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And, we can test this claim! This says that we have a horizontal force from the pulley on the large block of ##T##. (And, likewise, a vertical component of ##T## downwards.) This would give the horizontal acceleration of the large block and side block of:jbriggs444 said:The interaction with the pulley is more interesting. The cord exerts a uniform radial force along a 90 degree arc of the ideal pulley's surface. The net of this is a force whose magnitude is ##\sqrt{2}## times tension at an angle 45 degrees below the horizontal.

$$\ddot X = -\frac{T}{M+m} =-\frac{mg}{2M + 3m}$$As expected.