Question

A system of 3 blocks of masses of $10kg,20kg,30kg$ respectively, having flat surfaces of contact with coefficients of friction given in the figure are at rest initially. If a constant horizontal force of $F=100N$ is acting on $10kg$ block as shown in figure, then the magnitude of acceleration of the $20kg$ block with respect to the ground is (Take $g=10m/s^2)$

• A. $1m/s^2$
• B. $2m/s^2$
• C. $3m/s^2$
• D. $5m/s^2$

Answer 1

The correct option is A $1m/s^2$

Let as assume the blocks are moving together as one system with common acceleration $a_c$.

FBD of $10kg$ block:


From figure (i), we get
$R_1=10g=100N$
$100-f_1=10a_c-\left(1\right)$

FBD of $20kg+30kg$:


From figure (ii), we get
$R_1^{\prime }=R_1+50g$ and
$f_1=50a_c-\left(2\right)$

Solving (1) & (2),
$100-50a_c=10a_c$
$a_c=\frac{100}{60}=1.66m/s^2$

$\Rightarrow f_1=50\times 1.66=83.33N$
But, $f_{1max}={\mu }_1{R}_1=0.5\times 100=50N$

Since required $f_1>(f_1{)}_{max},10kg$ block slips on the system of the other two blocks and the kinetic friction is only $50N$.

Therefore, accelaration of $10kg$ block
$a_1=\frac{F-f_1}{m}=\frac{100-50}{10}=5m/s^2$.

To check if there is slip between $20kg$ and $30kg$ blocks, draw the FBDs


From (iii), $R_2=R_1+20g=300N$ and
$f_1-f_2=50-f_2=20a--\left(3\right)$


From (iv), $R_3=300+R_2=600N$ and
$f_2=30a--\left(4\right)$

Solving (3) & (4), we get
$50=50a\Rightarrow a=1m/s^2$
$\Rightarrow f_2=30a=30N$
$f_{2max}={\mu }_2{R}_2=0.25\times 300=75N$

Since, $f_2<(f_2{)}_{max}$, there is no slipping between $20kg$ & $30kg$ blocks
$\therefore$ They both move together with an acceleration $a=1m/s^2$ w.r.t ground.