The force is just large enough to make it on the verge of sliding out from between the upper block and table. Determine the force F at this instant and common acceleration of each block
$f_s$ max for the lower block and ground is $100N$ while between the two blocks is $15N$ So the total force required to pull it out is $115N$. But the answer given is $140N$.
What am I doing wrong?
The appropriate physics tag has been used to indicate it’s a question for the math part of physics$
Apply the Newton's law to the upper and lower blocks, respectively,
$$ \mu_1 m_1 g = m_1a$$ $$ F-\mu_1m_1 g -\mu_2(m_1 +m_2)g= m_2a $$
Solve for for the common acceleration $a=\mu_1g$ and the force,
$$ F=(\mu_1+\mu_2)(m_1 +m_2)g $$
Plug in the givens, $\mu_1=0.3$, $\mu_2=0.4$, $m_1=5kg$, $m_2=15kg$ and $g=10m/s^2$ to obtain $F=140N$.