Question

Two blocks A and B are connected by a light inextensible string passing over a fixed smooth pulley as shown in Fig. The coefficient of friction between the block A and B the horizontal table is $\mu =0.5$. If the block A is just to slip, find the ratio of masses of the blocks.

Answer 1

From FBD of A and B, shown in Fig. the block A is just to slip, the friction will reach to its limiting value
$f_{lim}=\mu N$ (i)

From FBD of A,

$N+Tsin\theta =m_{A}g$ (ii)

$Tcos\theta =f_{max}=\mu N$ (iii)

From (ii), $N=m_{A}g-Tsin\theta$ (iv)

From (iii) and (iv),

$Tcos\theta =\mu m_{A}g-Tsin\theta$ (v)

$T(cos\theta +\mu sin\theta )=\mu m_{A}g$ (vi)

From FBD of B, T= $m_{B}g$ (vii)

Taking ratio of (vi) and (vii),

$\frac{\mu }{Cos\theta +\mu sin\theta }=\frac{m_B}{m_A}$

$\therefore \frac{m_A}{m_B}=\frac{Cos\theta +\mu Sin\theta }{\mu }=\frac{Cos{37}^0+\left(0.5\right)Sin{37}^0}{\left(0.5\right)}=\frac{11}{5}$