Question

47.
Find the maximum value of (M/m) in the
situation shown in figure so that the system remains
at rest. Friction coefficient of both the contacts is ,
string is massless and pulley is friction less
sin 0
cos 0
sin 0-ucos
(B) sin 0-picos
picos 0
sin 0-icos 0
D) sin 0-ucos​

Answer 1

Answer:

The maximum value of (M/m) is μ(sinθ−μcosθ)

Explanation:

According to the problem  the mass of the first block is M which is inclined plane

For that block,

Ma=T−Mgsinθ−μMgcosθ [ where g is the gravitational force and μ friction coefficient]

As the mass is at rest,therefore the acceleration , a=0

Therefore,

T=Mgsinθ−μMgcosθ

T=Mg(sinθ−μcosθ)...(i)

Now the mass of the second block is m

and for the second block

ma=T−μmgas

as a=0,

T=μmg...(ii)

μmg=Mg(sinθ−μcosθ)

M/m=μ(sinθ−μcosθ)