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

Answer:

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.

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

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

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

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