Question

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 mu
string is massless and pulley is friction less.
cos
sin 0-ucose
sino
sin 0-ucos
ucose
sin 8-ucos
sin -ucos​

Answer 1

Answer:

The answer will be  mu/(sinθ−mucosθ)

Explanation:

According to the diagram, if the mass of the block  of the inclined plane M,

Therefore,

Ma=T−Mgsinθ−muMgcosθ

Now according to the problem the mass is at rest, therefore the acceleration is zero

Therefore,

T = Mgsinθ−muMgcosθ

=> T = Mg(sinθ−mucosθ)

Now for the second block the mass is m

ma=T−mumg

As the acceleration is zero,

Therefore,

T=mumg

Now putting the value of T here

Mg(sinθ−mucosθ) = mumg

=> M/m =  mu/(sinθ−mucosθ)