Question

A block of mass M1 is resting on an incline of angle θ and is attached to a second block of mass m by a cord that passes over a smooth peg. Mass m hangs vertically. The coefficient of static friction between the block and the incline is us. Find the range of possible values for m for which the system will be in static equilibrium.

Answer 1

Step-by-step explanation:

Given

mass of block resting on wedge is
`M_1`

mass of block hanging vertically is m

`\mu`coefficient of static friction

Let T be the tension in rope

thus for Equilibrium T=mg

`M_1g\sin \theta -T-f_r=0`

`f_r` is friction force

`f_r=M_1g\sin \theta -T`

substitute the value of T

`f_r=M_1g\sin \theta -mg`

Also
`f_r=\mu M_1g\cos \theta`

Direction of Friction can either be directing upward or downward depending upon which side system goes after it is released from equilibrium or about to release.

`-f_r<M_1g\sin \theta -mg<f_r`

`-\mu M_1g\cos \theta <M_1g\sin \theta -mg<\mu M_1g\cos \theta`

`M_1g(\mu \cos \theta -\sin \theta )<mg<M_1g(\mu \cos \theta +\sin \theta )`

`M_1(\mu \cos \theta -\sin \theta )<m<M_1(\mu \cos \theta +\sin \theta )`