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Q. Two masses $ A $ and $ B $ of $ 15\, kg $ and $ 10\, kg $ are connected with a string passing over a frictionless pulley fixed at the corner of a table (as shown in figure). The coefficient of friction between the table and block is $ 0.4 $ . The minimum mass of $ C $ , that may be placed on $ A $ to prevent it from moving is :Physics Question Image

UPSEEUPSEE 2005

Solution:

For blocks to be stationary, net force on each is zero.
The following free body diagram shows the various forces acting on the system. Let $m$ be the minimum mass of block $C$ and $f_s$ be the maximum value of static friction
image
For block A :
$R=\left(m+m_{A}\right) g, f_{s}=T$
$\therefore \,\mu\left(m+m_{A}\right) g=T\,\,\,\,\,\dots(i)$
For block B :
$T=m_{B}\, g\,\,\,\,\,\dots(ii)$
From Eqs. (i) and (ii), we get
$m=\frac{m_{B}-\mu m_{A}}{\mu}$
$m=\frac{10-0.4 \times 15}{0.4}=10 \,kg$