Question

Find the maximum value of M m in the situation shown in figure so that system remain in equilibrium (friction coefficient at both contacts is √3/4)​

Answer 1

Answer:

ANSWER

From FBD of block on inclined , for maximum ratio block should at the condition of just starting the motion

therefore : Mgsinθ=T+f

1

here f

1

=μMgcosθ

also block on horizontal plane is in the condition of just start

therefore : T=f

2

and f

2

=μmg

using these we get: Mgsinθ=μmg+μMgcosθ

Mg(sinθ−μcosθ)=μmg

m

M

=

sinθ−μcosθ

μ

for maximum value of M/m ,sinθ−μcosθ should be minimum whih is equal to −

1+μ

2

so minimum value of M/m=

1+μ

2

μ

If tanθ<μ

⇒Mgsinθ<μMgcosθ

This implies that block on inclined plane will not move as limiting frictional force is greater than component of gravitational force along inclination , therefore T=0 , for any ratio of M/m block will not move and frictional force acting on body on inclined will be Mgsinθ and block on horizontal will be 0.

Answer 2

Explanation:

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