Answer
$F_3= 146.6N$
Work Step by Step
Total mass ( $m_t$ ) = $m_1 +m_2 + m_3$
$m_t = 30kg+10kg + 20kg$
$m_t=60kg$
Finding the acceleration from the whole system:
$F_{net} = m_ta$
$F -μ_k F_N= m_ta$
$F -μ_k (m_tg)= m_ta$
$a = \frac{F -μ_k m_tg}{m_t}$
$a = \frac{440N -(0.7 \times 60kg \times 9.8m/s^2)}{60kg}$
$a = 0.47m/s^2$
Finding the force applied on crate 3 from crate 2:
$F_3 -μ_k (m_3g)= m_3a$
$F_3= m_3a + μ_k m_3g$
$F_3= m_3(a + μ_k g)$
$F_3= 20kg (0.47m/s^2 +(0.7 \times 9.8m/s^2))$
$F_3= 146.6N$
