Solution:
Force of friction on mass m2=μm2g
Force of friction on mass m3=μm3g
Let α be common acceleration of the system.
∴a=m1g−μm2g−μm3gm1+m2+m3
Here, m1=m2=m3=m
Hence, the downward acceleration of mass m1 is g(1−2μ)3
A system consists of three masses m1, m2 and m3 connected by a string passing over a pulley P. The mass m1 hangs freely and m2 and m3 are on a rough horizontal table (the coefficient of friction= μ)
The pulley is frictionless and of negligible mass. The downward acceleration of mass m1 is
(Assume m1=m2=m3=m)
Force of friction on mass m2=μm2g
Force of friction on mass m3=μm3g
Let α be common acceleration of the system.
∴a=m1g−μm2g−μm3gm1+m2+m3
Here, m1=m2=m3=m
Hence, the downward acceleration of mass m1 is g(1−2μ)3