91Ó°ÊÓ

According to Newton’s second law, the value of the force applied to an object can be calculated by multiplying the object’s mass with the object’s acceleration.

The mass of block A is $m_A$.

The mass of block B is $m_B$.

(a)

The free-body diagram for each block is shown as follows:

Here, $F_{NA}$is the normal force acting on block A, $F_T$is the tension in the cord, $a$is the acceleration of the system, and $g$is the acceleration due to gravity.

As there is no motion in the vertical direction, the normal force acting on block A is given as

$\begin{array}{c}{F}_{NA}-m_{A}g=0\\ F_{NA}=m_{A}g\end{array}$

Applying the equilibrium condition along the horizontal direction for block A,

$\begin{array}{c}∑{\left(F_A\right)}_x=F_T\\ =m_A{\left(a_A\right)}_x\end{array}$

Applying the equilibrium condition along the vertical direction for block B,

role="math" localid="1645505697802" $\begin{array}{c}∑{\left(F_B\right)}_y=m_{B}g-F_T\\ m_B{\left(a_B\right)}_y=m_{B}g-m_A\left(a_A\right)…\left(i\right)\end{array}$

As the two blocks are connected, the magnitudes of their acceleration will be the same. So,

${\left(a_B\right)}_y={\left(a_A\right)}_x=a$.

Substituting this value in equation (i),

$\begin{array}{c}{m}_{B}a=m_{B}g-m_{A}a\\ m_{B}a+m_{A}a=m_{B}g\\ a=\frac{m_{B}g}{\left(m_B+m_A\right)}\end{array}$

Thus, the formula for the acceleration of the system is $\frac{m_{B}g}{\left(m_B+m_A\right)}$.

The formula for the tension in the cord can be calculated as

$\begin{array}{c}{F}_T=m_{A}a\\ F_T=m_A\left[\frac{m_{B}g}{\left(m_B+m_A\right)}\right]\\ F_T=g\left[\frac{m_A{m}_B}{\left(m_B+m_A\right)}\right]\end{array}$

Thus, the formula for the tension in the cord is $g\left[\frac{m_A{m}_B}{\left(m_B+m_A\right)}\right]$.