6. (10 points) Two weights with masses \( m_{1}=1 \) kg and \( m_{2}=10 \) kg are placed on two horizontal shelves, connected by a weightless, inextensible string that passes over a weightless, frictionless pulley. The coefficients of friction between the weights and the shelves are \( \mu_{1}=0.3 \) and \( \mu_{2}=0.1 \). A horizontally directed force \( F=20 \) N is applied to the center of the pulley. Determine the accelerations of the weights and the center of mass of this system of bodies. The acceleration due to gravity \( g=10 \) m/s\(^2\).

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Answer: $a_{1}=7 \mathrm{M} / \mathbf{c}^{2}, a_{2}=0 \mathrm{M} / \mathrm{c}^{2}, a=0.64 \mathrm{M} / \mathrm{c}^{2}$.

Solution. Since the pulley is weightless, the tension in the string is: $T=\frac{F}{2}=10 \mathrm{H}$.

The sliding friction force for the first body:

$F_{\text {fr 1 }}=\mu_{1} N_{1}=\mu_{1} m_{1} g=0.3 \cdot 1 \cdot 10=3 \mathrm{H}$.

The sliding friction force for the second body:

$F_{\text {fr } 2}=\mu_{2} N_{2}=\mu_{2} m_{2} g=0.1 \cdot 10 \cdot 10=10 \mathrm{H}$.

Therefore, the second body remains stationary: $a_{2}=0 \mathrm{~m} / \mathrm{c}^{2}$.

Newton's second law for the first body: $T-F_{\text {fr } 1}=m_{1} a_{1}$,

the acceleration of the first body: $a_{1}=\frac{T-F_{\text {fr } 1}}{m_{1}}=\frac{10-3}{1}=7 \mathrm{M} / \mathrm{c}^{2}$.

The acceleration of the center of mass $a=\frac{m_{1} a_{1}+m_{2} a_{2}}{m_{1}+m_{2}}=\frac{1 \cdot 7+0}{1+10}=0.64 \mathrm{M} / \mathrm{c}^{2}$.