Answer:
[tex]( \nwarrow^{ + }) \sum \: F_{{y_{A}}_{2kg}} = F_{{y_{N_A}}_{2kg}} - m_{{y_{A}}_{2kg}}g \cos( \theta_{{y_{A}}_{2kg}}) = 0 \\ F_{{y_{N_A}}_{2kg}} = m_{{y_{A}}_{2kg}}g \cos( \theta_{{y_{A}}_{2kg}})...(1) \\ ( \nearrow^{ + }) \sum \: F_{{x_{A}}_{2kg}} = F_{{x_{T}}} - m_{{y_{A}}_{2kg}}g \sin( \theta_{{y_{A}}_{2kg}}) -F_{{Fr}_{A2kg}}= m_{{y_{A}}_{2kg}}a...(2) \\ \\ ( \nearrow^{ + }) \sum \: F_{{y_{B}}_{5kg}} = F_{{y_{N_B}}_{2kg}} - m_{{y_{B}}_{5kg}}g \cos( \theta_{{y_{B}}_{5kg}}) = 0 \\ F_{{y_{N_B}}_{5kg}} = m_{{y_{B}}_{5kg}}g \cos( \theta_{{y_{B}}_{5kg}})...(3) \\ ( \searrow^{ + }) \sum \: F_{{x_{B}}_{5kg}} = m_{{y_{B}}_{5kg}}g \sin( \theta_{{y_{B}}_{5kg}})-F_{{x_{T}}} -F_{{Fr}_{B5kg}}= m_{{y_{B}}_{5kg}}a...(4)[/tex]
Explanation:
All you need to do is plug in the values and solve the simultaneous equations to find the acceleration a, and T.
Always draw the free body diagram...
