106Acceleration of the rope = (Mg - Mg/2)/M = g/2
At the midpoint the forces acting are T towards right (say) and friction = 1/2*(M/2*g) = Mg/4
So, T-Mg/4 = (M/2*g/2)
T = Mg/2
A rope of length L and mass M is being pulled on a rough horizontal floor by a constant horizantal force F=Mg.The force is acting at one nd of the rope in the same direction as the length of the rope .Te coefficient of kinetic friction between rope and floor is 1/2.Then,the tension at the mid-pt of the rope is
(a)Mg/4
(b)2Mg/5
(c)Mg/8
(d)Mg/2
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