagram of the rod so formed will be the Net torque about the centre of mass being zero
Thus, according to the formula -
T1 × L/2 cosФ = T2 × L/2 cosФ
T1 = T2 = Mg/2
Therefore, the tension in the left string of length L is Mg/2
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