Q.
Two masses $A$ and $B$ of 10 kg and 5 kg respectively, 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 of $A$ with the table is 0.2. The minimum mass of $C$ that may be placed on $A$ to prevent it front moving is
A) 15 kg
B) 5 kg
C) 10 kg
D) Zero
Explanation
(a) Since the given system is in equilibrium therefore tension in the string is 5$g$
When we consider the combination of $A$ and $C$ , then
[image:5_image267.emf]
$T = \mu R$
or $T = 0.2(10 + mass\ of\ C)g$
or $5g = 0.2(10 + mass\ of\ C)g$
or mass of $C = 15$ kg
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.