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 [image:5_image71.jpeg]

Question

Dr. Abubacker · Accepted Answer

(a) 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:5_image216.jpeg]
For block A
$R = \left( m + m_{A} ight)g,f_{s} = T$
$	herefore\ \mu\left( m + m_{A} ight)g = T\ \ \ $…(i)
For block B
$T = m_{B}g$ …(ii)
From Eqs. (i) and (ii), we get
$m = \frac{m_{B} - \mu{\ m}_{A}}{\mu}$
$m = \ \frac{10 - 0.4 	imes 15}{0.4} = 10\ kg$

Explanation

Reviewed by Dr. Abubacker

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