Đáp án:
$\begin{array}{l}
a = 0,832m/{s^2}\\
T = 1,832N
\end{array}$
Giải thích các bước giải:
Gia tốc và lực căng dây là:
$\begin{array}{l}
F = P\sin \alpha + {F_{ms}} \Leftrightarrow F = P\sin \alpha + P\cos \alpha \mu \\
\Leftrightarrow 600 = 1000.\sin {60^o} + 1000.\cos {60^o}\mu \\
\left\{ \begin{array}{l}
{N_2} = {P_2}\\
T - {F_{m{s_2}}} = {m_2}a
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{N_2} = {m_2}g\\
T - {N_2}k = {m_2}a
\end{array} \right.\\
\Leftrightarrow T - {m_2}gk = {m_2}a \Rightarrow T = {m_2}\left( {a + gk} \right)\left( 1 \right)\\
\left\{ \begin{array}{l}
F\sin \alpha + {N_1} = {P_1}\\
F\cos \alpha - T - {F_{m{s_1}}} = {m_1}{a_1}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{N_1} = {m_1}g - F\sin \alpha \\
F\cos \alpha - T - {N_1}k = {m_1}{a_1}
\end{array} \right.\\
\Leftrightarrow F\cos \alpha - T - \left( {{m_1}g - F\sin \alpha } \right)k = {m_1}a\\
\Rightarrow T = F\cos \alpha - {m_1}a - {m_1}gk + F\sin \alpha \\
\Leftrightarrow T = F\left( {\cos \alpha + \sin \alpha k} \right) - {m_1}\left( {a + gk} \right)\left( 2 \right)\\
\left( 1 \right),\left( 2 \right) \Rightarrow F\left( {\cos \alpha + \sin \alpha k} \right) - {m_1}\left( {a + gk} \right) = {m_2}\left( {a + gk} \right)\\
\Leftrightarrow \left( {a + gk} \right)\left( {{m_1} + {m_2}} \right) = F\left( {\cos \alpha + \sin \alpha k} \right)\\
\Leftrightarrow a = \dfrac{{F\left( {\cos \alpha + \sin \alpha k} \right)}}{{{m_1} + {m_2}}} - gk\\
\Leftrightarrow a = \dfrac{{6.\left( {\cos {{30}^o} + \sin {{30}^o}.0,1} \right)}}{{1 + 2}} - 10.0,1 = 0,832m/{s^2}\\
\Rightarrow T = {m_2}\left( {a + gk} \right) = 1.\left( {0,832 + 10.0,1} \right) = 1,832N
\end{array}$
