FrictionMedium
Question
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. The coefficient of static friction of A with table is 0.2. The minimum mass of C that may be placed on A to prevent it from moving is
Options
A.15 kg
B.10 kg
C.5 kg
D.12kg
Solution
Apply Newton′s law for system along the string
mB g = m(mA + mC) × g
⇒ mC =
− mA = 
− 10 = 15 kg
mB g = m(mA + mC) × g
⇒ mC =
− mA = − 10 = 15 kgCreate a free account to view solution
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