Answer
(a) There are three forces acting on block $m_A$. These forces are the force of gravity (mg), the normal force from the table, and the tension force from the cord.
There are two forces acting on block $m_B$. These forces are the force of gravity (mg) and the tension force from the cord.
(b) $a = \frac{(m_B)(g)}{m_A + m_B}$
$F_T = \frac{(m_A)(m_B)(g)}{m_A + m_B}$
Work Step by Step
(a) There are three forces acting on block $m_A$. These forces are the force of gravity (mg), the normal force from the table, and the tension force from the cord.
There are two forces acting on block $m_B$. These forces are the force of gravity (mg) and the tension force from the cord.
(b) The force equation for block $m_A$:
$\sum F = (m_A) a$
$F_T = (m_A) a$
The force equation for block $m_B$:
$\sum F = (m_B) a$
$(m_B)(g) - F_T = (m_B) a$
$(m_B)(g) - m_A a = (m_B) a$
$a = \frac{(m_B)(g)}{m_A + m_B}$
We can replace a in the force equation for $m_A$:
$F_T = \frac{(m_A)(m_B)(g)}{m_A + m_B}$
