Question

Solve this question please

Answer 1

`( \\warrow^( + )) \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) \\ ( \\earrow^( + )) \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) \\ \\ ( \\earrow^( + )) \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)`

Step-by-step 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...