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**1. Homework Statement**

**2. Homework Equations**

F = ma

**3. The Attempt at a Solution**

What we want is for m2 to be at rest

**relative to m1**because this implies m2 will not be sliding across the surface of m2 and this immediately implies m3 must be at rest in the vertical direction (in the lab frame and frame of m1 because there will be no pseudo forces in the vertical direction on m3) because if m3 was not at rest vertically then m2 would be slipping along m1 while m1 was moving which is a contradiction if we assume m2 is at rest with respect to m1 while m1 is moving. Boosting to a frame co - moving with m1 we have that in this frame, [itex]m_{2}a_{2} = F_{apparent} = T - m_{2}a = 0[/itex] where [itex]a = \frac{F}{m_{1} + m_{2} + m_{3}}[/itex] is the acceleration of the entire apparatus (and consequently m2 since they all move together under F). As noted above, this immediately implies [itex]m_{3}a_{3} = T - m_{3}g = 0[/itex] so combining this together we have that [itex]m_{3}g = m_{2}a[/itex] so [itex]F = (m_{1} + m_{2} + m_{3})(\frac{m3}{m2})g[/itex]. If I take m1 = m2 = m3 = m then F = 3mg as stated in the ans. clue. I also checked the limiting cases m3 = 0 which gives F = 0 as it should because if there is no m3 then m2 will be at rest even in the lab frame so we don't need any force on m1. If m2 = 0 then F = infinity which makes sense since if m2 = 0, m3 will go into free fall and the only way to stop it from going into free fall using a

**horizontal**force would be an infinite one that accelerates it in the horizontal direction so fast that it doesn't get a chance to fall (this is unphysical of course). Is the solution right? Thanks!

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