Question

In the diagram shown, the blocks $\text{A, B}$ and $\text{C}$ weight $3\text{kg}$, $4\text{kg}$ and $5\text{kg}$ respectively. The coefficient of sliding friction between any two surfaces is $0.25$. $\text{A}$ is held at rest by a massless rigid rod fixed to the wall while $\text{B}$ and $\text{C}$ are connected by a light flexible cord passing round a frictionless pulley. The force $F$ necessary to drag $\text{C}$ along the horizontal surface to the left at constant speed is (Assume that the arrangement shown in the diagram, $\text{B}$ on $\text{C}$ and $\text{A}$ on $\text{B}$ is maintained all through and $g=10{\text{m/s}}^2$)

Answer 1

Given,
mass of block $\text{A}$, $M_A=3\text{kg}$,
mass of block $\text{B}$, $M_B=4\text{kg}$
mass of block $\text{C}$, $M_C=5\text{kg}$
$\mu =0.25$ [ any two surfaces ]

When force $F$ is applied on $\text{C}$, the block $\text{C}$ will move towards left.
The free body diagram for $\text{C}$:

As $\text{C}$ is moving with constant speed
$F=f_1+f_2+T.....\left(1\right)$
Where $f_1$ is friction between floor and block $\text{C}$ and $f_2$ between blocks $\text{B}$ and $\text{C}$.

FBD for mass $\text{B}$:


As $\text{B}$ moves with constant speed
$T=f_2+f_3.....\left(2\right)$

From equation $\left(1\right)$ and $\left(2\right)$, we get
$F=f_1+f_2+f_2+f_3$
$\Rightarrow F=f_1+2f_2+f_3$

Substituting the values of $f_1$, $f_2$ and $f_3$, we get

$\Rightarrow F=f_1+2f_2+f_3$

$\Rightarrow F=\mu (M_A+M_B+M_C)g+\mu (M_A+M_B)g+\mu \left(M_{A}g\right)$

Substituting the values, we get
$\Rightarrow F=0.25\times (3+4+5)10+0.25\times 2(3+4)10+0.25\times \left(3\right)\times 10$

$\therefore F=72.5\text{N}$

Accepted answers : $72.5,72.50$