Question

problem - solving activity
forces (friction): two boxes ($m_1 = 2$ kg, $m_2 = 4$ kg) are stacked as
shown in the figure below. the bottom block is on a frictionless
surface. the coefficient of static friction between the blocks is 0.35.
what is the maximum horizontal force that can be applied to the
lower block without the upper block slipping?

Explanation:

Step1: Find the maximum acceleration of the upper block

The maximum static - friction force on the upper block \(m_1\) is \(f_{s,\text{max}}=\mu_sm_1g\). According to Newton's second law \(F = ma\), for the upper block \(f_{s,\text{max}}=m_1a_{\text{max}}\). So \(a_{\text{max}}=\mu_sg\). Given \(\mu_s = 0.35\) and \(g = 9.8\ m/s^{2}\), then \(a_{\text{max}}=0.35\times9.8\ m/s^{2}=3.43\ m/s^{2}\)

Step2: Consider the two - block system

Treat the two blocks (\(m_1 + m_2\)) as a single system. According to Newton's second law \(F=(m_1 + m_2)a_{\text{max}}\). Given \(m_1 = 2\ kg\) and \(m_2=4\ kg\), then \(m_1 + m_2=(2 + 4)\ kg=6\ kg\)

Substitute \(a_{\text{max}} = 3.43\ m/s^{2}\) and \(m_1 + m_2 = 6\ kg\) into \(F=(m_1 + m_2)a_{\text{max}}\)

\(F=6\times3.43\ N = 20.58\ N\)

Answer:

\(20.58\ N\)