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Two blocks connected by a rope of negligible mass are being dragged by a force at a 25° angle above horizontal (see figure below). Suppose F = 105.0 N, m1 = 10.0 kg, m2 = 22.0 kg, and the coefficient of kinetic friction between each block and the surface is 0.150. Find the acceleration of the system and the tension in the rope
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First treat the two-block system as one system Horizontally Fcos(theta) - (mu)(m1+m2)(N)=(m1+m2)a and vertically (m1+m2)g = N + Fsin(theta)
eliminate N and find a
get it?
For Body 1: \[Fcos \Theta-\mu(mg - Fsin \Theta)-T = ma\] For Body 2 : \[T - \mu mg = ma \]
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Just Substitute The Given Info..And u have it..)
@stroodles77 Did u understand..)
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