Question

If the system is initially at rest. Find the acceleration of $2kg$ and $4kg$ masses, where it is given that $2kg$ mass does not slide on $4kg$ mass for the figure shown. The external force $F$ acting on the mass $2kg$ at an angle $\theta ={37}^{\circ }$. (Take $g=10m/s^2$)

• A. $a_A=a_B=\frac{4}{3}m/s^2$
• B. $a_A=a_B=\frac{8}{3}m/s^2$
• C. $a_A=a_B=\frac{3}{4}m/s^2$
• D. $a_A=a_B=\frac{10}{3}m/s^2$

Answer 1

The correct option is B $a_A=a_B=\frac{8}{3}m/s^2$

The FBDs of the masses are as shown


from the FBD we have
$N_1=F\sin {37}^{\circ }+2g=(\frac{3F}{5}+20)\text{N}$
As per the given condition i.e. mass $2kg$ will not slide on the mass $4kg$, we have
Limiting friction $f_{\text{max}}=\mu N_1=0.5(\frac{3F}{5}+20)\text{N}$

Also, $\text{Horizontal applied force}\le f_{\text{max}}$
$\Rightarrow F\cos {37}^{\circ }\le 0.5(\frac{3F}{5}+20)$
$\Rightarrow \frac{4F}{5}\le 0.5(\frac{3F}{5}+20)$
$\Rightarrow (\frac{4F}{5}-\frac{3F}{10})\le 10$
$\Rightarrow F\le 20\text{N}$
Thus, the maximum value of applied external force can be $F_{\text{max}}=20\text{N}$

$\therefore$ common acceleration of the masses is
$a_c=\frac{F\cos {37}^{\circ }}{m_A+m_B}=\frac{20\times 0.8}{6}=\frac{16}{6}=\frac{8}{3}m/s^2$