Considering the \(y\) component of Newton’s Second Law for box 2 (the top box), we can find the value of the normal force exerted by box 1: \[\begin{aligned} \sum F_y &= N_2 - F_{2g} = 0\\ \therefore ...Considering the \(y\) component of Newton’s Second Law for box 2 (the top box), we can find the value of the normal force exerted by box 1: \[\begin{aligned} \sum F_y &= N_2 - F_{2g} = 0\\ \therefore N_2 &= m_2 g\end{aligned}\] The maximal magnitude of the force of static friction, \(f_{2s}\), between the two boxes is given by: \[\begin{aligned} f_{2s} = \mu_sN_2 = \mu_s m_2g\end{aligned}\] This is the maximal magnitude of the force that can accelerate box 1.
Considering the \(y\) component of Newton’s Second Law for box 2 (the top box), we can find the value of the normal force exerted by box 1: \[\begin{aligned} \sum F_y &= N_2 - F_{2g} = 0\\ \therefore ...Considering the \(y\) component of Newton’s Second Law for box 2 (the top box), we can find the value of the normal force exerted by box 1: \[\begin{aligned} \sum F_y &= N_2 - F_{2g} = 0\\ \therefore N_2 &= m_2 g\end{aligned}\] The maximal magnitude of the force of static friction, \(f_{2s}\), between the two boxes is given by: \[\begin{aligned} f_{2s} = \mu_sN_2 = \mu_s m_2g\end{aligned}\] This is the maximal magnitude of the force that can accelerate box 1.
