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 =
Create a free account to view solution
View Solution FreeMore Friction Questions
A horizontal force of 10 N is necessary to just hold a block stationary against a wall. The coefficient of friction betw...A contact force exerted by one body on horizontal surface is equal to the normal force (≠ 0) between them. It can ...In the arrangement shown in the figure mass of the block B and A are 2 m,, 8 m respectively. Surface between B and floor...In the figure, a ladder of mass m is shown leaning against a wall. It is in static equilibrium making an angle θ wi...In the given figure the value(s) of mass m for which the 100 kg block remains in static equilibrium is (g = 10 m/s2)...