Because the blocks (each of mass m) don't slide across the cut, they have a common horizontal acceleration a to the left that follows from Newton's second law: (2m) a = Fnet = 2F - F = F a = F/(2m) Now consider only the left block. On it, two horizontal forces act: the applied force F to the right and the horizontal component of the reaction force that B exerts on A, F_BA (which is normal to the cut line). Its horizontal component is F_BA sin(30) , i.e.
F_BA * 1/2. So Newton's second law applied to block A alone gives m a = F_BA / 2 - F. But we already found that a = F/(2m), so F/2 = F_BA/2 -F Therefore F_BA = 3 F Of course, Newton's third law implies that F_AB = 3 F as well.
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