FrictionHard
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
View Solution FreeMore Friction Questions
A block of mass M is pulled along a horizontal friction surface by a rope of mass m. If a force P is applied at the free...Block of mass m is on an inclined plane of angle θ. The coefficient of friction between the block and the plane is ...Two identical blocks of same masses are placed on a fixed wedge as shown in figure. Coefficient of friction between all ...A marble block of mass 2 kg lying on ice when given a velocity of 6 m/s is stopped by friction in 10 s. then the coeffic...A horizontal force of 10 N is necessary to just hold a block stationary against a wall. The coefficient of friction betw...