Which Acceleration Formula is Correct for the Larger Block in the Pulley System?

AI Thread Summary
The discussion focuses on determining the correct acceleration formula for a larger block in a pulley system, with two proposed formulas: 4mg/(M+17m) and 4mg/(M+16m). The user derived the acceleration as a = 4mg/(M + 16m) using free body equations for both the larger and smaller blocks. Despite this calculation, the textbook provides a different answer. The consensus among participants supports the user's derived formula as correct, indicating a potential error in the textbook. The conversation emphasizes the importance of showing work in problem-solving for clarity and accuracy.
dk_ch
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the acceleration of the larger block of mass M on frictionless floor of the pully system as in (fig given bmp attachment) will be
(i) 4mg/(M+17m) (ii) 4mg/(M+16m)

which answer is correct and how?
 

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Welcome to PF. Note that you must show some work before we can help you in the homework forums: i.e. fill in parts 2 and 3 on the homework template!
 


I tried to solve the problem using following two equations taking large and small block
respectively as freebodies

Ma = 4T ------(1)
mg-T = 4ma -------(2)

solving these two i get acceleration of the larger block a = 4mg/(M + 16m)

the answer given in the text is the other one. please provide me the right answer.
 


I agree with your answer.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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