Answer to Question #181148 in Classical Mechanics for sune

Question

Two boxes of fruits moving with constant velocity on a rough surface are connected by a light string, as shown in the figure, with m1 = 20.0 kg and m2 = 30.0 kg. A force of 70.0 N is applied to the 30.0-kg box at an angle of  = 30 with the horizontal. The coefficient of kinetic friction between each box and the surface is the same. Calculate the coefficient of kinetic friction between each box and the surface, and the magnitude of the tension in the string.

Answer 1

$\begin{cases} \vec{T}+\vec{F}+ \vec{F_{fr2}}+m_2\vec{g}+\vec{N_2}=\vec{0}\\ \vec{T}+\vec{F_{fr1}}+m_1\vec{g}+\vec{N_1}=\vec{0} \end{cases}$

$\begin{cases} -T+Fcos\alpha-\mu N_2=0 \\ -m_2g+Fsin\alpha+N_2=0\\ T-\mu N_1=0\\ -m_1g+N_1=0 \end{cases} \implies$

$\mu=\frac{Fcos\alpha}{m_1g+m_2g-Fsin\alpha}=0.13,$

$T=\mu m_1g=26~N.$