Question

an
le
in
360
59. Three blocks with masses m, 2m and 3m are
connected by strings, as shown in the figure.
After an upward force F is applied on block
m, the masses move upward at constant speed
v. What is the net force on the block of mass
2m? (g is the acceleration due to gravity)
F
(2013 NEET)
9
m
2m
Зm​

Answer 1

Let $\sf{T_1}$ be the tension of the string connecting blocks of masses m and 2m, and $\sf{T_2}$ be that of the string connecting blocks of masses 2m and 3m.

The free body diagram of each block of masses m, 2m and 3m are given below.

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\put(0,0){\framebox(5,5){\sf{m}}}\put(2.5,0){\vector(0,-1){10}}\put(2.5,5){\vector(0,1){10}}\put(-3.5,-14){ \sf{T_1+mg} }\put(2,17){\sf{F}}\end{picture}$           $\setlength{\unitlength}{1mm}\begin{picture}(5,5)\put(0,0){\framebox(6,6){\sf{2m}}}\put(3,0){\vector(0,-1){10}}\put(3,6){\vector(0,1){10}}\put(-3,-14){ \sf{T_2+2mg} }\put(2,18){ \sf{T_1} }\end{picture}$             $\setlength{\unitlength}{1mm}\begin{picture}(5,5)\put(0,0){\framebox(6,6){\sf{3m}}}\put(3,0){\vector(0,-1){10}}\put(3,6){\vector(0,1){10}}\put(0,-14){ \sf{3mg} }\put(1,18){ \sf{T_2} }\end{picture}$

And the free body diagram of the whole system is,

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\put(0,0){\framebox(6,6){\sf{6m}}}\put(3,0){\vector(0,-1){10}}\put(3,6){\vector(0,1){10}}\put(0,-14){ \sf{6mg} }\put(2,18){ \sf{F} }\end{picture}$

Since the system moves up with constant velocity, the system has no net acceleration, nor net force.

$\longrightarrow\sf{F-6mg=0}$

$\longrightarrow\sf{F=6mg}$

The net force acting on the block of mass 2m is (let upward),

$\longrightarrow\sf{F_{net}=T_1-T_2-2mg\quad\quad\dots(1)}$

But from the free body diagram of the block of mass m,

$\longrightarrow\sf{T_1=F-mg}$

$\longrightarrow\sf{T_1=6mg-mg}$

$\longrightarrow\sf{T_1=5mg}$

And from the free body diagram of the block of mass 3m,

$\longrightarrow\sf{T_2=3mg}$

Then (1) becomes,

$\longrightarrow\sf{F_{net}=5mg-3mg-2mg}$

$\longrightarrow\sf{\underline{\underline{F_{net}=0}}}$

Therefore the net force acting on the block of mass 2m is zero.

Answer 2

Answer:

Net Force, F = 0

Explanation:

[Refer to the above user's solution if you love toiling]

Given;-

           Velocity = Constant

∴ Acceleration = 0

We know that, F = ma

∴                     F = 2m × 0

                        F = 0

[Note; In order to understand my solution, you need a very competitive approach towards the problem. You can only understand this if you have all concepts clear regarding the case]

Hope it helps! ;-))