Question

What minimum value of M is required to move block A with constant velocity?

Answer 1

This scenario is clearly impossible. I'll prove it to you. Either the problem is over specified or some crucial information is missing. Some obvious missing info is-nature of pulley and strings-are they massless and frictionless? But I'll make this assumption anyway.

Also, where is this friction:between table and the blocks or only between B and C or both? I'll take the latter case.Now, apply laws of motion to each element in the system one by one.

A

Because A is to be moved and that to with constant speed, kinetic friction comes into play

$T_5=7\mu_k Mg$

Pulley 1

$T_3=T_4=\frac{T_5}{2}=\frac{7}{2}\mu_k Mg$

B

Thus, block B is also moving with constant speed, then

$T_2=T_3+F_{BC}$

where

$F_{BC}=\mu_kMg$

is the friction acting between B and C.

$T_2=\frac{9}{2}\mu_kMg\ldots(\textup{I})$

C

Maximum friction available for C is

$F_c(\textup{max})=4\mu_s Mg$

Clearly,

$F_C(\textup{max})>T_4-F_{BC}=2.5\mu_kMg$

Thus which means C can't move.

Pulley 2

$T_1=T_2\ldots\textup{II}$

D

D's motion is directly dependent on B. Since the latter is unaccelerated so is D.

$T_1=8Mg\ldots\textup{III}$

Equations I,II and III give

$\frac{9}{2}\mu_kMg=8Mg \Rightarrow ??$

M gets eliminated from the problem instead we get to obtain the value of friction coefficient!!!