Question

In the system shown below, $M_A=3\text{kg},M_B=4\text{kg}$ and $M_C=8\text{kg}$. Co-efficient of friction between any two surfaces is 0.25. Pulley is frictionless and string is massless. If $A$ is connected to the wall through a massless rigid rod, then choose the correct option(s)

• A. Value of $F$ to keep $C$ moving with constant speed is $80\text{N}$
• B. Value of $F$ to keep $C$ moving with constant speed is $120\text{N}$
• C. If $F$ is $200\text{N}$, then acceleration of $B$ is $10\text{m}/{\text{s}}^2$
• D. To slide $C$ towards left, $F$ should be at least $50\text{N}$

Answer 1

Answer: (A), (C)

If $F$ is $200\text{N}$, then acceleration of $B$ is $10\text{m}/{\text{s}}^2$

Given, mass $M_A=3\text{kg}$, $M_B=4\text{kg}$, $M_C=8\text{kg}$
Co-efficient of friction $\mu =0.25$
Let us suppose, tension in the string is $T$ and constant speed of $C$ is $u$. As $C$ moves forward, then friction on it will act backward.


Here, $N_1=(M_A+M_B+M_C)g=150\text{N}$
$N_2=(M_A+M_B)g=70\text{N}$
$N_3=\left(M_A\right)g=30\text{N}$

FBD of $C$:


$\sum F=ma$ along horizontal
$\Rightarrow F-f_2-T-f_1=0$ [as speed is constant, $a=0$]
$\Rightarrow F=f_1+f_2+T....\left(1\right)$

FBD of $B$ :


$\sum F=ma$ along horizontal
$\Rightarrow f_3+f_2-T=0$ [as speed is constent, $a=0$]
$\Rightarrow T=f_3+f_2...\left(2\right)$

On putting equation $\left(2\right)$ in $\left(1\right)$
$F=f_1+f_2+f_3+f_2$
$F=f_1+2f_2+f_3$
$F=\mu N_1+2\mu N_2+\mu N_3$
$F=\mu [N_1+2N_2+N_3]=0.25[150+2\times 70+30]$
$F=80\text{N}$
$\therefore$ option A is correct and B is wrong.

Let us again assume that tension is string is $T$ and acceleration is $a$ forward, therefore friction on $C$ will act backward.
FBD of $C$ :


$\sum F=ma$ along horizontal
$\Rightarrow F-f_2-f_1-T=M_{C}a$
$\Rightarrow 200-f_2-f_1-T=8a...\left(1\right)$
[$F=200\text{N}$ given]

FBD of $B$:


$\sum F=ma$ along horizontal
$\Rightarrow T-f_3-f_2=M_{B}a$
$\Rightarrow T-f_3-f_2=4a----\left(2\right)$
[Due to string constraint, acceleration of blocks C and B are same]

By adding equation $\left(1\right)$ and $\left(2\right)$
$200-2f_2-f_1-f_3=12a$
$\Rightarrow 200-[f_1+2f_2+f_3]=12a$
$\Rightarrow 200-\mu [N_1+2N_2+N_3]=12a$
$\Rightarrow 200-0.25[150+2\times 70+30]=12a$
$\Rightarrow a=10\text{m}/{\text{s}}^2$
$\therefore$ option C is correct.

Maximum frictional Force on $C$ is $f_1+f_2=\mu N_1+\mu N_2=0.25[150+70]=55\text{N}$ and total force towards right is $(55+T)\text{N}$.
So minimum $F$ required is $(55+T)\text{N}$.
Option D is wrong.