Question

Find the maximum value of $F$ such that the system is moving together. Take $g=10{\text{m/s}}^2$.

• A. $200\text{N}$
• B. $150\text{N}$
• C. $100\text{N}$
• D. $50\text{N}$

Answer 1

The correct option is B $150\text{N}$

$m_1=20\text{kg};m_2=10\text{kg}$
Taking system as both blocks, we know that by the formula,
$F=(m_1+m_2)({\mu }_{1}g+a_s)=(m_1+m_2)a_s$ and we know that $a_s={\mu }_{2}g$
Since $({\mu }_1=0)$
$F_{\text{max}}=(m_1+m_2){\mu }_{2}g=(10+20)\left(0.5\right)10=150\text{N}$

Alternately:
At just sliding condition, limiting friction $f=0.5\times 10\times 10=50N$ is acting.

$F-50=20a\left(1\right)$
$f=10a\left(2\right)$
$50=10a$
$\therefore a=5{\text{m/s}}^2$
Hence, $F=50+20\times 5=150\text{N}$
$\therefore F_{\text{max}}=150\text{N}$.